Find an equation of the line of intersection of the two given planes.

$x=4$ and $3x-5y+2z=-3$

Consider the two planes $x=4$ and $3x-5y+2z=-3$

The goal is to figure out what the equation is for the line of intersection of the two planes.

The normal vector to the plane $x=4$is ${\mathbf{N}}_1=⟨1,0,0⟩$and the normal vector to the plane $3x-5y+2z=-3$is ${\mathbf{N}}_2=⟨3,-5,2⟩$

Because vectors are not scalar multiples of each other, the normal vectors are not parallel. As a result, the planes must cross.

The line of intersection must be orthogonal to each of the normal vectors. To find the direction vector, $\mathbf{d}$, for the line of intersection, use the cross-product.

The direction vector $\mathbf{d}$is given by:

$=\left|\begin{array}{c}ijk\\ 100\\ 3-52\end{array}\right|$

$=⟨0,-2,-5⟩$

To find the point of intersection, set $z=0$ in each of the equations $x=4$ and $3x-5y+2z=-3$

The solution of the following equations

Therefore, the point that lies on the both planes is $(4,3,0)$

The line of intersection is:

Therefore, line of intersection is $\mathbf{r}\left(t\right)=⟨4,3+2t,5t⟩$