Let p be the plane passing through the intersection of the planes $$\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 5$$ and $$\overrightarrow r \,.\,\left( {2\widehat i - \widehat j + \widehat k} \right) = 3$$, and the point (2, 1, $-$2). Let the position vectors of the points X and Y be $\widehat i - 2\widehat j + 4\widehat k$ and $5\widehat i - \widehat j + 2\widehat k$ respectively. Then the points :
Solution
<p>Let the equation of required plane</p>
<p>$\pi :(x + 3y - z - 5) + \lambda (2x - y + z - 3) = 0$</p>
<p>$\because$ (2, 1, $-$2) lies on it so, $2 + \lambda ( - 2) = 0$</p>
<p>$\Rightarrow \lambda = 1$</p>
<p>Hence, $\pi :3x + 2y - 8 = 0$</p>
<p>$\because$ ${\pi _x} = - 9$, ${\pi _y} = 5$, ${\pi _{x + y}} = 4$</p>
<p>${\pi _{x - y}} = - 22$ and ${\pi _{y - x}} = 6$</p>
<p>Clearly X and Y are on opposite sides of plane $\pi$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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