Parker Bird

Answered

2022-07-23

I have got this line

$\mathrm{g}:\phantom{\rule{thickmathspace}{0ex}}\overrightarrow{\mathrm{x}\phantom{\rule{thickmathspace}{0ex}}}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\left(\begin{array}{c}-1\\ 2\\ 0\end{array}\right)\phantom{\rule{thickmathspace}{0ex}}+\phantom{\rule{thickmathspace}{0ex}}\mathrm{s}\cdot \left(\begin{array}{c}-1\\ 4\\ -1\end{array}\right)$

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:

$t(-1-s)+(2+4s)+t(-2)=0$

This is the same as:

$2-t+s(4-2t)=0$

$=>s={\textstyle \frac{t-2}{4-2t}}=-{\textstyle \frac{1}{2}}$

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?

$\mathrm{g}:\phantom{\rule{thickmathspace}{0ex}}\overrightarrow{\mathrm{x}\phantom{\rule{thickmathspace}{0ex}}}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\left(\begin{array}{c}-1\\ 2\\ 0\end{array}\right)\phantom{\rule{thickmathspace}{0ex}}+\phantom{\rule{thickmathspace}{0ex}}\mathrm{s}\cdot \left(\begin{array}{c}-1\\ 4\\ -1\end{array}\right)$

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:

$t(-1-s)+(2+4s)+t(-2)=0$

This is the same as:

$2-t+s(4-2t)=0$

$=>s={\textstyle \frac{t-2}{4-2t}}=-{\textstyle \frac{1}{2}}$

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?

Answer & Explanation

kartonaun

Expert

2022-07-24Added 14 answers

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.

As for t=2 the vector (−1,2,0) lies in the plane the whole line lies in that plane.

Taniya Burns

Expert

2022-07-25Added 4 answers

Calling

$p=(x,y,z{)}^{\u2020}\phantom{\rule{0ex}{0ex}}\overrightarrow{n}=(t,1,t{)}^{\u2020}\phantom{\rule{0ex}{0ex}}{p}_{0}=(-1,2,0{)}^{\u2020}\phantom{\rule{0ex}{0ex}}\overrightarrow{v}=(-1,4,-1{)}^{\u2020}$

we have the plane $\mathrm{\Pi}\to p\cdot \overrightarrow{n}=0$ and the line $L\to p={p}_{0}+s\overrightarrow{v}$ . Now if $L\in \mathrm{\Pi}$ we have

$({p}_{0}+s\overrightarrow{v})\cdot \overrightarrow{n}=0\Rightarrow \{\begin{array}{l}{p}_{0}\cdot \overrightarrow{n}=0\\ \overrightarrow{v}\cdot \overrightarrow{n}=0\end{array}$

From the second equation we have $-t+4-t=0\Rightarrow t=2$ and now making ${p}_{0}\cdot \overrightarrow{n}=2\times (-1)+1\times 2+2\times 0=0$ hence for $t=2\Rightarrow L\in \mathrm{\Pi}$

$p=(x,y,z{)}^{\u2020}\phantom{\rule{0ex}{0ex}}\overrightarrow{n}=(t,1,t{)}^{\u2020}\phantom{\rule{0ex}{0ex}}{p}_{0}=(-1,2,0{)}^{\u2020}\phantom{\rule{0ex}{0ex}}\overrightarrow{v}=(-1,4,-1{)}^{\u2020}$

we have the plane $\mathrm{\Pi}\to p\cdot \overrightarrow{n}=0$ and the line $L\to p={p}_{0}+s\overrightarrow{v}$ . Now if $L\in \mathrm{\Pi}$ we have

$({p}_{0}+s\overrightarrow{v})\cdot \overrightarrow{n}=0\Rightarrow \{\begin{array}{l}{p}_{0}\cdot \overrightarrow{n}=0\\ \overrightarrow{v}\cdot \overrightarrow{n}=0\end{array}$

From the second equation we have $-t+4-t=0\Rightarrow t=2$ and now making ${p}_{0}\cdot \overrightarrow{n}=2\times (-1)+1\times 2+2\times 0=0$ hence for $t=2\Rightarrow L\in \mathrm{\Pi}$

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