A line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the…
- 1
- $\sqrt{2}$
- $\sqrt{3}$
- 2
Solution

$\therefore$ Equation of line are $ \begin{gathered} \frac{x-2}{1 / \sqrt{3}}=\frac{y+1}{1 / \sqrt{3}}=\frac{z-2}{1 / \sqrt{3}} \\ \Rightarrow \quad x-2=y+1=z-2=r \text { (say) } \end{gathered} $ $\therefore$ Any point on the line is $ Q \equiv(r+2 r-1, r+2) $ $\because Q$ lies on the plane $2 x+y+z=9$ $ \begin{aligned} & \therefore \quad 2(r+2)+(r-1)+(r+2)=9 \\ & \Rightarrow 4 r+5=9 \Rightarrow r=1 \Rightarrow Q(3,0,3) \\ & \therefore P Q=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2} \\ & =\sqrt{3} \end{aligned} $
Asked in: JEE Advanced 2009 (Paper 2)