If $\overrightarrow a = \alpha \widehat i + \beta \widehat j + 3\widehat k$,
$\overrightarrow b = - \beta \widehat i - \alpha \widehat j - \widehat k$ and
$\overrightarrow c = \widehat i - 2\widehat j - \widehat k$
such that $\overrightarrow a \,.\,\overrightarrow b = 1$ and $\overrightarrow b \,.\,\overrightarrow c = - 3$, then $${1 \over 3}\left( {\left( {\overrightarrow a \times \overrightarrow b } \right)\,.\,\overrightarrow c } \right)$$ is equal to _____________.
Answer (integer)
2
Solution
$$\overrightarrow a .\overrightarrow b = 1 \Rightarrow - \alpha \beta - \alpha \beta - 3 = 1$$<br><br>$\Rightarrow \alpha \beta = - 2$ .... (i)<br><br>$$\overrightarrow b .\overrightarrow c = - 3 \Rightarrow - \beta + 2\alpha + 1 = - 3$$<br><br>$2\alpha - \beta = - 4$ ..... (ii)<br><br>Solving (i) & (ii) $\alpha$ = $-$1, $\beta$ = 2,<br><br>$${1 \over 3}((\overrightarrow a \, \times \overrightarrow b )\,.\,\overrightarrow c ) = {1 \over 3}\left| {\matrix{
{ - 1} & 2 & 3 \cr
{ - 2} & 1 & { - 1} \cr
1 & { - 2} & { - 1} \cr
} } \right| = 2$$
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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