Question

Determine whether the planes are parallel, perpendicular or neither. If neither find the angle between them.

x+4y-3z=1, -3x-12y+6z=1

Answer 1

The normal vectors are
\[n_1=(1;4;-3),\quad n_2=(-3;-12;6).\]
The dot product of n_1 and n_2
\[n_1\cdot n_2=1\cdot(-3)+4\cdot(-12)+(-3)\cdot 6=-69.\]
The cosine of the angle between them is
\[\cos\theta=\frac{n_1\cdot n_2}{|n_1|\cdot|n_2|}=\frac{-69}{\sqrt{26}\sqrt{189}}\approx -0.984309\]
So
\[\theta\approx 169.84^\circ\]