Let the image of the point $\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ in the plane $x-2 y+z-2=0$ be P. If the distance of the point $Q(6,-2, \alpha), \alpha > 0$, from $\mathrm{P}$ is 13 , then $\alpha$ is equal to ___________.
Answer (integer)
15
Solution
Image of point $\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$
<br/><br/>$$
\begin{aligned}
& \frac{x-\frac{5}{3}}{1}=\frac{y-\frac{5}{3}}{-2}=\frac{\mathrm{z}-\frac{8}{3}}{1}=\frac{-2\left(1 \times \frac{5}{3}+(-2) \times \frac{8}{3}+1 \times \frac{8}{3}-2\right)}{1^2+2^2+1^2} =\frac{1}{3} \\\\
& \therefore x=2, y=1, \mathrm{z}=3 \\\\
&PQ^2 = 13^2=(6-2)^2+(-2-1)^2+(\alpha-3)^2 \\\\
& \Rightarrow(\alpha-3)^2=144 \Rightarrow \alpha=15(\because \alpha>0)
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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