Discussion Forum

Let P(1,-2,5) be the foot of the perpendicular drawn from the origin to the plane $\pi_{1}$  and  the same P be the foot of the perpendicular from (1,2,-1) to the plane $\pi_{2}$ . then the acute   angle between the planes $\pi_{1}$ and  $\pi_{2}$ is 

Answer:

Explanation:

According to given informations, the direction ratios of normal to the plane $\pi_{1}$ are 1,-0, -2-0, 5-0 ie  1,-2,5

 And similarly direction ratios of normal to the plane $\pi_{2} $ are

     1-1, 2-(-2), -1-5 i.e, 0,4,-6

 so, the acute angle between planes $\pi_{1} $  and $\pi_{2}$  is 

 $cos^{-1}|\left(\frac{(1)(0)+(-2)(4)+(5)(-6)}{\sqrt{1+4+25}\sqrt{0+16+36}}\right)|$

  =$cos^{-1}|\frac{-8-30}{\sqrt{30}\sqrt{52}}|=cos^{-1}\left(\frac{38}{2\sqrt{30\times 13}}\right)$

   = $\cos ^{-1}\left(\frac{19}{\sqrt{390}}\right)$

 hence, option (a) is correct