Let P be a plane passing through the points (1, 0, 1), (1, $-$2, 1) and (0, 1, $-$2). Let a vector $\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$ be such that $\overrightarrow a$ is parallel to the plane P, perpendicular to $(\widehat i + 2\widehat j + 3\widehat k)$ and $\overrightarrow a \,.\,(\widehat i + \widehat j + 2\widehat k) = 2$, then ${(\alpha - \beta + \gamma )^2}$ equals ____________.
Answer (integer)
81
Solution
Equation of plane :<br><br>$$\left| {\matrix{
{x - 1} & {y - 0} & {z - 1} \cr
{1 - 1} & 2 & {1 - 1} \cr
{1 - 0} & {0 - 1} & {1 + 2} \cr
} } \right| = 0$$<br><br>$\Rightarrow 3x - z - 2 = 0$<br><br>$\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$ || to 3x $-$ z $-$ 2 = 0<br><br>$\Rightarrow 3\alpha - 8 = 0$ ..... (1)<br><br>$\overrightarrow a \bot \widehat i + \widehat j + 3\widehat k$<br><br>$\Rightarrow \alpha + 2\beta + 3\gamma = 0$ ...... (2)<br><br>$\overrightarrow a .(\widehat i + \widehat j + 2\widehat k) = 0$<br><br>$\Rightarrow$ $\alpha$ + $\beta$ + 2$\gamma$ = 2 ........ (3)<br><br>On solving 1, 2 & 3<br><br>$\alpha$ = 1, $\beta$ = $-$5, $\gamma$ = 3<br><br>So, ($\alpha$ $-$ $\beta$ + $\gamma$)<sup>2</sup> = 81
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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