Esercizi di geometria affine nello spazio

ABCD\mathbb{A}^3(\mathbb{R})

ABC\in\mathbb{A}^3(\mathbb{R})\pi\subset\mathbb{A}^3(\mathbb{R})

D\in\mathbb{A}^3(\mathbb{R})\piABC\in\mathbb{A}^3(\mathbb{R})\overrightarrow{AB}\overrightarrow{AC}\in \mathbb{R}^3

k\in\mathbb{R}xyz\in\mathbb{R}\pi\piABC

D

\piABC

D

\piABC

D

\piABC

D

\piABC

D

ABC\mathbb{A}^3(\mathbb{R})

A=(3,0,-2),\; \qquad B=(4,-1,-4), \qquad\; C=(2,1,0);A=(2,2,2), \qquad\quad B=(1,1,-1),\quad \qquad C=(-1,2,3);A=(1,0,0), \qquad\quad B=(1,0,-\pi),\quad \qquad C=(0,0,1);A=(3,4,2), \qquad\quad B=(5,-1,1),\quad \qquad C=(2,-3,-2);A=(-3,2,1), \;\qquad B=(-4,2,3), \quad\qquad C=(-5,2,-1).\overrightarrow{AB}r\subset \mathbb{A}^3(\mathbb{R})A

t\in\mathbb{R}\piABC

\overrightarrow{AB}\overrightarrow{AC}k=-1rt\in\mathbb{R}

\piABC

\piABC

\piABC

\piABC

\mathbb{A}^3(\mathbb{R})

A=(4,2,0), \quad \qquad B=(1,3,1), \qquad\quad \quad C=(1,3,5);A=(1,-2,1), \qquad\; B=(2,-1,2),\quad \qquad\; C=(1,-1,2);A=(2,1,0), \qquad\quad B=(2,0,3),\quad \quad \qquad C=(2,2,-3);A=(1,-2,1), \qquad\; B=(2,1,0),\qquad \qquad C=(0,-5,2);A=(1,-2,1), \;\qquad B=(2,0,3), \qquad\qquad C=(1,-3,4).\pi

A\overrightarrow{AB}v\overrightarrow{AB}

t_1t_2\in\mathbb{R}\piABC

\piABC

\piABC

\overrightarrow{AB}\overrightarrow{AC}k=-1rt\in\mathbb{R}

rv=(1,0,0)\overrightarrow{AB}\pit_1t_2\in\mathbb{R}

\piABC

\overrightarrow{AB}\overrightarrow{AC}k=-1rt\in\mathbb{R}

rv=(1,0,0)\overrightarrow{AB}\pit_1t_2\in\mathbb{R}

\piABC

\mathcal{R}=\{O,e_1,e_2,e_3\}\mathbb{A}^3(\mathbb{R})\{e_1,e_2,e_3\}\mathbb{R}^3

\mathbb{A}^3(\mathbb{R})\mathbb{A}^3(\mathbb{R})\mathbb{R}^3O\mathcal{R}’v_1, v_2, v_3v_1, v_2, v_3\mathcal{E}=\{e_1, e_2, e_3\}

\mathcal{R}’\textbf{x}=(x_1,x_2,x_3)\in\mathbb{R}^3\mathbb{A}^3(\mathbb{R})\mathcal{R}\textbf{x}’=(x’_1,x’_2,x’_3)\in\mathbb{R}^3\mathcal{R}’

\text{M}

v\in \mathbb{R}^3A\in \mathbb{A}^3(\mathbb{R})

v=(-1,2,3),\qquad\quad\; A=(0,-3,-2);v=(1,1,0),\qquad \qquad A=(1,1,0);v=\left(-\dfrac{1}{2},1,\dfrac{2}{3}\right),\qquad \; A=(-2,1,1).t

x

t

t

AB\mathbb{A}^3(\mathbb{R})

A=(-1,-1,0),\qquad B=(-3,1,0);A=(2,1,2),\qquad\quad\;\; B=(1,1,-1);A=\left(\dfrac{3}{2},\dfrac{8}{5},1\right), \qquad\; B=\left(-\dfrac{1}{2},\dfrac{4}{5},3\right).v=B-A

rs

r: \begin{cases} x+y+z=11 \\ 2x+y-z=0 \end{cases},\qquad\qquad s: \begin{cases} x+z+1=0 \\ 3x+4y+2=0 \end{cases};r: \begin{cases} x-y+z=1 \\ 2y-z=0 \end{cases},\qquad\qquad\; s: \begin{cases} x=t-3\\ y=-t \\ z=-2t+6 \end{cases}, \quad t \in \mathbb{R};r: \begin{cases} x=t+2\\ y=-t-1 \\ z=3t+4 \end{cases},\quad t \in \mathbb{R} \qquad\; s: \begin{cases} x=u+3\\ y=u+3 \\ z=u+4 \end{cases}, \quad u \in \mathbb{R};r: \begin{cases} x+y-z=0 \\ 2x+y-z=1 \end{cases},\qquad\qquad s: \begin{cases} x=2-t\\ y=-1+2t\\ z=-1+3t \end{cases}, \quad t \in \mathbb{R};r: \begin{cases} x-y-z+2=0 \\ 3y+z-1=0 \end{cases},\qquad\quad s: \begin{cases} x=1+2t\\ y=1-t\\ z=3t \end{cases}, \quad t \in \mathbb{R};r: 1-x=y = z-1,\qquad\quad\qquad s: \begin{cases} x= 2t \\ y = -2t + 1 \\ z=-2t + 2 \end{cases}, \quad t \in \mathbb{R};r: \begin{cases} x+y+z+1=0 \\ y+3z+2=0 \end{cases},\qquad\quad s: \begin{cases} 2x+8z+1=0 \\ x+y+7z-a=0 \end{cases}.n_1,n_2\in\mathbb{R}^3n_1=(a_1,b_1,c_1)n_2=(a_2,b_2,c_2)d_1,d_2\in\mathbb{R}v_r\in\mathbb{R}^3

434

342\infty^124rs344

43

*** QuickLaTeX cannot compile formula:
. 
	Poiché <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-7ad7ef6519eb686f8874208340de8543_l3.svg" class="ql-img-inline-formula " alt="\text{rang}(A) \neq \text{rang}(A^*)" title="Rendered by QuickLaTeX.com" height="19" width="154" style="vertical-align: -5px;"/> il sistema è incompatibile, inoltre la matrice incompleta ha rango massimo e quindi le rette sono <strong>sghembe</strong>.
	<strong>Approccio parametrico</strong>
Per trovare le equazioni parametriche di <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/>, iniziamo da:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-b8a1eaf390de38a2bff2bfbafaa184cd_l3.svg" height="54" width="152" class="ql-img-displayed-equation " alt="\[\begin{cases}  x + y + z = 11 \\ 2x + y - z = 0 \end{cases}.\]" title="Rendered by QuickLaTeX.com"/>
Poniamo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-cbf8c0c1b51f70b7f46ec5512ac8509a_l3.svg" class="ql-img-inline-formula " alt="x = t" title="Rendered by QuickLaTeX.com" height="12" width="40" style="vertical-align: 0px;"/> come parametro. Sostituiamo questa scelta nelle due equazioni per ricavare <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d0342e16a1ea68370f19082b5525ff78_l3.svg" class="ql-img-inline-formula " alt="y" title="Rendered by QuickLaTeX.com" height="12" width="9" style="vertical-align: -4px;"/> e <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-70714939c882c5b3e8818b6feceeca80_l3.svg" class="ql-img-inline-formula " alt="z" title="Rendered by QuickLaTeX.com" height="8" width="9" style="vertical-align: 0px;"/>:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-684ed2184907d71bcc642776f7ec586b_l3.svg" height="43" width="441" class="ql-img-displayed-equation " alt="\[\begin{aligned} x + y + z &= 11 \implies t + y + z = 11 \implies y + z = 11 - t, \\ 2x + y - z &= 0 \implies 2t + y - z = 0 \implies y = z - 2t. \end{aligned}\]" title="Rendered by QuickLaTeX.com"/>
Sostituendo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-9c8e6176548eff8e55e3eda4928a63cd_l3.svg" class="ql-img-inline-formula " alt="y = z - 2t" title="Rendered by QuickLaTeX.com" height="16" width="79" style="vertical-align: -4px;"/> nell'equazione <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-2e1249184932045eca16ef8bee1e8c76_l3.svg" class="ql-img-inline-formula " alt="y + z = 11 - t" title="Rendered by QuickLaTeX.com" height="16" width="109" style="vertical-align: -4px;"/>:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-40d9f50d1dc44eaced9ffd7c3d38e07e_l3.svg" height="36" width="581" class="ql-img-displayed-equation " alt="\[(z - 2t) + z = 11 - t \implies 2z - 2t = 11 - t \implies 2z = 11 + t \implies z = \dfrac{11 + t}{2}.\]" title="Rendered by QuickLaTeX.com"/>
Ora che abbiamo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-70714939c882c5b3e8818b6feceeca80_l3.svg" class="ql-img-inline-formula " alt="z" title="Rendered by QuickLaTeX.com" height="8" width="9" style="vertical-align: 0px;"/>, possiamo ricavare <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d0342e16a1ea68370f19082b5525ff78_l3.svg" class="ql-img-inline-formula " alt="y" title="Rendered by QuickLaTeX.com" height="12" width="9" style="vertical-align: -4px;"/>:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-f19d568c42a22c756e111bc244df992a_l3.svg" height="36" width="385" class="ql-img-displayed-equation " alt="\[y = z - 2t = \dfrac{11 + t}{2} - 2t = \dfrac{11 + t - 4t}{2} = \dfrac{11 - 3t}{2}.\]" title="Rendered by QuickLaTeX.com"/>
Quindi, le equazioni parametriche di <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-082dafacb3bd9698c23fc3694e7c702c_l3.svg" height="110" width="223" class="ql-img-displayed-equation " alt="\[r: \begin{cases} x = t \\[5pt] y = \dfrac{11 - 3t}{2} \\[7pt] z = \dfrac{11 + t}{2} \end{cases}, \quad t \in \mathbb{R}.\]" title="Rendered by QuickLaTeX.com"/>
Per <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/>, partiamo da:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-15fc4b246476a3edcf8ea750527f2e08_l3.svg" height="54" width="144" class="ql-img-displayed-equation " alt="\[\begin{cases}  x + z = -1 \\ 3x + 4y = -2 \end{cases}.\]" title="Rendered by QuickLaTeX.com"/>
Poniamo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-eced17ab2b8bce7837b0b39e7ca71638_l3.svg" class="ql-img-inline-formula " alt="x = u" title="Rendered by QuickLaTeX.com" height="8" width="44" style="vertical-align: 0px;"/> come parametro. Sostituiamo questa scelta nelle due equazioni per ricavare <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d0342e16a1ea68370f19082b5525ff78_l3.svg" class="ql-img-inline-formula " alt="y" title="Rendered by QuickLaTeX.com" height="12" width="9" style="vertical-align: -4px;"/> e <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-70714939c882c5b3e8818b6feceeca80_l3.svg" class="ql-img-inline-formula " alt="z" title="Rendered by QuickLaTeX.com" height="8" width="9" style="vertical-align: 0px;"/>:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-eaf62921efc7bd3a87b6d466175fecb9_l3.svg" height="68" width="411" class="ql-img-displayed-equation " alt="\[\begin{aligned} x + z &= -1 \implies u + z = -1 \implies z = -1 - u, \\[5pt] 3x + 4y &= -2 \implies 3u + 4y = -2 \implies y = \dfrac{-2 - 3u}{4}. \end{aligned}\]" title="Rendered by QuickLaTeX.com"/>
Quindi, le equazioni parametriche di <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-7882cfa328e98578423da1b887a6930d_l3.svg" height="96" width="237" class="ql-img-displayed-equation " alt="\[s: \begin{cases} x = u \\[5pt] y = \dfrac{-2 - 3u}{4} \\[5pt] z = -1 - u \end{cases}, \quad u \in \mathbb{R}.\]" title="Rendered by QuickLaTeX.com"/>
Ora che abbiamo le equazioni parametriche, possiamo confrontare i vettori direttori delle due rette:
- Il vettore direttore della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> è <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-68eaf98c2a635dde07cfa5a579ea724e_l3.svg" class="ql-img-inline-formula " alt="v_r = \left(1, -\dfrac{3}{2}, \dfrac{1}{2}\right)" title="Rendered by QuickLaTeX.com" height="43" width="127" style="vertical-align: -17px;"/>.
- Il vettore direttore della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> è <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-1f6413992ee4492c1b4221858a339e20_l3.svg" class="ql-img-inline-formula " alt="v_s = \left(1, -\dfrac{3}{4}, -1\right)" title="Rendered by QuickLaTeX.com" height="43" width="137" style="vertical-align: -17px;"/>.
Poiché <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c834e3ad875823ae754b64207f68fee3_l3.svg" class="ql-img-inline-formula " alt="v_r" title="Rendered by QuickLaTeX.com" height="11" width="15" style="vertical-align: -3px;"/> e <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d3fc806216c65f4958d882f75b0caaa2_l3.svg" class="ql-img-inline-formula " alt="v_s" title="Rendered by QuickLaTeX.com" height="11" width="15" style="vertical-align: -3px;"/> non sono proporzionali, le rette non sono parallele.
Confrontiamo ora le equazioni parametriche di <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> e <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> ponendo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-0fa42d737b075e162f86e99423e25730_l3.svg" class="ql-img-inline-formula " alt="r(t) = s(u)" title="Rendered by QuickLaTeX.com" height="19" width="83" style="vertical-align: -5px;"/>:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-932115641615e12f8f4930bac6a10845_l3.svg" height="110" width="161" class="ql-img-displayed-equation " alt="\[\begin{cases} t = u \\[5pt] \dfrac{11 - 3t}{2} = \dfrac{-2 - 3u}{4} \\[7pt] \dfrac{11 + t}{2} = -1 - u. \end{cases}\]" title="Rendered by QuickLaTeX.com"/>
Sostituendo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-ef2919a2fff99ca053888917721f31cc_l3.svg" class="ql-img-inline-formula " alt="t = u" title="Rendered by QuickLaTeX.com" height="12" width="40" style="vertical-align: 0px;"/> nelle altre due equazioni:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-985a06d805cd13a9aa094b3b8f580a29_l3.svg" height="79" width="170" class="ql-img-displayed-equation " alt="\[\begin{cases} \dfrac{11 - 3u}{2} = \dfrac{-2 - 3u}{4}, \\[7pt] \dfrac{11 + u}{2} = -1 - u. \end{cases}\]" title="Rendered by QuickLaTeX.com"/>
La prima equazione diventa:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-f8dad01ed9fd662ee93da9765cc02453_l3.svg" height="19" width="565" class="ql-img-displayed-equation " alt="\[2(11 - 3u) = -2 - 3u \implies 22 - 6u = -2 - 3u \implies 24 = 3u \implies u = 8.\]" title="Rendered by QuickLaTeX.com"/>
Ora sostituiamo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-7922f67ac0552e93076fa893bd920595_l3.svg" class="ql-img-inline-formula " alt="u = 8" title="Rendered by QuickLaTeX.com" height="12" width="43" style="vertical-align: 0px;"/> nella seconda equazione:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-a9304c39108c8ed399792e4d4571b40a_l3.svg" height="36" width="247" class="ql-img-displayed-equation " alt="\[\dfrac{11 + 8}{2} = -1 - 8 \implies \dfrac{19}{2} = -9,\]" title="Rendered by QuickLaTeX.com"/>
che è un'assurdità, quindi il sistema è incompatibile.
 Poiché il sistema parametrico è incompatibile e i vettori direttori non sono proporzionali, le rette <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> e <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono <strong>sghembe</strong>.
[/learn_more]
[learn_more caption="Svolgimento punto 2."]
Dato che il testo fornisce l'equazione parametrica di <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/>, utilizziamo il secondo approccio.
Per ricavare l'equazione parametrica della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/>, partiamo da:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-b20f18737bc69ba12b607da0a3cbb156_l3.svg" height="54" width="143" class="ql-img-displayed-equation " alt="\[\begin{cases}  x - y + z = 1 \\ 2y - z = 0 \end{cases}.\]" title="Rendered by QuickLaTeX.com"/>
Poniamo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-607d161560ee83bae7cf115ca4c5192f_l3.svg" class="ql-img-inline-formula " alt="y = t" title="Rendered by QuickLaTeX.com" height="16" width="39" style="vertical-align: -4px;"/> come parametro e sostituiamo nelle due equazioni:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-2ba60674d2e41a27b2ab80a2118b92ec_l3.svg" height="39" width="255" class="ql-img-displayed-equation " alt="\[\begin{aligned} x - t + z &= 1 \implies x + z = t + 1, \\ z &= 2t. \end{aligned}\]" title="Rendered by QuickLaTeX.com"/>
Sostituendo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-8ce8e55042288c0beac384b8d3088e47_l3.svg" class="ql-img-inline-formula " alt="z = 2t" title="Rendered by QuickLaTeX.com" height="12" width="48" style="vertical-align: 0px;"/> nella prima equazione:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-3a78f18edd7fb5f60f9e5c9b196f0a13_l3.svg" height="14" width="230" class="ql-img-displayed-equation " alt="\[x + 2t = t + 1 \implies x = 1 - t.\]" title="Rendered by QuickLaTeX.com"/>
Quindi, le equazioni parametriche di <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-26d88c4aee1d45046c499a2f4850cd1d_l3.svg" height="75" width="203" class="ql-img-displayed-equation " alt="\[r: \begin{cases} x = 1 - t \\ y = t \\ z = 2t \end{cases}, \quad t \in \mathbb{R}.\]" title="Rendered by QuickLaTeX.com"/>
Le equazioni parametriche di <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-6e5b25038526b089c7304c9d14842fd6_l3.svg" height="75" width="231" class="ql-img-displayed-equation " alt="\[s: \begin{cases} x = u - 3 \\ y = -u \\ z = -2u + 6 \end{cases}, \quad u \in \mathbb{R}.\]" title="Rendered by QuickLaTeX.com"/>
Ora confrontiamo i vettori direttori delle due rette:
- Il vettore direttore della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> è <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-e6d632935874ad3a57ea462c0ec8ae97_l3.svg" class="ql-img-inline-formula " alt="v_r = (-1, 1, 2)" title="Rendered by QuickLaTeX.com" height="19" width="109" style="vertical-align: -5px;"/>.
- Il vettore direttore della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> è <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-12b21f3acbce659131f2d5d7d519c0db_l3.svg" class="ql-img-inline-formula " alt="v_s = (1, -1, -2)" title="Rendered by QuickLaTeX.com" height="19" width="123" style="vertical-align: -5px;"/>.
Poiché i due vettori sono proporzionali, le rette sono parallele o coincidenti.
Studiamo la differenza fra le due equazioni 
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-8e04f89c447a02801fc997a64b02e3d1_l3.svg" height="19" width="347" class="ql-img-displayed-equation " alt="\[r(t)-s(u)=(-t-u+4, t-u, 2t+2u-6).\]" title="Rendered by QuickLaTeX.com"/>
Tale vettore non è proporzionale a

*** Error message:
Two \documentclass or \documentstyle commands.
leading text: \documentclass{
Missing $ inserted.
leading text: ...atex.com-7ad7ef6519eb686f8874208340de8543_
Please use \mathaccent for accents in math mode.
leading text: ...le="vertical-align: -5px;"/> il sistema è
Bad math environment delimiter.
leading text: ...class="ql-img-displayed-equation " alt="\[
Missing \endgroup inserted.
leading text: ... + z = 11 \\ 2x + y - z = 0 \end{cases}.\]
\begin{document} ended by \end{equation*}.
leading text: ... + z = 11 \\ 2x + y - z = 0 \end{cases}.\]
Missing $ inserted.
leading text: ... + z = 11 \\ 2x + y - z = 0 \end{cases}.\]
Extra \endgroup.
leading text: ... + z = 11 \\ 2x + y - z = 0 \end{cases}.\]
Missing $ inserted.
leading text: ...atex.com-cbf8c0c1b51f70b7f46ec5512ac8509a_

v_rut

*** QuickLaTeX cannot compile formula:
e quindi le rette sono parallele.
Per determinare l'equazione del piano contenente le due rette, scegliamo i seguenti punti:
1. Punto <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-91955a944e8fd6731d7bdb79a1205009_l3.svg" class="ql-img-inline-formula " alt="P" title="Rendered by QuickLaTeX.com" height="12" width="14" style="vertical-align: 0px;"/> su <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/>: per <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-e711842f756bc8041fec162610de5d76_l3.svg" class="ql-img-inline-formula " alt="t = 0" title="Rendered by QuickLaTeX.com" height="12" width="39" style="vertical-align: 0px;"/>, <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-6b6fd593cb21fdcb193791c2999e3fc6_l3.svg" class="ql-img-inline-formula " alt="P = (1, 0, 0)" title="Rendered by QuickLaTeX.com" height="19" width="93" style="vertical-align: -5px;"/>.
2. Punto <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-a25d5c938b835615f6af55864c72fd03_l3.svg" class="ql-img-inline-formula " alt="Q" title="Rendered by QuickLaTeX.com" height="16" width="14" style="vertical-align: -4px;"/> su <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/>: per <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-3592d09c9b2fe4c52299a3a64e4352c4_l3.svg" class="ql-img-inline-formula " alt="u = 0" title="Rendered by QuickLaTeX.com" height="12" width="43" style="vertical-align: 0px;"/>, <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-42506636253ddeab1f51090223afaa0f_l3.svg" class="ql-img-inline-formula " alt="Q = (-3, 0, 6)" title="Rendered by QuickLaTeX.com" height="19" width="107" style="vertical-align: -5px;"/>.
Costruiamo due vettori di giacitura:
1. Il vettore direttore di <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/>: <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-e6d632935874ad3a57ea462c0ec8ae97_l3.svg" class="ql-img-inline-formula " alt="v_r = (-1, 1, 2)" title="Rendered by QuickLaTeX.com" height="19" width="109" style="vertical-align: -5px;"/>.
2. Un vettore differenza tra <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-91955a944e8fd6731d7bdb79a1205009_l3.svg" class="ql-img-inline-formula " alt="P" title="Rendered by QuickLaTeX.com" height="12" width="14" style="vertical-align: 0px;"/> e <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-a25d5c938b835615f6af55864c72fd03_l3.svg" class="ql-img-inline-formula " alt="Q" title="Rendered by QuickLaTeX.com" height="16" width="14" style="vertical-align: -4px;"/>: <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-a660d80bee62be3939892681ee195b39_l3.svg" class="ql-img-inline-formula " alt="P - Q = (1 - (-3), 0 - (-3), 0 - 6) = (4, 0, -6)" title="Rendered by QuickLaTeX.com" height="19" width="369" style="vertical-align: -5px;"/>.
Il piano è generato da questi due vettori. L'equazione cartesiana del piano è data da \ref{piano}:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-5dd9f50f97b50de3896b1f89a0181316_l3.svg" height="64" width="201" class="ql-img-displayed-equation " alt="\[\det \begin{pmatrix} x - 1 & y & z \\ -1 & 1 & 2 \\ 4 & 0 & -6 \end{pmatrix} = 0.\]" title="Rendered by QuickLaTeX.com"/>
Calcoliamo il determinante:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-674f2089661ccb70c279f817ac3008f9_l3.svg" height="43" width="478" class="ql-img-displayed-equation " alt="\[(x - 1)  \det \begin{pmatrix} 1 & 2 \\ 0 & -6 \end{pmatrix} - y  \det \begin{pmatrix} -1 & 2 \\ 4 & -6 \end{pmatrix} + z  \det \begin{pmatrix} -1 & 1 \\ 4 & 0 \end{pmatrix} = 0,\]" title="Rendered by QuickLaTeX.com"/>
che si semplifica in:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-21c6f92d1fae09309ac422b8f9f2824d_l3.svg" height="19" width="278" class="ql-img-displayed-equation " alt="\[(x - 1)(- 6) - y(6 - 8) + z(- 4) = 0.\]" title="Rendered by QuickLaTeX.com"/>
Espandendo, otteniamo l'equazione del piano:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d6370e768150650fe70bf4ce35cf5989_l3.svg" height="16" width="170" class="ql-img-displayed-equation " alt="\[-3x + y - 2z + 3 = 0.\]" title="Rendered by QuickLaTeX.com"/>
Le rette sono quindi <strong>parallele</strong> e complanari, e l'equazione del piano che le contiene è <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c0d5ea9cd9db6a8a9262570a34f03ccd_l3.svg" class="ql-img-inline-formula " alt="-3x + y - 2z + 3 = 0" title="Rendered by QuickLaTeX.com" height="16" width="166" style="vertical-align: -4px;"/>.
[/learn_more]
[learn_more caption="Svolgimento punto 3."]
Utilizziamo l'approccio parametrico. Ricaviamo quindi le equazioni parametriche delle due rette e procediamo all'analisi della loro posizione reciproca.
Le equazioni parametriche della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-b2cda48be50ee2e151415d1749334f93_l3.svg" height="75" width="216" class="ql-img-displayed-equation " alt="\[r: \begin{cases} x = t + 2 \\ y = -t - 1 \\ z = 3t + 4 \end{cases}, \quad t \in \mathbb{R}.\]" title="Rendered by QuickLaTeX.com"/>
Le equazioni parametriche della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d70534cef6440d573555cf4d925b0e0d_l3.svg" height="75" width="211" class="ql-img-displayed-equation " alt="\[s: \begin{cases} x = u + 3 \\ y = u + 3 \\ z = u + 4 \end{cases}, \quad u \in \mathbb{R}.\]" title="Rendered by QuickLaTeX.com"/>
I vettori direttori delle due rette sono ricavati, come di consueto, dai coefficienti dei parametri <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-34054c9e6c78692a8eca30f551d72449_l3.svg" class="ql-img-inline-formula " alt="t" title="Rendered by QuickLaTeX.com" height="12" width="6" style="vertical-align: 0px;"/> e <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-671f6952c2f8cbc47a47623a478981b6_l3.svg" class="ql-img-inline-formula " alt="u" title="Rendered by QuickLaTeX.com" height="8" width="10" style="vertical-align: 0px;"/> nelle rispettive equazioni parametriche:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-4f73886d17a0275f150ace92880d2d5a_l3.svg" height="64" width="201" class="ql-img-displayed-equation " alt="\[v = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}, \quad v_s = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.\]" title="Rendered by QuickLaTeX.com"/>
Confrontando i vettori direttori <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c834e3ad875823ae754b64207f68fee3_l3.svg" class="ql-img-inline-formula " alt="v_r" title="Rendered by QuickLaTeX.com" height="11" width="15" style="vertical-align: -3px;"/> e <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d3fc806216c65f4958d882f75b0caaa2_l3.svg" class="ql-img-inline-formula " alt="v_s" title="Rendered by QuickLaTeX.com" height="11" width="15" style="vertical-align: -3px;"/>, osserviamo che non sono proporzionali, quindi le rette non sono parallele. Ora verifichiamo se sono incidenti, ovvero se esiste un punto di intersezione comune.
Per verificare se le rette sono incidenti, uguagliamo le loro equazioni parametriche, ossia poniamo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-0fa42d737b075e162f86e99423e25730_l3.svg" class="ql-img-inline-formula " alt="r(t) = s(u)" title="Rendered by QuickLaTeX.com" height="19" width="83" style="vertical-align: -5px;"/>:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d0836e26a40d34beafb3bb2ae7d9b55e_l3.svg" height="75" width="154" class="ql-img-displayed-equation " alt="\[\begin{cases} t + 2 = u + 3 \\ -t - 1 = u + 3 \\ 3t + 4 = u + 4 \end{cases}.\]" title="Rendered by QuickLaTeX.com"/>
Dalla prima equazione otteniamo:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-96ecdc8082f7dba1d15dc32edff52361_l3.svg" height="14" width="225" class="ql-img-displayed-equation " alt="\[t + 2 = u + 3 \implies t = u + 1.\]" title="Rendered by QuickLaTeX.com"/>
Sostituendo questo risultato nella seconda equazione:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-10d4d1157453c7a5012c7f6d43bcf78a_l3.svg" height="37" width="556" class="ql-img-displayed-equation " alt="\[-(u + 1) - 1 = u + 3 \implies -u - 2 = u + 3 \implies -2u = 5 \implies u = -\dfrac{5}{2}.\]" title="Rendered by QuickLaTeX.com"/>
Ora sostituiamo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-55e1694a36d973492ba87381a072a2c3_l3.svg" class="ql-img-inline-formula " alt="u = -\dfrac{5}{2}" title="Rendered by QuickLaTeX.com" height="37" width="59" style="vertical-align: -12px;"/> nella relazione <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-3a15de103027c0e3d9b9de527390993f_l3.svg" class="ql-img-inline-formula " alt="t = u + 1" title="Rendered by QuickLaTeX.com" height="14" width="70" style="vertical-align: -2px;"/>:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-dd9cc434825661c80937484cf2bea259_l3.svg" height="37" width="141" class="ql-img-displayed-equation " alt="\[t = -\dfrac{5}{2} + 1 = -\dfrac{3}{2}.\]" title="Rendered by QuickLaTeX.com"/>
Infine proviamo a verificare la terza equazione ed otteniamo
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-75a11a2e9343666b1a9da4f285032cfd_l3.svg" height="43" width="774" class="ql-img-displayed-equation " alt="\[3t + 4 = u + 4 \implies 3\left(-\dfrac{3}{2}\right) + 4 = -\dfrac{5}{2} + 4 \implies -\dfrac{9}{2} + 4 = -\dfrac{5}{2} + 4 \implies \dfrac{-9 + 8}{2} = \dfrac{-5 + 8}{2} \implies -\dfrac{1}{2} = \dfrac{3}{2},\]" title="Rendered by QuickLaTeX.com"/>
che è un'assurdità. Quindi, il sistema è incompatibile.
Poiché il sistema è incompatibile e i vettori direttori non sono proporzionali, concludiamo che le rette <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> e <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono </strong>sghembe</strong>.
[/learn_more]
[learn_more caption="Svolgimento punto 4."]
Seguiamo l'approccio basato sulle equazioni parametriche.
Per trovare le equazioni parametriche della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/>, partiamo dalle sue equazioni cartesiane:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-bfa7018bf66c08c401b2abab9622222b_l3.svg" height="54" width="152" class="ql-img-displayed-equation " alt="\[\begin{cases}  x + y - z = 0 \\  2x + y - z = 1 \end{cases}.\]" title="Rendered by QuickLaTeX.com"/>
Sottraiamo la prima equazione dalla seconda per eliminare <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-4060e98331a15f1e33d174f2e1f6c9a5_l3.svg" class="ql-img-inline-formula " alt="y - z" title="Rendered by QuickLaTeX.com" height="12" width="40" style="vertical-align: -4px;"/>:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-74c3c42dac4e39b992cadb97d98a755d_l3.svg" height="19" width="358" class="ql-img-displayed-equation " alt="\[(2x + y - z) - (x + y - z) = 1 - 0 \implies x = 1.\]" title="Rendered by QuickLaTeX.com"/>
Sostituendo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d4f2472099ca1ad9256611efeae0fce2_l3.svg" class="ql-img-inline-formula " alt="x = 1" title="Rendered by QuickLaTeX.com" height="12" width="42" style="vertical-align: 0px;"/> nella prima equazione di <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/>:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-a7e0d8a533eebd63ab504c39cf7ef8b4_l3.svg" height="16" width="227" class="ql-img-displayed-equation " alt="\[1 + y - z = 0 \implies y = z - 1.\]" title="Rendered by QuickLaTeX.com"/>
Quindi, ponendo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c32dd62231e4b7dc929627878b023f69_l3.svg" class="ql-img-inline-formula " alt="z=t" title="Rendered by QuickLaTeX.com" height="12" width="39" style="vertical-align: 0px;"/>, abbiamo che le equazioni parametriche della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-bb7d3673c00c48fab60a7124ba280e7e_l3.svg" height="75" width="203" class="ql-img-displayed-equation " alt="\[r: \begin{cases} x = 1 \\ y = t - 1 \\ z = t \end{cases}, \quad t \in \mathbb{R}.\]" title="Rendered by QuickLaTeX.com"/>
Le equazioni parametriche della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono già fornite nel testo:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-ef70409de4f249130d1050f69ff035ae_l3.svg" height="75" width="233" class="ql-img-displayed-equation " alt="\[s: \begin{cases} x = 2 - t' \\ y = -1 + 2t' \\ z = -1 + 3t' \end{cases}, \quad t' \in \mathbb{R}.\]" title="Rendered by QuickLaTeX.com"/>
I vettori direttori delle due rette non sono proporzionali, infatti abbiamo
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-0d7b9a18243c7edc6e33eff785319d87_l3.svg" height="19" width="253" class="ql-img-displayed-equation " alt="\[v_r=(0,1,1) \quad \text{e}\quad v_s=(-1,2,3).\]" title="Rendered by QuickLaTeX.com"/>
Verifichiamo se le rette <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> ed <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono incidenti, cercando di trovare un punto di intersezione risolvendo il sistema parametrico:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-bdd32ac4311dc4092598604feadcb598_l3.svg" height="75" width="164" class="ql-img-displayed-equation " alt="\[\begin{cases} 1 = 2 - t' \\ t - 1 = -1 + 2t' \\ t = -1 + 3t' \end{cases}.\]" title="Rendered by QuickLaTeX.com"/>
Dalla prima equazione abbiamo:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-9c5c241fc8c51361ecf43e3dbebab1ff_l3.svg" height="16" width="47" class="ql-img-displayed-equation " alt="\[t' = 1.\]" title="Rendered by QuickLaTeX.com"/>
Sostituendo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-8f12b5722a6476a9a3abc5bf342a5cc2_l3.svg" class="ql-img-inline-formula " alt="t' = 1" title="Rendered by QuickLaTeX.com" height="14" width="43" style="vertical-align: 0px;"/> nella seconda e terza equazione:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-91d5994b2277a55a446211849148027d_l3.svg" height="41" width="259" class="ql-img-displayed-equation " alt="\[\begin{aligned} t - 1 &= -1 + 2 \cdot 1 = 1 \implies t = 2, \\ t &= -1 + 3 \cdot 1 = 2. \end{aligned}\]" title="Rendered by QuickLaTeX.com"/>
Quindi il sistema è compatibile e le due rette si intersecano nel punto <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-540f94d66fa01a568103267d6bcf1ea0_l3.svg" class="ql-img-inline-formula " alt="P = (1, 1, 2)" title="Rendered by QuickLaTeX.com" height="19" width="93" style="vertical-align: -5px;"/>.
Poiché le rette <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> ed <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> si intersecano, esiste un piano <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-b6fd00bfadcac2047aa504e8766a84f1_l3.svg" class="ql-img-inline-formula " alt="\pi" title="Rendered by QuickLaTeX.com" height="8" width="11" style="vertical-align: 0px;"/> che le contiene. Per determinare l'equazione cartesiana del piano <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-b6fd00bfadcac2047aa504e8766a84f1_l3.svg" class="ql-img-inline-formula " alt="\pi" title="Rendered by QuickLaTeX.com" height="8" width="11" style="vertical-align: 0px;"/>, possiamo utilizzare il punto <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-540f94d66fa01a568103267d6bcf1ea0_l3.svg" class="ql-img-inline-formula " alt="P = (1, 1, 2)" title="Rendered by QuickLaTeX.com" height="19" width="93" style="vertical-align: -5px;"/> e i vettori direttori delle due rette.
Il vettore direttore di <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> è:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-5417fa1267f6a28373a975d123c9eaf8_l3.svg" height="19" width="101" class="ql-img-displayed-equation " alt="\[v_r = (0, 1, 1),\]" title="Rendered by QuickLaTeX.com"/>
mentre il vettore direttore di <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> è:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-7e33b33d33fd345a931120258ef893ab_l3.svg" height="19" width="113" class="ql-img-displayed-equation " alt="\[v_s = (-1, 2, 3).\]" title="Rendered by QuickLaTeX.com"/>
Per trovare l'equazione del piano, calcoliamo il prodotto vettoriale tra <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c834e3ad875823ae754b64207f68fee3_l3.svg" class="ql-img-inline-formula " alt="v_r" title="Rendered by QuickLaTeX.com" height="11" width="15" style="vertical-align: -3px;"/> e <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d3fc806216c65f4958d882f75b0caaa2_l3.svg" class="ql-img-inline-formula " alt="v_s" title="Rendered by QuickLaTeX.com" height="11" width="15" style="vertical-align: -3px;"/>, che ci darà il vettore ortogonale al piano:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-1faf2501a199138990bc5ae6414255d3_l3.svg" height="65" width="685" class="ql-img-displayed-equation " alt="\[n = v_r \times v_s = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ -1 & 2 & 3 \end{vmatrix} = \mathbf{i}(1 \cdot 3 - 1 \cdot 2) - \mathbf{j}(0 \cdot 3 - 1 \cdot (-1)) + \mathbf{k}(0 \cdot 2 - 1 \cdot (-1)) = \mathbf{i}(1) - \mathbf{j}(1) + \mathbf{k}(1).\]" title="Rendered by QuickLaTeX.com"/>
Quindi, il tale vettore è <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-3791da540e6144d5f336037166d220c6_l3.svg" class="ql-img-inline-formula " alt="n = (1, -1, 1)" title="Rendered by QuickLaTeX.com" height="19" width="103" style="vertical-align: -5px;"/>.
L'equazione cartesiana del piano passa per il punto <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-540f94d66fa01a568103267d6bcf1ea0_l3.svg" class="ql-img-inline-formula " alt="P = (1, 1, 2)" title="Rendered by QuickLaTeX.com" height="19" width="93" style="vertical-align: -5px;"/> ed è ortogonale al vettore <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-e414b8c4088eb4ea67d5c6d5854da2ba_l3.svg" class="ql-img-inline-formula " alt="n" title="Rendered by QuickLaTeX.com" height="8" width="11" style="vertical-align: 0px;"/> è
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-aaa7ab719593ee59fa98f973f521fcb8_l3.svg" height="19" width="267" class="ql-img-displayed-equation " alt="\[1(x - 1) - 1(y - 1) + 1(z - 2) = 0,\]" title="Rendered by QuickLaTeX.com"/>
che si semplifica in:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-db910d9b269bf8f2692384344209112f_l3.svg" height="16" width="139" class="ql-img-displayed-equation " alt="\[x - y + z - 2 = 0.\]" title="Rendered by QuickLaTeX.com"/>
Quindi, l'equazione del piano <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-b6fd00bfadcac2047aa504e8766a84f1_l3.svg" class="ql-img-inline-formula " alt="\pi" title="Rendered by QuickLaTeX.com" height="8" width="11" style="vertical-align: 0px;"/> che contiene le rette <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> ed <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> è:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-db910d9b269bf8f2692384344209112f_l3.svg" height="16" width="139" class="ql-img-displayed-equation " alt="\[x - y + z - 2 = 0.\]" title="Rendered by QuickLaTeX.com"/>
[/learn_more]
[learn_more caption="Svolgimento punto 5."]
Seguiamo nuovamente l'approccio basato sulle equazioni parametriche. Per trovare le equazioni parametriche della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/>, partiamo dalle sue equazioni cartesiane:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-1082333d814516ca33477445de5f477f_l3.svg" height="54" width="152" class="ql-img-displayed-equation " alt="\[\begin{cases}  x + y - z = 1 \\  2x - y + z = 3 \end{cases}.\]" title="Rendered by QuickLaTeX.com"/>
Sommiamo queste due equazioni per eliminare <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d0342e16a1ea68370f19082b5525ff78_l3.svg" class="ql-img-inline-formula " alt="y" title="Rendered by QuickLaTeX.com" height="12" width="9" style="vertical-align: -4px;"/> e <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-70714939c882c5b3e8818b6feceeca80_l3.svg" class="ql-img-inline-formula " alt="z" title="Rendered by QuickLaTeX.com" height="8" width="9" style="vertical-align: 0px;"/>:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c0fa89873359e3ac912fc696f2115fd6_l3.svg" height="36" width="462" class="ql-img-displayed-equation " alt="\[(x + y - z) + (2x - y + z) = 1 + 3 \implies 3x = 4 \implies x = \dfrac{4}{3}.\]" title="Rendered by QuickLaTeX.com"/>
Sostituendo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-30caa40b9d659ccc449e13dda9c014af_l3.svg" class="ql-img-inline-formula " alt="x = \dfrac{4}{3}" title="Rendered by QuickLaTeX.com" height="36" width="45" style="vertical-align: -12px;"/> nella prima equazione:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-30126e19226820c668213115ce522609_l3.svg" height="36" width="315" class="ql-img-displayed-equation " alt="\[\dfrac{4}{3} + y - z = 1 \implies y - z = 1 - \dfrac{4}{3} = -\dfrac{1}{3}.\]" title="Rendered by QuickLaTeX.com"/>
Ora esprimiamo <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d0342e16a1ea68370f19082b5525ff78_l3.svg" class="ql-img-inline-formula " alt="y" title="Rendered by QuickLaTeX.com" height="12" width="9" style="vertical-align: -4px;"/> in funzione di <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-70714939c882c5b3e8818b6feceeca80_l3.svg" class="ql-img-inline-formula " alt="z" title="Rendered by QuickLaTeX.com" height="8" width="9" style="vertical-align: 0px;"/>:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-e267cf058d069f354d95da100a7c0b2c_l3.svg" height="36" width="81" class="ql-img-displayed-equation " alt="\[y = z - \dfrac{1}{3}.\]" title="Rendered by QuickLaTeX.com"/>
Quindi, le equazioni parametriche della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/>, posto <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c32dd62231e4b7dc929627878b023f69_l3.svg" class="ql-img-inline-formula " alt="z=t" title="Rendered by QuickLaTeX.com" height="12" width="39" style="vertical-align: 0px;"/>, sono:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d052cf8f730622dba47e12edaee85793_l3.svg" height="97" width="207" class="ql-img-displayed-equation " alt="\[r: \begin{cases} x = \dfrac{4}{3} \\ y = t - \dfrac{1}{3} \\ z = t \end{cases}, \quad t \in \mathbb{R}.\]" title="Rendered by QuickLaTeX.com"/>
Le equazioni parametriche della retta <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono già fornite nel testo:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-5387b49e6865d20376202b1221782396_l3.svg" height="75" width="225" class="ql-img-displayed-equation " alt="\[s: \begin{cases} x = 1 + 2t' \\ y = -2 + t' \\ z = 1 + 3t' \end{cases}, \quad t' \in \mathbb{R}.\]" title="Rendered by QuickLaTeX.com"/>
I vettori direttori delle due rette non sono proporzionali, infatti
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-798288e81f1ae50105df578555e97d3a_l3.svg" height="19" width="239" class="ql-img-displayed-equation " alt="\[v_r=(0,1,1) \quad \text{e}\quad v_s=(2,1,3).\]" title="Rendered by QuickLaTeX.com"/>
Ora verifichiamo se le rette <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-17c57195a2db07101fd3ffe719cdddfd_l3.svg" class="ql-img-inline-formula " alt="r" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> ed <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-074dbb879776f38c99499c615fe936be_l3.svg" class="ql-img-inline-formula " alt="s" title="Rendered by QuickLaTeX.com" height="8" width="8" style="vertical-align: 0px;"/> sono incidenti, cercando di trovare un punto di intersezione risolvendo il sistema parametrico:
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-0ff582c3a118ddb56f9ea018c966223f_l3.svg" height="97" width="159" class="ql-img-displayed-equation " alt="\[\begin{cases} \dfrac{4}{3} = 1 + 2t' \\ t - \dfrac{1}{3} = -2 + t' \\ t = 1 + 3t' \end{cases}.\]" title="Rendered by QuickLaTeX.com"/>
Dalla prima equazione si ricava

*** Error message:
Two \documentclass or \documentstyle commands.
leading text: \documentclass{
Missing $ inserted.
leading text: ...atex.com-91955a944e8fd6731d7bdb79a1205009_
Please use \mathaccent for accents in math mode.
leading text: Il piano è
Please use \mathaccent for accents in math mode.
leading text: ...ttori. L'equazione cartesiana del piano è
Bad math environment delimiter.
leading text: ...class="ql-img-displayed-equation " alt="\[
Missing \endgroup inserted.
leading text: ...& 1 & 2 \\ 4 & 0 & -6 \end{pmatrix} = 0.\]
\begin{document} ended by \end{equation*}.
leading text: ...& 1 & 2 \\ 4 & 0 & -6 \end{pmatrix} = 0.\]
Missing $ inserted.
leading text: ...& 1 & 2 \\ 4 & 0 & -6 \end{pmatrix} = 0.\]
Extra \endgroup.
leading text: ...& 1 & 2 \\ 4 & 0 & -6 \end{pmatrix} = 0.\]

t’

xs

AA^*

a\in\mathbb{R}

rr’P \notin r \cup r’sPrr’

Pr\piPr’\sigmaPrr’sPs\pir\sigmar’rP=(1,-1,0)\pi: 2x-y+4=0

r\pi

r: \begin{cases*} x+y+z=11 \\ 2x+y-z= 0 \end{cases*},\qquad \qquad \pi : 3x+4y +2=0;r: \begin{cases*} x-y+z=1 \\ 2y-z= 0 \end{cases*},\qquad\qquad \pi : \begin{cases*} x=t’+t-3 \\ y=-t’-2t\\ z=2t’-2t+6 \end{cases*},\; t,t’ \in \mathbb{R};r: \begin{cases*} x= u +2 \\ y=-u -1 \\ z=3u + 4 \end{cases*},\; u \in \mathbb{R}\qquad \pi : \begin{cases*} x=t’-t+3 \\ y=-t’+t+3\\ z=2t’-3t+4 \end{cases*},\; t,t’ \in \mathbb{R};r: \begin{cases*} x-y+z=2 \\ 2y-z= 0 \end{cases*},\qquad \qquad\pi : \begin{cases*} x=2-t +2t’ \\ y=-1+2t-t’\\ z=-1+3t+2t’ \end{cases*},\; t,t’ \in \mathbb{R};r: \begin{cases*} x-y-z+2=0 \\ 3y+z-1= 0 \end{cases*},\qquad \;\pi : \begin{cases*} x=1+2t -t’ \\ y=1-t+t’\\ z=3t \end{cases*},\; t,t’ \in \mathbb{R};r: 1-x=y=z-1,\qquad \quad\; \pi : \begin{cases*} x=2t -t’-3 \\ y=t-2t’+1\\ z=3t+t’ \end{cases*},\; t,t’ \in \mathbb{R};r: \begin{cases*} x-y+z+1=0 \\ y+3z+2= 0 \end{cases*},\qquad\quad \pi :2x-2y+2z+1=0.

*** QuickLaTeX cannot compile formula:
</li>
</ol>
</div>
[learn_more caption="Svolgimento."]
Una retta ed un piano nello spazio possono essere 
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-27e49cccda278470ae7436bace68813e_l3.svg" height="1" width="1" class="ql-img-displayed-equation " alt="\[\quad\]" title="Rendered by QuickLaTeX.com"/>
<ul>
        <li> <strong>Retta incidente al piano</strong>: in tal caso si intersecano in un unico punto.</li>
        <li> <strong>Retta parallela al piano</strong>: in tal caso retta e piano non si intersecano in alcun punto.</li>
        <li> <strong>Retta giacente sul piano</strong>: in tal caso la retta è interamente contenuta nel piano.</li>
</ul>
Notiamo che in

*** Error message:
Error: get_image_size(): -1

3

2\infty^1

\pi\pi

*** QuickLaTeX cannot compile formula:
e il direttore della retta:
	<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c1e443f0a2bab4b9a50c489d2f8e6176_l3.svg" height="66" width="161" class="ql-img-displayed-equation " alt="\[\begin{vmatrix} 	-1&1&2\\1&-1&2\\1&-2&-2 	\end{vmatrix}=-4.\]" title="Rendered by QuickLaTeX.com"/>
	Dunque <strong>la retta è incidente al piano</strong>.
[/learn_more]
[learn_more caption="Svolgimento punto 3."]
Le due varietà affini in tal caso sono date entrambe in forma parametrica, procediamo quindi con lo studio della dipendenza lineare fra le loro giaciture, determinando il vettore direttore della retta e i vettori di giacitura del piano per ispezione diretta dei coefficienti dei parametri. Si ha
	<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-fa91f642e748f59bd8b3fc9afd901547_l3.svg" height="66" width="147" class="ql-img-displayed-equation " alt="\[\begin{vmatrix} 	1&-1&3\\1&-1&2\\-1&1&-3 	\end{vmatrix}=0,\]" title="Rendered by QuickLaTeX.com"/>
	poché le prime due colonne della matrice coincidono. Dato che si verifica agevolmente che il punto

*** Error message:
Error: get_image_size(): -1

(3,3,4)\in\pir

*** QuickLaTeX cannot compile formula:
, <strong>la retta è parallela al piano</strong>.
[/learn_more]
[learn_more caption="Svolgimento punto 4."]
Come nel secondo caso abbiamo la retta in forma parametrica e il piano in forma cartesiana. Ricaviamo il vettore direttore della retta, che è identico al caso precedente:
	<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-f593e9abdc5ca0dcd3db4a4a855d87f8_l3.svg" height="66" width="317" class="ql-img-displayed-equation " alt="\[v_r=n_1\times n_2=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-1&1\\0&2&-1\end{vmatrix}=(-1,1,2),\]" title="Rendered by QuickLaTeX.com"/>
	e studiamo la sua dipendenza dalla giacitura del piano ricavandone i generatori per ispezione diretta delle equazioni parametriche di quest'ultimo. Si ha
	<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-69578b567e1d712fdffabd97492d3451_l3.svg" height="66" width="147" class="ql-img-displayed-equation " alt="\[\begin{vmatrix} 	-1&1&2\\-1&2&3\\2&-1&2 	\end{vmatrix}=-5.\]" title="Rendered by QuickLaTeX.com"/>
	Quindi <strong>la retta è incidente al piano</strong>.
[/learn_more]
[learn_more caption="Svolgimento punto 5."]
Sono fornite equazioni cartesiane per la retta e parametriche per il piano. Ricaviamo il vettore direttore della retta tramite il prodotto vettoriale fra le normali ai due piani in forma cartesiana che la definiscono.
	<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-0cb34dcc28999849aefbaacef8887e85_l3.svg" height="66" width="317" class="ql-img-displayed-equation " alt="\[v_r=n_1\times n_2=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-1&-1\\0&3&1\end{vmatrix}=(2,-1,3).\]" title="Rendered by QuickLaTeX.com"/>
	Studiamo la sua dipendenza dalla giacitura del piano ricavandone i generatori per ispezione diretta delle equazioni parametriche di quest'ultimo:
	<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-b05450c40f27f435b24f0418894bde2c_l3.svg" height="66" width="133" class="ql-img-displayed-equation " alt="\[\begin{vmatrix} 	2&-1&3\\2&-1&3\\-1&1&0 	\end{vmatrix}=0,\]" title="Rendered by QuickLaTeX.com"/>
	poiché il vettore direttore della retta coincide con uno dei due vettori che generano la giacitura del piano.
	Dato che si verifica agevolmente che il punto

*** Error message:
Error: get_image_size(): -1

(1,1,0)\in\pir

2r33

*** QuickLaTeX cannot compile formula:
:
		<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-2b3498986b086397dc6a040946763ea9_l3.svg" height="66" width="413" class="ql-img-displayed-equation " alt="\[\begin{vmatrix} 	1&-1&-1\\0&1&-2\\2&-2&-1 	\end{vmatrix}=\begin{vmatrix} 	1&-2\\-2&-1 	\end{vmatrix}+2\begin{vmatrix}-1&-1\\1&-2\end{vmatrix}=-5+6=1.\]" title="Rendered by QuickLaTeX.com"/>
	Il sistema quindi è incompatibile e <strong>la retta è parallela al piano</strong>.
[/learn_more]
<div style="padding: 10px; background-color: #bef5ad;">
<strong style="color: #000000;">Esercizio 11</strong> <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-47c5ad7bd42823dbf18c33db4aa83798_l3.svg" class="ql-img-inline-formula " alt=" (\bigstar\bigstar\largewhitestar\largewhitestar\largewhitestar)" title="Rendered by QuickLaTeX.com" height="20" width="124" style="vertical-align: -5px;"/>. Determinare la posizione reciproca tra i piani

*** Error message:
Error: get_image_size(): -1

\pi\sigma

\pi: x+y+z=1, \qquad\qquad \qquad\qquad \;\; \sigma: 3x+4y+2=0;\pi : x-3y+2z=1, \qquad\qquad \qquad \quad\;\; \sigma : 2x-6y +4z = 5;\pi: x – y + z = 8, \qquad\qquad \qquad\qquad \; \sigma: \begin{cases*} x = 2t-t’-3\\ y= -2t + t’\\ z=2t’+6 \end{cases*}, \; t,t’\in \mathbb{R};\pi: \begin{cases*} x = t+t’+2\\ y= -2 -2 t’-1\\ z=-3t+t’+4 \end{cases*}, \; t,t’\in \mathbb{R}, \qquad \sigma: \begin{cases*} x = u+3 \\ y= u+3u’+3\\ z=u+4 \end{cases*}, \; u,u’\in \mathbb{R};\pi: \begin{cases*} x = t+t’+2\\ y= -t -2 t’-1\\ z=-3t+t’+4 \end{cases*}, \; t,t’\in \mathbb{R}, \qquad \sigma: \begin{cases*} x = 2u-u’+3 \\ y= -3u+2u’-2\\ z=-2u -u’ +1 \end{cases*}, \; u,u’\in \mathbb{R}.

*** QuickLaTeX cannot compile formula:
</li>
</ol>
</div>
[learn_more caption="Svolgimento."]
Due piani nello spazio affine tridimensionale possono essere 
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-27e49cccda278470ae7436bace68813e_l3.svg" height="1" width="1" class="ql-img-displayed-equation " alt="\[\quad\]" title="Rendered by QuickLaTeX.com"/>
<ul>
        <li> <strong>Piani incidenti</strong>: in tal caso i due piani si intersecano lungo una retta.</li>
        <li> <strong>Piani paralleli</strong>: in tal caso i due piani non si intersecano.</li>
        <li> <strong>Piani coincidenti</strong>: in tal caso i due piani condividono tutti i punti.</li>
</ul>
Notiamo che due piani in tre dimensioni non possono essere sghembi.
 Per lo studio di tale problema geometrico proponiamo nuovamente due metodologie, una basata sulle equazioni parametriche ed una basata su quelle cartesiane.
 L'approccio parametrico si focalizza sullo studio della dipendenza fra le giaciture dei piani, ottenibili per ispezione diretta dai coefficienti dei parametri.
<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-27e49cccda278470ae7436bace68813e_l3.svg" height="1" width="1" class="ql-img-displayed-equation " alt="\[\quad\]" title="Rendered by QuickLaTeX.com"/>
<ul>
        <li> <strong>Piani incidenti</strong>: in questo caso l'intersezione fra gli spazi di giacitura dei due piani è uno spazio vettoriale di dimensione uno, corrispondente allo spazio direttore della retta in cui i piani si intersecano. </li>
        <li> <strong>Piani paralleli</strong>: in questo caso gli spazi di giacitura coincidono, ma i piani non condividono alcun punto, quindi scelto un qualsiasi punto di

*** Error message:
Error: get_image_size(): -1

\pi\sigma\pi\sigma2

\pi\sigma\pi\sigma

2\infty^1211

2

1\sigma

(2,-1,4)\in\pi\sigmau=u’=-1r:\begin{cases}x=t\\y=t\\z=1\end{cases},\quad t\in\mathbb{R}\quad\qquad s:\begin{cases} x=-4t’\\y=-4t’\\z= -4\end{cases},\quad t’\in\mathbb{R}r:\begin{cases}x=t+1\\y=2t+3,\\z=t\end{cases},\quad t\in\mathbb{R}\quad s:\begin{cases} x=2t’\\y=-t’\\z= 0\end{cases},\quad t’\in\mathbb{R}r:\begin{cases}x=2t\\y=-t+12,\\z=1-t\end{cases},\quad t\in\mathbb{R}\quad s:\begin{cases} x=26t’\\y=-t’\\z= -12\end{cases},\quad t’\in\mathbb{R}7

-4rz=1sz=-4

*** QuickLaTeX cannot compile formula:
deduciamo che non esiste un punto appartenente ad entrambe le rett, che quindi sono <strong>parallele</strong>.</li>
<li> Per prima cosa ricaviamo per ispezione diretta dalle equazioni parametriche i vettori direttori delle due rette:
		<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-e9738cb75f2a0eff2863c04b17b96a9f_l3.svg" height="19" width="253" class="ql-img-displayed-equation " alt="\[v_r=(1,2,1),\qquad v_s=(2,-1,0).\]" title="Rendered by QuickLaTeX.com"/>
		I due vettori non sono proporzionali, quindi le rette saranno incidenti o sghembe. Per determinare l'intersezione uguagliamo componente per componente le due equazioni parametriche. Si ricava il seguente sistema:
		<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-1bf506f151060fab98bd548e537b59f3_l3.svg" height="97" width="415" class="ql-img-displayed-equation " alt="\[\begin{cases} 		t+1=2t'\\2t+3=-t'\\t=0 		\end{cases}\iff \quad 		\begin{cases} 		1=2t'\\3=-t'\\t=0 		\end{cases}\iff  		\begin{cases} 		t'=\dfrac{1}{2}\\[7pt]t'=-3\\t=0, 		\end{cases}\]" title="Rendered by QuickLaTeX.com"/>
		dove nel primo passaggio la terza equazione è stata sostituita nelle altre due. Il sistema risulta incompatibile poiché

*** Error message:
Error: get_image_size(): -1

t’

trst=13t’=1P=(26,-1,-12)\piP_1=(1,1,0)P_2=(1,3,7)P_3=(2,4,1)\overrightarrow{P_2P_1}=P_2-P_1\overrightarrow{P_3P_1}=P_3-P_1P_1

uvuv

\pi

*** QuickLaTeX cannot compile formula:
passante per l'origine e parallelo alle seguenti rette date in forma cartesiana:
		<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-7fd8609d8f6eb6afd9b7d2d748001203_l3.svg" height="54" width="347" class="ql-img-displayed-equation " alt="\[r\colon\begin{cases} 		x - 2z =0 \\ 		y + z - 1 = 0 		\end{cases}  \mbox{ e } \; \; s:\begin{cases} 		x - 3z + 2 =0 \\ 		y + 2z + 4 = 0. 		\end{cases}\]" title="Rendered by QuickLaTeX.com"/>
</div>
[learn_more caption="Svolgimento."]
	Lo studio delle equazioni cartesiane e parametriche di una retta nello spazio è trattato, rispettivamente, nelle soluzioni dell'esercizio

*** Error message:
Error: get_image_size(): -1

2713rs

\piP=(0, 1, 2)r

rr

h,k\in\mathbb{R}hkP

k\neq0

*** QuickLaTeX cannot compile formula:
. Il piano cercato è:
	<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c506558c186126ff784386ea1ddec1f2_l3.svg" height="19" width="391" class="ql-img-displayed-equation " alt="\[\pi\colon k(2x+y-z+1)=0 \iff 2x+y-z+1=0.\]" title="Rendered by QuickLaTeX.com"/>
	D'altronde, notando che il punto

*** Error message:
Error: get_image_size(): -1

P\pi

*** QuickLaTeX cannot compile formula:
cercato.
[/learn_more]
<div style="padding: 10px; background-color: #bef5ad;">
<strong style="color: #000000;">Esercizio 16</strong> <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-563061f31f78294986510095e759d908_l3.svg" class="ql-img-inline-formula " alt=" (\bigstar\largewhitestar\largewhitestar\largewhitestar\largewhitestar)" title="Rendered by QuickLaTeX.com" height="20" width="124" style="vertical-align: -5px;"/>. Si considerino il piano <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-abb8538a4c6254f5c11d17fef2ea7a3f_l3.svg" height="16" width="137" class="ql-img-displayed-equation " alt="\[\pi\colon 2x-y+z=0\]" title="Rendered by QuickLaTeX.com"/> e, per ogni

*** Error message:
Error: get_image_size(): -1

a\in\mathbb{R}r_a

a \in \mathbb{R}\pir_aar_a \subset \pi10

*** QuickLaTeX cannot compile formula:
.  Come prescritto, il modo più semplice per affrontare il problema è sostituire le equazioni parametriche della retta nell'equazione cartesiana del piano e studiare la dipendenza del valore

*** Error message:
Error: get_image_size(): -1

ta

a=0a=1a\neq 0,1

*** QuickLaTeX cannot compile formula:
retta e piano sono <strong>incidenti</strong> nel punto corrispondente al valore del parametro <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-21d7e32cc5976ac31fc9cc2e5fbf0467_l3.svg" height="36" width="78" class="ql-img-displayed-equation " alt="\[t=\dfrac{1}{1-a},\]" title="Rendered by QuickLaTeX.com"/> che sostituito nell'equazione parametrica della retta determina le coordinate del punto di intersezione:
		<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-35f288c21b937f6bab7cf2fbd7b10b38_l3.svg" height="44" width="222" class="ql-img-displayed-equation " alt="\[P=\left(\dfrac{1}{1-a},\dfrac{2+a-a^2}{1-a},a\right).\]" title="Rendered by QuickLaTeX.com"/> </li>
</ul>
[/learn_more]
<div style="padding: 10px; background-color: #bef5ad;">
<strong style="color: #000000;">Esercizio 17</strong> <img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-563061f31f78294986510095e759d908_l3.svg" class="ql-img-inline-formula " alt=" (\bigstar\largewhitestar\largewhitestar\largewhitestar\largewhitestar)" title="Rendered by QuickLaTeX.com" height="20" width="124" style="vertical-align: -5px;"/>. 	Scrivere le equazioni parametriche per la retta

*** Error message:
Error: get_image_size(): -1

r

z=t

*** QuickLaTeX cannot compile formula:
. La liceità di tale posizione sarà verificata a posteriori.
	<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-343f970cfbb1f6b9d5fc4ebdaa6ed24e_l3.svg" height="75" width="406" class="ql-img-displayed-equation " alt="\[\begin{cases} 	x-y=-t-4\\ 	x+y=5t-2\\ 	z=t 	\end{cases} \iff \quad 	\begin{cases} 	x=2t-3\\ 	y=3t+1\\ 	z=t\end{cases},\quad t\in\mathbb{R}.\]" title="Rendered by QuickLaTeX.com"/>
	dove nel secondo passaggio abbiamo esplicitato

*** Error message:
Error: get_image_size(): -1

xyttz=tzrx=tr\pit

PP=\left(-\dfrac{18}{7},\dfrac{23}{14},\dfrac{3}{14}\right).a\in \mathbb{R}

r=\alpha \cap \beta\alpha\beta

\beta

r

y=u

\left(\mathbb{N},\, \mathbb{Z},\, \mathbb{Q}\right)\mathbb{R}$

Tutte le cartelle di Analisi Matematica

Tutti gli esercizi di geometria

In questa sezione vengono raccolti molti altri esercizi che coprono tutti gli argomenti di geometria proposti all’interno del sito con lo scopo di offrire al lettore la possibilità di approfondire e rinforzare le proprie competenze inerenti a tali argomenti.

Risorse didattiche aggiuntive per approfondire la matematica