91Ó°ÊÓ

The two planes $y-5z=3$ and $6x-7y=5$

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 $y-5z=3$is ${\mathbf{N}}_1=⟨0,1,-5âŸ\copyright$and the normal vector to the plane $6x-7y=5$is ${\mathbf{N}}_2=⟨6,-7,0âŸ\copyright$

The normal vectors are not parallel since they are not scalar multiples of each other. 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 for the line of intersection.

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

$$\begin{gathered} \mathbf{d}={\mathbf{N}}_1×{\mathbf{N}}_2 \\ =⟨0,1,-5âŸ\copyright ×⟨6,-7,0âŸ\copyright \end{gathered}$$

$=\left|\begin{array}{c}ijk\\ 01-5\\ 6-70\end{array}\right|$

$=⟨-35,-30,-6âŸ\copyright$

To find the point of intersection, set $z=0$ in each of the equations $y-5z=3$ and $6x-7y=5$

The solution of the following equations

Therefore, the point that lies on the both planes is $\left(\frac{13}{3},3,0\right)$

The line of intersection is:

Therefore, line of intersection is