JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 13)
If vectors $$\overrightarrow {{a_1}} = x\widehat i - \widehat j + \widehat k$$ and $$\overrightarrow {{a_2}} = \widehat i + y\widehat j + z\widehat k$$ are collinear, then a possible unit vector parallel to the vector $$x\widehat i + y\widehat j + z\widehat k$$ is :
$${1 \over {\sqrt 3 }}\left( {\widehat i - \widehat j + \widehat k} \right)$$
$${1 \over {\sqrt 2 }}\left( { - \widehat j + \widehat k} \right)$$
$${1 \over {\sqrt 2 }}\left( {\widehat i - \widehat j} \right)$$
$${1 \over {\sqrt 3 }}\left( {\widehat i + \widehat j - \widehat k} \right)$$
Explanation
$$\overrightarrow {{a_2}} = \lambda \overrightarrow {{a_1}} $$
$$\widehat i + y\widehat j + z\widehat k = \lambda (x\widehat i - \widehat j + \widehat k)$$
$$1 = \lambda x,y = - \lambda ,z = \lambda $$
$$x\widehat i + y\widehat j + z\widehat k = {1 \over \lambda }\widehat i - \lambda \widehat j + \lambda \widehat k$$
Unit vector $$ = {{{1 \over \lambda }\widehat i - \lambda \widehat j + \lambda \widehat k} \over {\sqrt {{1 \over {{\lambda ^2}}} + {\lambda ^2} + {\lambda ^2}} }}$$
$$ = {{\widehat i - {\lambda ^2}\widehat j + {\lambda ^2}\widehat k} \over {\sqrt {1 + 2{\lambda ^4}} }}$$
Let $${\lambda ^2} = 1$$, possible unit vector $$ = {{\widehat i - \widehat j + \widehat k} \over {\sqrt 3 }}$$
$$\widehat i + y\widehat j + z\widehat k = \lambda (x\widehat i - \widehat j + \widehat k)$$
$$1 = \lambda x,y = - \lambda ,z = \lambda $$
$$x\widehat i + y\widehat j + z\widehat k = {1 \over \lambda }\widehat i - \lambda \widehat j + \lambda \widehat k$$
Unit vector $$ = {{{1 \over \lambda }\widehat i - \lambda \widehat j + \lambda \widehat k} \over {\sqrt {{1 \over {{\lambda ^2}}} + {\lambda ^2} + {\lambda ^2}} }}$$
$$ = {{\widehat i - {\lambda ^2}\widehat j + {\lambda ^2}\widehat k} \over {\sqrt {1 + 2{\lambda ^4}} }}$$
Let $${\lambda ^2} = 1$$, possible unit vector $$ = {{\widehat i - \widehat j + \widehat k} \over {\sqrt 3 }}$$
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