The vector equation of the plane passing through the intersection
of the planes $\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$ and $\overrightarrow r .\left( {\widehat i - 2\widehat j} \right) = - 2$, and the point (1, 0, 2) is :
Solution
Given, point (1, 0, 2)<br/><br/>Equation of plane = <br/><br/>$\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) = 1$ and $\overrightarrow r\,.\,(\widehat i - 2\widehat j) = - 2$<br/><br/>Equation of plane passing through the intersection of given planes is<br/><br/>$$[\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) - 1] + \lambda [\overrightarrow r\,.\,(\widehat i - 2\widehat j) + 2] = 0$$<br/><br/>$\because$ This plane passes through point (1, 0, 2) i.e., <br/><br/>vector $(\widehat i + 2\widehat k)$<br/><br/>$\therefore$ $$[(\widehat i + 2\widehat k)\,.\,(\widehat i + \widehat j + \widehat k) - 1] + \lambda [(\widehat i + 2\widehat k)\,.\,(\widehat i - 2\widehat j) + 2] = 0$$<br/><br/>$\Rightarrow (3 - 1) + \lambda (1 + 2) = 0$<br/><br/>$\Rightarrow 2 + \lambda \times 3 = 0$<br/><br/>$\Rightarrow \lambda = - 2/3$<br/><br/>Hence, equation of required plane is<br/><br/>$$[\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) - 1] + \left( {{{ - 2} \over 3}} \right)[\overrightarrow r\,.\,(\widehat i - 2\widehat j) + 2] = 0$$<br/><br/>$\Rightarrow$ $$3[\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) - 1] - 2[\overrightarrow r\,.\,(\widehat i - 2\widehat j) + 2] = 0$$<br/><br/>$\Rightarrow$ $\overrightarrow r\,.\,(\widehat i + 7\widehat j + 3\widehat k) = 7$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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