Question

$A(0,0,0)$ , $B(1,1,1)$ , $C(3,2,1)$ and $D(2,3,1)$ be four points. The angle between the planes through the points $A,B,C$ and through the points $A,B,D$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is D $\frac{\pi }{3}$

Let $n_1$ and $n_2$ be the vectors normal to the planes $ABC$ and respectively.

$n_1=AB\times AC=-i+2j-k$

$n_2=AB\times AD=i+j-2k$

Let $\theta$ be the acute angle between the plane, then $\theta$ is the acute angle between their normals $n_1$ and $n_2$

$\therefore \cos \theta =\frac{|-1+2+2|}{\sqrt{6}.\sqrt{6}}=\frac{3}{6}=\frac{1}{2}=\cos \frac{\pi }{3}$

$\Rightarrow \theta =\frac{\pi }{3}.$