Question

Angle between the planes $2x-y+z=6$ and $x+y+2z=3$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{2}$
• D. 0

Answer 1

Answer: (A)

For the plane $2x-y+z=6,$
$a_1=2,b_1=-1,c_1=1$ and for the plane
$x+y+2z=3,$ $a_2=1,b_2=1,c_2=2$
$\therefore$ $\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$=\frac{2\times 1+1\times -1+2\times 1}{\sqrt{2^2+{(-1)}^2+1^2}\sqrt{1^2+1^2+2^2}}$
$=\frac{2-1+2}{\sqrt{4+1+1}\sqrt{1+1+4}}$
$=\frac{3}{6}=\frac{1}{2}$
$\Rightarrow$ $\cos \theta =\frac{1}{2}=\cos \frac{\pi }{3}$
$\Rightarrow$ $\theta =\frac{\pi }{3}$