Medium MCQ +4 / -1 PYQ · JEE Mains 2022

Let the plane $P:\overrightarrow r \,.\,\overrightarrow a = d$ contain the line of intersection of two planes $$\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 6$$ and $$\overrightarrow r \,.\,\left( { - 6\widehat i + 5\widehat j - \widehat k} \right) = 7$$. If the plane P passes through the point $\left( {2,3,{1 \over 2}} \right)$, then the value of ${{|13\overrightarrow a {|^2}} \over {{d^2}}}$ is equal to :

  1. A 90
  2. B 93 Correct answer
  3. C 95
  4. D 97

Solution

<p>${P_1}:x + 3y - z = 6$</p> <p>${P_2}: - 6x + 5y - z = 7$</p> <p>Family of planes passing through line of intersection of P<sub>1</sub> and P<sub>2</sub> is given by $x(1 - 6\lambda ) + y(3 + 5\lambda ) + z( - 1 - \lambda ) - (6 + 7\lambda ) = 0$</p> <p>It passes through $\left( {2,3,{1 \over 2}} \right)$</p> <p>So, $$2(1 - 6\lambda ) + 3(3 + 5\lambda ) + {1 \over 2}( - 1 - \lambda ) - (6 + 7\lambda ) = 0$$</p> <p>$$ \Rightarrow 2 - 12\lambda + 9 + 15\lambda - {1 \over 2} - {\lambda \over 2} - 6 - 7\lambda = 0$$</p> <p>$\Rightarrow {9 \over 2} - {{9\lambda } \over 2} = 0 \Rightarrow \lambda = 1$</p> <p>Required plane is</p> <p>$- 5x + 8y - 2z - 13 = 0$</p> <p>Or $\overrightarrow r .\,( - 5\widehat i + 8\widehat j - 2\widehat k) = 13$</p> <p>$${{|13\overrightarrow a {|^2}} \over {|d{|^2}}} = {{{{13}^2}} \over {{{(13)}^2}}}.\,|\overrightarrow a {|^2} = 93$$</p>

About this question

Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios

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