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

$x-2z=7$ and $x-3y-4z=0$

Consider the two planes $x-2z=7$ and $x-3y-4z=0$

The goal is to determine the equation for the line of intersection between the two supplied planes.

The normal vector to the plane $x-2z=7$is ${\mathbf{N}}_1=⟨1,0,-2⟩$and the normal vector to the plane $x-3y-4z=0$is ${\mathbf{N}}_2=⟨1,-3,-4⟩$

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

Each of the normal vectors must be orthogonal to the line of intersection. Use the cross-product to obtain the direction vector, $\mathbf{d}$for the line of intersection.

The direction vector dis given by:

=\left|\begin{array}{ccc}

i j k \\

1 0 -2 \\

1 -3 -4

\end{array}\right| \\

$=⟨-6,2,-3⟩$

To find the point of intersection, set $z=0$in each of the equations $x-2z=7$and $x-3y-4z=0$

The solution of the following equations

$$\begin{gathered} x=7(\text{Set}z=0) \\ x-3y=0(\text{Set}z=0) \end{gathered}$$

is $x=7$and $y=\frac{7}{3}$

Therefore, the point that lies on the both planes is $\left(7,\frac{7}{3},0\right)$

The line of intersection is:

Therefore, line of intersection is