Q82 / 155 · WBJEE 2016 · Mathematics

Angle between the planes $x+y+2 z=6$ and $2 x-y+z=9$ is

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

Solution

Equations of planes are $x+y+2 z-6=0$ and $\quad 2 x-y+z-9=0$ $\cos \theta=\frac{a_{1} a_{2}+b_{1}b_{2}+c _{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$ $=\frac{1 \times 2+1 \times(-1)+2 \times 1}{\sqrt{1+1+4} \sqrt{4+1+1}}$ $=\frac{2-1+2}{\sqrt{6} \times \sqrt{6}}=\frac{3}{6}=\frac{1}{2}$ $\Rightarrow \quad \cos \theta=\cos \left(\frac{\pi}{3}\right)$ $\therefore \quad \theta=\frac{\pi}{3}$

Hence, the correct answer is option: 'C'.

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