A line with positive direction cosines passes through the point $\mathrm{P}(2,-1,2)$ and makes equal angles…

A line with positive direction cosines passes through the point $\mathrm{P}(2,-1,2)$ and makes equal angles with the co-ordinate axes. The line meets the plane $2 x+y+z=9$ at point $\mathrm{Q}$. The length of the line segment $P Q$ equals
  1. $3$
  2. $\sqrt{2}$
  3. $\sqrt{3}$
  4. $2$

Solution

Since direction cosines of PQ are equal and positive. $\therefore \quad$ The d.r.s. of PQ are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ $\therefore \quad$ The equation of the line $\mathrm{PQ}$ is $\begin{aligned} & \frac{x-2}{\frac{1}{\sqrt{3}}}=\frac{y+1}{\frac{1}{\sqrt{3}}}=\frac{\mathrm{z}-2}{\frac{1}{\sqrt{3}}} \\ & \Rightarrow x-2=y+1=\mathrm{z}-2=\mathrm{k}, \text { say } \end{aligned}$ $\therefore \quad$ Co-ordinates of the point $\mathrm{Q}$ are $(\mathrm{k}+2, \mathrm{k}-1, \mathrm{k}+2)$ The point $\mathrm{Q}$ lies on the plane $2 x+y+z=9$ $\begin{array}{ll} \therefore & 2(\mathrm{k}+2)+\mathrm{k}-1+\mathrm{k}+2=9 \\ & \Rightarrow 4 \mathrm{k}+5=9 \quad \Rightarrow \mathrm{k}=1 \\ \therefore & \mathrm{Q} \equiv(3,0,3) \\ \therefore & \mathrm{PQ}=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2} \\ & =\sqrt{1+1+1}=\sqrt{3} \end{array}$

Asked in: MHT CET 2023 (14 May Shift 1)

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