Question

The angle between the plane passing through the points $A(0,0,0),B(1,1,1),C(3,2,1)$ & the plane passing through $A(0,0,0),B(1,1,1),D(3,1,2)$ is

• A. ${90}^o$
• B. ${45}^o$
• C. ${120}^o$
• D. ${30}^o$

Answer 1

The correct option is C ${120}^o$

$\begin{array}{c}{\pi }_1=ax+by+cz=0\\ a+b+c=0\\ 3a+2b+c=0\\ \frac{a}{-1}=\frac{-b}{1-3}=\frac{c}{2-3}\\ \frac{a}{-1}=\frac{b}{2}=\frac{c}{-1}\\ -x+2y-z=0\\ {\pi }_1:x-2y+z=0\\ and,\\ {\pi }_2=ax+by+cz=0\\ a+b+c=0\\ 3a+b+2c=0\\ \frac{a}{1}=\frac{-b}{2-3}=\frac{c}{1-3}\\ \frac{a}{1}=\frac{b}{1}=\frac{c}{-2}\\ {\pi }_2:x+y-2z=0\\ Now,\\ \cos \theta =\frac{\left(1-2-2\right)}{\sqrt{6}\sqrt{6}}=\frac{-3}{6}=\frac{-1}{2}\\ \therefore \theta ={120}^{\circ }\\ Hence,theoptionCisthecorrectanswer.\end{array}$