Find an equation for the plane that contains the line $x=-1+3t$ , $y=1+2t$, $z=2+4t$ and is perpendicular to the plane $2x+y-3z=-4$.

My procedure:

The vector $\perp$ to the given plane and the vector $\parallel$ the given line must be parallel, therefore, by letting: $$\begin{align} \text{Vector parallel to line: }\quad\vec{u}&=3\hat{i}+2\hat{j}+4\hat{k}\\ \text{Vector perpendicular to plane: }\quad\vec{v}&=2\hat{i}+\hat{j}-3\hat{k}\\ \end{align}$$ the vector product of the two must be equal to $0$: $$\vec{u}\times\vec{v}=0 \\\ \vec{u}\times\vec{v}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 3 & 2 & 4 \\ 2 & 1 & -3 \end{vmatrix}=-10\hat{i}+17\hat{j}-\hat{k}$$ This shows that they are not parallel, in fact they have an angle of:$$\alpha=\cos^{-1}\left(\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}\right)=\cos^{-1}\left(-2\sqrt{\frac{2}{203}}\right)\approx 101.45°$$ Since the plane we are looking for must contain the line, and the line is not perpendicular to the given plane, then there cannot exist such plane.

Would this be correct? Because I was told by the professor that the plane exists. Thank you for any help!

For visualization purposes:

From the plot below, we can clearly see that the given line and plane are not perpendicular. enter image description here enter image description here

  • $\begingroup$ What do you mean by “the vector of the plane?” $\endgroup$
    – amd
    Sep 19, 2017 at 0:31
  • $\begingroup$ @amd the vector perpendicular to the plane $\endgroup$
    – DMH16
    Sep 19, 2017 at 1:01

1 Answer 1


For starters, the vector perpendicular to the unknown plane must be parallel to the given line.

  • $\begingroup$ Shouldn't it be perpendicular? If we define the unknown plane as: $Ax+By+Cz=d$, then $\langle A,B,C\rangle\cdot\langle3,2,4\rangle =0$. Since the unknown plane must contain the line, then the two vectors must be $\perp$ $\endgroup$
    – DMH16
    Sep 19, 2017 at 14:49
  • $\begingroup$ oops, sorry, perpendicular was what i meant. But you must write two equations involving the unknown plane. The other one is $(A,B,C)\cdot(2,1,-3)=0$. $\endgroup$
    – Philip Roe
    Sep 20, 2017 at 2:25
  • $\begingroup$ Yes! Which means that $\langle 3,2,4 \rangle \times \langle 2,1,-3\rangle =0$ but this is not verified! Which means that there cannot exist a plane that contains the given line and is perpendicular to the given plane. Try it! $\endgroup$
    – DMH16
    Sep 20, 2017 at 4:58
  • $\begingroup$ I explained this in my answer above. The line MUST be perpendicular to the give plane in order to find the unknown plane. $\endgroup$
    – DMH16
    Sep 20, 2017 at 5:07
  • 1
    $\begingroup$ Aha, now I think I understand where you are going wrong. For plane A to be perpendicular to plane B, it is not necessary that every line contained in plane A be perpendicular to plane B. Open a book at right angles. Draw a diagonal line on one page. $\endgroup$
    – Philip Roe
    Sep 20, 2017 at 8:58

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