Let a line with direction ratios $a,-4 a,-7$ be perpendicular to the lines with direction ratios $3,-1,2 b$ and $b, a,-2$. If the point of intersection of the line $\frac{x+1}{a^{2}+b^{2}}=\frac{y-2}{a^{2}-b^{2}}=\frac{z}{1}$ and the plane $x-y+z=0$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to _________.
Answer (integer)
10
Solution
<p>Given $a\,.\,3 + ( - 4a)( - 1) + ( - 7)2b = 0$ ...... (1)</p>
<p>and $ab - 4{a^2} + 14 = 0$ ....... (2)</p>
<p>$\Rightarrow {a^2} = 4$ and ${b^2} = 1$</p>
<p>$\therefore$ $L \equiv {{x + 1} \over 5} = {{y - 2} \over 3} = {z \over 1} = \lambda$ (say)</p>
<p>$\Rightarrow$ General point on line is $(5\lambda - 1,\,3\lambda + 2,\,\lambda )$ for finding point of intersection with $x - y + z = 0$ we get $(5\lambda - 1) - (3\lambda + 2) + (\lambda ) = 0$</p>
<p>$\Rightarrow 3\lambda - 3 = 0 \Rightarrow \lambda = 1$</p>
<p>$\therefore$ Point at intersection $(4,\,5,\,1)$</p>
<p>$\therefore$ $\alpha + \beta + \gamma = 4 + 5 + 1 = 10$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
This question is part of PrepWiser's free JEE Main question bank. 182 more solved questions on Three Dimensional Geometry are available — start with the harder ones if your accuracy is >70%.