Find a plane parallel to the following line: $$x=3t+2, y=-t+1, z=t-1$$ and perpendicular to:
$$5x-2y+z=3.$$
I've tried the following:
The normal vector to the plane above is <5,-2,1>.
The normal/direction vector to line is <3, -1, 1>.
I did the dot product on these two normal vectors, and I got 18. Now I'm not sure where to go from here to get the equation of the plane I'm supposed to be finding.
Best Answer
If a plane is given by $Ax + By + Cz = D$, then the direction of the vector normal to a plane is $\begin{bmatrix} A \\ B \\ C\end{bmatrix}$.
If a line is given by $x = M_1t + B_1, y = M_2t + B_2, z = M_3t + B_3$, then the direction of the line is $\begin{bmatrix} M_1 \\ M_2 \\ M_3\end{bmatrix}$
A line is parallel to a plane (I'm assuming this is what you mean) if it never intersects the plane or always intersects the plane. That means that the normal vector of the plane and the direction of the line are perpendicular.
Two vectors, $\begin{bmatrix} x_1 \\ y_1 \\ z_1\end{bmatrix}$ and $\begin{bmatrix} x_2 \\ y_2 \\ z_2\end{bmatrix}$, are perpendicular if and only if $x_1x_2 + y_1y_2 + z_1z_2 = 0$.