The planes `px+2y+2z-3=0 and 2x-y+z+2=0` intersect at an angle `pi/4`. What is the value of `p^2`?

Options

A. 24
B. 12
C. 6
D. 3

Answer

A

Explanation

We know that the between the planes `a_1x+b_1y +c_1z+d_1=0and a_2x+b_2y+c_z=d_2=0` is given by
`costheta=abs((a_1a_2+b_1b_2+c_1c_2)/sqrt(a_1^2+b_1^2+c_1^2sqrt(a_2^2+b_2^2+c_2^2)))`
Given equation of planes are `px+2y+2z-3=0 and 2x-y+z+2=0` ltbRgt On comparing with standard equations, we get `a_1=p,a_2=2,b_2=-1, c_1=2,c_2=1`
Also, `theta=pi/4` (given)
`:. cos.pi/4=abs((pxx2+2xx(-1)+2xx1)/(sqrt(p^2+4+4)sqrt(4+1+1)))`
`rArr 1/sqrt3=(2p)/(sqrt(p^2+8)sqrt6)rArr1/2=(4p^2)/((p^2+8)6)`
`rArr3/4=p^2/(p^2+8)`
`rArr 3p^2+24=4p^2rArr p^2=24`