## anonymous 5 years ago Determine whether the planes are parallel, perpendicular or neither. If neither find the angle between them. x+4y-3z=1, -3x-12y+6z=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 $\theta\approx 169.84^\circ$