Let $\overrightarrow a = \widehat i - 2\widehat j + \widehat k$ and $\overrightarrow b = \widehat i - \widehat j + \widehat k$ be two vectors. If $\overrightarrow c$ is a vector such that $$\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a $$ and $\overrightarrow c .\overrightarrow a = 0$, then $\overrightarrow c .\overrightarrow b$ is equal to
Solution
$\overrightarrow a = \widehat i - 2\widehat j + \widehat k$
<br><br>$\overrightarrow b = \widehat i - \widehat j + \widehat k$
<br><br>$\left| {\overrightarrow a } \right|$ = $\sqrt 6$, $\left| {\overrightarrow b } \right|$ = $\sqrt 3$
<br><br>and ${\overrightarrow a .\overrightarrow b }$ = 4
<br><br>Given $$\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a $$
<br><br>$\Rightarrow$ ${\left( {\overrightarrow b \times \overrightarrow c } \right)}$ - ${\left( {\overrightarrow b \times \overrightarrow a } \right)}$ = 0
<br><br>$\Rightarrow$ $${\overrightarrow b \times \left( {\overrightarrow c - \overrightarrow a } \right)}$$ = 0
<br><br>$\therefore$ $${\overrightarrow b \parallel \left( {\overrightarrow c - \overrightarrow a } \right)}$$
<br><br>$\Rightarrow$ $${\left( {\overrightarrow c - \overrightarrow a } \right) = \lambda \overrightarrow b }$$
<br><br>$\Rightarrow$ ${\overrightarrow c = \overrightarrow a + \lambda \overrightarrow b }$
<br><br>$\Rightarrow$ $${\overrightarrow c .\overrightarrow a = \overrightarrow a .\overrightarrow a + \lambda \overrightarrow a .\overrightarrow b }$$
<br><br>$\Rightarrow$ 0 = $${{{\left| {\overrightarrow a } \right|}^2} + \lambda \left( {\overrightarrow a .\overrightarrow b } \right)}$$
<br><br>$\Rightarrow$ $\lambda$ = $${{{ - {{\left| {\overrightarrow a } \right|}^2}} \over {\overrightarrow a .\overrightarrow b }}}$$ = ${{ - 6} \over 4} = - {3 \over 2}$
<br><br>$\therefore$ $\overrightarrow c$ = ${\overrightarrow a - {3 \over 2}\overrightarrow b }$
<br><br>$\Rightarrow$ $\overrightarrow c$ = ($\widehat i - 2\widehat j + \widehat k$) - ${3 \over 2}$($\widehat i - \widehat j + \widehat k$)
<br><br>$\Rightarrow$ $\overrightarrow c$ = $- {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right)$
<br><br>$\therefore$ $\overrightarrow c .\overrightarrow b$ = $- {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right)$($\widehat i - \widehat j + \widehat k$)
<br><br>$\therefore$ $\overrightarrow c .\overrightarrow b$ = $- {1 \over 2}$
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%.