QJEEAdv16Ma28aa: JEE Advanced 2016 Physics - Class 11 JEEAdv16Ma28aa Solution: Let be a unit vector in and Given that there exists a vector in …

We have $\widehat u = {u_1}\widehat i + {u_2}\widehat j + {u_3}\widehat k$

That is, $\left| {\widehat u} \right| = 1 = \sqrt {u_1^2 + u_2^2 + u_3^2}$

$\Rightarrow u_1^2 + u_2^2 + u_3^2 = 1$

Also, it is given that

$\widehat \omega = {1 \over {\sqrt 6 }}(\widehat i + \widehat j + 2\widehat k)$

That is, $\left| {\widehat \omega } \right| = 1$

Now, $\left| {\widehat u \times \overrightarrow v } \right| = 1$

That is, $\left| {\widehat u} \right|\left| {\overrightarrow v } \right|\sin \theta = 1 \Rightarrow \left| {\overrightarrow v } \right| = {1 \over {\sin \theta }}$

which shows that there are infinitely many possible values exist for $\overrightarrow v$ (here $\theta$ is angle between the vectors $\overrightarrow v$ and $\widehat u$ ).

Hence, option (B) is correct.

Now, $\widehat \omega \,.\,(\widehat u \times \overrightarrow v ) = 1$

$\left| {\widehat \omega \,.\,(\widehat u \times \overrightarrow v )} \right| = 1$

That is, $\left| {\widehat \omega } \right|\left| {\widehat u \times \overrightarrow v } \right|\cos \alpha = 1$

where $\alpha$ is the angle between $\widehat \omega$ and $\widehat u \times \overrightarrow v$ .

Therefore, (1)(1)cos $\alpha$ = 1

$\Rightarrow$ a = 0

which means that $\widehat \omega$ and $\widehat u \times \overrightarrow v$ are parallel vector or $\widehat \omega$ is perpendicular vector to $\widehat u$ and $\overrightarrow v$ .

$\widehat u\,.\,\widehat \omega = 0$

$({u_1}\widehat i + {u_2}\widehat j + {u_3}\widehat k)\,.\,\left( {{{(\widehat i + \widehat j + 2\widehat k)} \over {\sqrt 6 }}} \right) = 0$

${u_1} + {u_2} + 2{u_3} = 0$

If $\widehat u$ lies in xy-plane then u₃ = 0. Therefore,

${u_1} + {u_2} = 0 \Rightarrow {u_1} = - {u_2} \Rightarrow \left| {{u_1}} \right| = \left| {{u_2}} \right|$

Hence, option (C) is correct.