Question

Let $A(0,0,0),B(1,1,1),C(3,2,1)$ and $D(3,1,2)$ 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

Answer: (D)

Let $n_1$ and $n_2$ be the vectors normal to the planes $ABC$ and $ABD$ 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 planes, then $\theta$ is the acute angle between their normals $n_1$ and $n_2$
$\therefore \cos \theta =\frac{|-1+2+2|}{\sqrt{6}\cdot \sqrt{6}}$
$=\frac{3}{2}=\frac{1}{2}=\cos \frac{\pi }{3}$
$\Rightarrow =\frac{\pi }{3}$