A plane $\pi$ contains the line r = i + $3$j - k +$\lambda(2$i - j + k). If $\pi$ is parallel to $3$i - $4$j, how to find the vector equation of $\pi$?
I know that parallel planes have the same perpendicular vector. But I don’t seem to see how to find any normal vectors to the plane. I can’t cross product two lines, can I? So, how should I solve this question? Thanks
Both the vectors $3i - 4j$ and $2i - j + k$ are parallel to the plane, so the normal vector of the plane is perpendicular to both of these vectors. We can find it using the cross product
$$r = \begin{array}{|c c c |} i & j & k \\ 3 & -4 & 0 \\ 2 & -1 & 1\end{array} = -4i-3j+5k$$
This gives us the equation of the plane as $-4x - 3y + 5z= c$ or $4x +3y -5z =d$. The point $i + 3j - k$ lies in this plane, so substituting these values gives us the value of $d$.
$$d = -4 -9 -5 = -18 $$
The equation of the plane is thus $ -4x -3y +5z = -18 $