JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 4)
A vector $$\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k\left( {\alpha ,\beta \in R} \right)$$ lies in the plane of the vectors, $$\overrightarrow b = \widehat i + \widehat j$$ and $$\overrightarrow c = \widehat i - \widehat j + 4\widehat k$$. If $$\overrightarrow a $$ bisects the angle between $$\overrightarrow b $$ and $$\overrightarrow c $$, then:
$$\overrightarrow a .\widehat i + 3 = 0$$
$$\overrightarrow a .\widehat k - 4 = 0$$
$$\overrightarrow a .\widehat i + 1 = 0$$
$$\overrightarrow a .\widehat k + 2 = 0$$
Explanation
Angle bisector $$\overrightarrow a = \lambda \left( {\widehat b + \widehat c} \right)$$
= $$\lambda \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }} + {{\widehat i - \widehat j + 4\widehat k} \over {3\sqrt 2 }}} \right)$$
$$ \Rightarrow $$ $$\overrightarrow a = {\lambda \over {3\sqrt 2 }}\left( {4\widehat i + 2\widehat j + 4\widehat k} \right)$$
comparing with $$\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k$$
$${{2\lambda } \over {3\sqrt 2 }}$$ = 2
$$ \Rightarrow $$ $$\lambda $$ = $${3\sqrt 2 }$$
$$ \therefore $$ $$\overrightarrow a = \left( {4\widehat i + 2\widehat j + 4\widehat k} \right)$$
Then $$\overrightarrow a .\widehat k - 4 $$
= 4 - 4 = 0
= $$\lambda \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }} + {{\widehat i - \widehat j + 4\widehat k} \over {3\sqrt 2 }}} \right)$$
$$ \Rightarrow $$ $$\overrightarrow a = {\lambda \over {3\sqrt 2 }}\left( {4\widehat i + 2\widehat j + 4\widehat k} \right)$$
comparing with $$\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k$$
$${{2\lambda } \over {3\sqrt 2 }}$$ = 2
$$ \Rightarrow $$ $$\lambda $$ = $${3\sqrt 2 }$$
$$ \therefore $$ $$\overrightarrow a = \left( {4\widehat i + 2\widehat j + 4\widehat k} \right)$$
Then $$\overrightarrow a .\widehat k - 4 $$
= 4 - 4 = 0
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