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

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Aicharya · Accepted Answer

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 PQ is $\begin{aligned} & \frac{x-2}{\frac{1}{\sqrt{3}}}=\frac{y+1}{\frac{1}{\sqrt{3}}}=\frac{z-2}{\frac{1}{\sqrt{3}}} \\ & \Rightarrow x-2=y+1=z-2=\mathrm{k}, \text { say } \end{aligned}$ $\therefore \quad$ Co-ordinates of the point Q are $(k+2, k-1 ; k+2)$
The point Q lies on the plane $2 x+y+z=9$ $\begin{array}{ll} \therefore & 2(k+2)+k-1+k+2=9 \\ & \Rightarrow 4 k+5=9 \quad \Rightarrow k=1 \\ \therefore & Q \equiv(3,0,3) \\ \therefore & P Q=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2} \\ & =\sqrt{1+1+1} \\ & =\sqrt{3} \text { units } \end{array}$

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

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