JEE MAIN - Mathematics (2020 - 8th January Evening Slot - No. 17)
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
$$ - {1 \over 2}$$
$$ - {3 \over 2}$$
$${1 \over 2}$$
-1
Explanation
$$\overrightarrow a = \widehat i - 2\widehat j + \widehat k$$
$$\overrightarrow b = \widehat i - \widehat j + \widehat k$$
$$\left| {\overrightarrow a } \right|$$ = $$\sqrt 6 $$, $$\left| {\overrightarrow b } \right|$$ = $$\sqrt 3 $$
and $${\overrightarrow a .\overrightarrow b }$$ = 4
Given $$\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a $$
$$ \Rightarrow $$ $${\left( {\overrightarrow b \times \overrightarrow c } \right)}$$ - $${\left( {\overrightarrow b \times \overrightarrow a } \right)}$$ = 0
$$ \Rightarrow $$ $${\overrightarrow b \times \left( {\overrightarrow c - \overrightarrow a } \right)}$$ = 0
$$ \therefore $$ $${\overrightarrow b \parallel \left( {\overrightarrow c - \overrightarrow a } \right)}$$
$$ \Rightarrow $$ $${\left( {\overrightarrow c - \overrightarrow a } \right) = \lambda \overrightarrow b }$$
$$ \Rightarrow $$ $${\overrightarrow c = \overrightarrow a + \lambda \overrightarrow b }$$
$$ \Rightarrow $$ $${\overrightarrow c .\overrightarrow a = \overrightarrow a .\overrightarrow a + \lambda \overrightarrow a .\overrightarrow b }$$
$$ \Rightarrow $$ 0 = $${{{\left| {\overrightarrow a } \right|}^2} + \lambda \left( {\overrightarrow a .\overrightarrow b } \right)}$$
$$ \Rightarrow $$ $$\lambda $$ = $${{{ - {{\left| {\overrightarrow a } \right|}^2}} \over {\overrightarrow a .\overrightarrow b }}}$$ = $${{ - 6} \over 4} = - {3 \over 2}$$
$$ \therefore $$ $$\overrightarrow c $$ = $${\overrightarrow a - {3 \over 2}\overrightarrow b }$$
$$ \Rightarrow $$ $$\overrightarrow c $$ = ($$\widehat i - 2\widehat j + \widehat k$$) - $${3 \over 2}$$($$\widehat i - \widehat j + \widehat k$$)
$$ \Rightarrow $$ $$\overrightarrow c $$ = $$ - {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right)$$
$$ \therefore $$ $$\overrightarrow c .\overrightarrow b $$ = $$ - {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right)$$($$\widehat i - \widehat j + \widehat k$$)
$$ \therefore $$ $$\overrightarrow c .\overrightarrow b $$ = $$ - {1 \over 2}$$
$$\overrightarrow b = \widehat i - \widehat j + \widehat k$$
$$\left| {\overrightarrow a } \right|$$ = $$\sqrt 6 $$, $$\left| {\overrightarrow b } \right|$$ = $$\sqrt 3 $$
and $${\overrightarrow a .\overrightarrow b }$$ = 4
Given $$\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a $$
$$ \Rightarrow $$ $${\left( {\overrightarrow b \times \overrightarrow c } \right)}$$ - $${\left( {\overrightarrow b \times \overrightarrow a } \right)}$$ = 0
$$ \Rightarrow $$ $${\overrightarrow b \times \left( {\overrightarrow c - \overrightarrow a } \right)}$$ = 0
$$ \therefore $$ $${\overrightarrow b \parallel \left( {\overrightarrow c - \overrightarrow a } \right)}$$
$$ \Rightarrow $$ $${\left( {\overrightarrow c - \overrightarrow a } \right) = \lambda \overrightarrow b }$$
$$ \Rightarrow $$ $${\overrightarrow c = \overrightarrow a + \lambda \overrightarrow b }$$
$$ \Rightarrow $$ $${\overrightarrow c .\overrightarrow a = \overrightarrow a .\overrightarrow a + \lambda \overrightarrow a .\overrightarrow b }$$
$$ \Rightarrow $$ 0 = $${{{\left| {\overrightarrow a } \right|}^2} + \lambda \left( {\overrightarrow a .\overrightarrow b } \right)}$$
$$ \Rightarrow $$ $$\lambda $$ = $${{{ - {{\left| {\overrightarrow a } \right|}^2}} \over {\overrightarrow a .\overrightarrow b }}}$$ = $${{ - 6} \over 4} = - {3 \over 2}$$
$$ \therefore $$ $$\overrightarrow c $$ = $${\overrightarrow a - {3 \over 2}\overrightarrow b }$$
$$ \Rightarrow $$ $$\overrightarrow c $$ = ($$\widehat i - 2\widehat j + \widehat k$$) - $${3 \over 2}$$($$\widehat i - \widehat j + \widehat k$$)
$$ \Rightarrow $$ $$\overrightarrow c $$ = $$ - {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right)$$
$$ \therefore $$ $$\overrightarrow c .\overrightarrow b $$ = $$ - {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right)$$($$\widehat i - \widehat j + \widehat k$$)
$$ \therefore $$ $$\overrightarrow c .\overrightarrow b $$ = $$ - {1 \over 2}$$
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