If the lines ${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1}$ and ${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1}$ intersect at the point P, then the distance of the point P from the plane $z = a$ is :
Solution
<p>${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1} = \lambda$ (say)</p>
<p>& ${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1} = \mu$ (say)</p>
<p>$\therefore$ $\lambda + 1 = 2\mu + a$ ...... (i)</p>
<p>$2\lambda + 2 = 3\mu - 2$ ..... (ii)</p>
<p>$\lambda - 3 = \mu + 3$ .... (iii)</p>
<p>By (i) & (ii)</p>
<p>$\Rightarrow 3\mu - 2 = 4\mu + 2a + 2$</p>
<p>$\mu=-2(1+a)$ & $\lambda=5-3a$</p>
<p>Put $\lambda$ & $\mu$ in (iii) we get</p>
<p>$a=-9$</p>
<p>$\mu=16$</p>
<p>$\lambda=22$</p>
<p>$\therefore$ Point of intersection $\equiv(23,46,19)$</p>
<p>Distance from $z=-9$ is 28</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%.