Question

Angle between the planes $2x+2y-3z-5=0$ and $3x-3y+5z-3=0$ is ${\cos }^{-1}\frac{k}{\sqrt{17}\sqrt{43}}$ then $k$ is

Answer 1

We have,

$P_1:2x+2y-3z=5$

$P_2:3x-3y+5z=3$
Angle between the normals is same as angle between the planes.

$\therefore \cos \theta =\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\cos \theta =\frac{\mid 6-6-15\mid }{\sqrt{4+4+9}\sqrt{9+9+25}}$

$\cos \theta =\frac{15}{\sqrt{17}\sqrt{43}}$

$\theta ={\cos }^{-1}\left(\frac{15}{\sqrt{17}\sqrt{43}}\right)$