If $\mathrm{P}(2, \beta, \alpha)$ lies on the plane $x+2 y-z-2=0$ an $\mathrm{Q}(\alpha,-1, \beta)$ lies on…

If $\mathrm{P}(2, \beta, \alpha)$ lies on the plane $x+2 y-z-2=0$ an $\mathrm{Q}(\alpha,-1, \beta)$ lies on the plane $2 x-y+3 z+6=0$ then 1 direction cosines of the line PQ are
  1. $\left(-\frac{4}{\sqrt{17}}, 0, \frac{1}{\sqrt{17}}\right)$
  2. $\left(+\frac{4}{\sqrt{17}}, 0, \frac{1}{\sqrt{17}}\right)$
  3. $\left(\frac{1}{\sqrt{17}}, 0, \frac{4}{\sqrt{17}}\right)$
  4. $\left(-\frac{1}{\sqrt{17}}, 0, \frac{4}{\sqrt{17}}\right)$

Solution

$\begin{aligned} & \text { } P(2, \beta, \alpha) \text { lies on } x+2 y-z-2=0 \\ & \Rightarrow 2+2 \beta-\alpha-2=0 \\ & \Rightarrow \alpha=2 \beta ...(i)\\ & Q(\alpha,-1, \beta) \text { lies on } 2 x-y+3 z+6=0 \\ & \Rightarrow 2 \alpha+3 \beta+7=0 \end{aligned}$...(ii)
Solving (i) and (ii), $\alpha=-2, \beta=-1$ $\overrightarrow{\mathrm{PQ}}=-4 \hat{i}+\hat{k}$
So, direction cosines are $\left(\frac{-4}{\sqrt{17}}, 0, \frac{1}{\sqrt{17}}\right)$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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