 Parker Bird

2022-07-23

I have got this line
$\mathrm{g}:\phantom{\rule{thickmathspace}{0ex}}\stackrel{\to }{\mathrm{x}\phantom{\rule{thickmathspace}{0ex}}}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\left(\begin{array}{c}-1\\ 2\\ 0\end{array}\right)\phantom{\rule{thickmathspace}{0ex}}+\phantom{\rule{thickmathspace}{0ex}}\mathrm{s}\cdot \left(\begin{array}{c}-1\\ 4\\ -1\end{array}\right)$
And this plane:
${E}_{t}:\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}tx+y+tz=0$
Now for which t is the line element of the plane?
My calculations: When I put the equation of the line in the equation of the plane i get: And this plane:
$t\left(-1-s\right)+\left(2+4s\right)+t\left(-2\right)=0$
This is the same as:
$2-t+s\left(4-2t\right)=0$
$=>s=\frac{t-2}{4-2t}=-\frac{1}{2}$
And now I do not know how to go on.
The solution is t=2
Can someone explain the last step to the solution please? kartonaun

Expert

The line and the plane are parallel iff the plane's normal vector (t,1,t) and the line's direction vector (−1,4,−1) are perpendicular; this gives t=2.
As for t=2 the vector (−1,2,0) lies in the plane the whole line lies in that plane. Taniya Burns

Expert

Calling
$p=\left(x,y,z{\right)}^{†}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{n}=\left(t,1,t{\right)}^{†}\phantom{\rule{0ex}{0ex}}{p}_{0}=\left(-1,2,0{\right)}^{†}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{v}=\left(-1,4,-1{\right)}^{†}$
we have the plane $\mathrm{\Pi }\to p\cdot \stackrel{\to }{n}=0$ and the line $L\to p={p}_{0}+s\stackrel{\to }{v}$ . Now if $L\in \mathrm{\Pi }$ we have
$\left({p}_{0}+s\stackrel{\to }{v}\right)\cdot \stackrel{\to }{n}=0⇒\left\{\begin{array}{l}{p}_{0}\cdot \stackrel{\to }{n}=0\\ \stackrel{\to }{v}\cdot \stackrel{\to }{n}=0\end{array}$
From the second equation we have $-t+4-t=0⇒t=2$ and now making ${p}_{0}\cdot \stackrel{\to }{n}=2×\left(-1\right)+1×2+2×0=0$ hence for $t=2⇒L\in \mathrm{\Pi }$

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