The two planes $4 x - 3 y + z = 6$ and $8 x + 7 y + 9 z = 1$ are perpendicular.
A)false
B)true

Question

Quizplus · Accepted Answer

The dot product of the normal vectors of the two planes must be zero for them to be perpendicular. The normal vector of the first plane is $4\mathbf{i} - 3\mathbf{j} + \mathbf{k}$ and for the second plane, it is $8\mathbf{i} + 7\mathbf{j} + 9\mathbf{k}$. The dot product is $4*8 - 3*7 + 1*9 = 32 - 21 + 9 = 20$, which is not zero, so the planes are not perpendicular.

The Two Planes 4x−3y+z=64 x - 3 y + z = 64x−3y+z=6

The Two Planes $4 x - 3 y + z = 6$