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