Let $\overrightarrow a$ = 2$\widehat i$ $-$ 3$\widehat j$ + 4$\widehat k$ and $\overrightarrow b$ = 7$\widehat i$ + $\widehat j$ $-$ 6$\widehat k$.
If $\overrightarrow r$ $\times$ $\overrightarrow a$ = $\overrightarrow r$ $\times$ $\overrightarrow b$, $\overrightarrow r$ . ($\widehat i$ + 2$\widehat j$ + $\widehat k$) = $-$3, then $\overrightarrow r$ . (2$\widehat i$ $-$ 3$\widehat j$ + $\widehat k$) is equal to :
Solution
$\overrightarrow a = (2, - 3,4)$, $\overrightarrow b = (7,1, - 6)$<br><br>$$\overrightarrow r \times \overrightarrow a - \overrightarrow r \times \overrightarrow b = 0$$<br><br>$\overrightarrow r \times (\overrightarrow a - \overrightarrow b ) = 0$<br><br>$\overrightarrow r = \lambda (\overrightarrow a - \overrightarrow b )$<br><br>$\overrightarrow r = \lambda ( - 5\widehat i - 4\widehat j + 10\widehat k)$<br><br>$\overrightarrow r \,.\,(2, - 3,1) = ?$<br><br>Given $\overrightarrow r \,.\,(1,2,1) = - 3$<br><br>$\lambda ( - 5 - 8 + 10) = - 3 \Rightarrow \lambda = 1$<br><br>$\therefore$ $( - 5, - 4,10)\,.\,(2, - 3,1)$<br><br>= - 10 + 12 + 10 = 12
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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