Let $\overrightarrow a = \widehat i + 5\widehat j + \alpha \widehat k$, $\overrightarrow b = \widehat i + 3\widehat j + \beta \widehat k$ and $\overrightarrow c = - \widehat i + 2\widehat j - 3\widehat k$ be three vectors such that, $\left| {\overrightarrow b \times \overrightarrow c } \right| = 5\sqrt 3$ and ${\overrightarrow a }$ is perpendicular to ${\overrightarrow b }$. Then the greatest amongst the values of ${\left| {\overrightarrow a } \right|^2}$ is _____________.
Answer (integer)
90
Solution
Since, $\overrightarrow a .\,\overrightarrow b = 0$<br><br>$1 + 15 + \alpha \beta = 0 \Rightarrow \alpha \beta = - 16$ .... (1)<br><br>Also, <br><br>$${\left| {\overrightarrow b \, \times \overrightarrow c } \right|^2} = 75 \Rightarrow (10 + {\beta ^2})14 - {(5 - 3\beta )^2} = 75$$<br><br>$\Rightarrow$ 5$\beta$<sup>2</sup> + 30$\beta$ + 40 = 0<br><br>$\Rightarrow$ $\beta$ = $-$4, $-$2<br><br>$\Rightarrow$ $\alpha$ = 4, 8<br><br>$$ \Rightarrow \left| {\overrightarrow a } \right|_{\max }^2 = {(26 + {\alpha ^2})_{\max }} = 90$$
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
This question is part of PrepWiser's free JEE Main question bank. 169 more solved questions on Vector Algebra are available — start with the harder ones if your accuracy is >70%.