MCQ
If the angle between the planes $\bar{r} \cdot(x \hat{i}+\hat{j}-\hat{k})=4$ and $\bar{r} \cdot(\hat{i}+x \hat{j}+\hat{k})=-1$ is $\frac{\pi}{3}$, then the value of $x$ is
  • 2
  • B
    $0$
  • C
    -2
  • D
    4

Answer

Correct option: A.
2
(A)
Here, $\overline{ n }_1=(x \hat{ i }+\hat{ j }-\hat{ k })$, and
$\overline{ n }_2=(\hat{ i }+x \hat{ j }-\hat{ k })$
$\therefore \quad \cos \theta=\left|\frac{\overline{ n }_1 \overline{ n }_2}{\left\|\overline{ n }_1\right\| \overline{ n }_2 \|}\right|$
$\Rightarrow \cos \frac{\pi}{3}=\left|\frac{(x \hat{ i }+\hat{ j }-\hat{ k }) \cdot(\hat{ i }+x \hat{ j }+\hat{ k })}{\sqrt{x^2+1+1} \cdot \sqrt{1+x^2+1}}\right|$
$\Rightarrow \frac{1}{2}= \pm\left(\frac{x+x-1}{x^2+2}\right)$
$\Rightarrow \frac{2 x-1}{x^2+2}=\frac{1}{2}$ ...(considering positive value)
$\Rightarrow x^2+2-4 x+2=0$
$\Rightarrow(x-2)^2=0$
$\Rightarrow x=2$

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