Discussion Forum

Let  $\hat{u}=u_{1}\hat{i}+u_{2}\hat{j}+u_{3}\hat{k}$ be a unit vector  R³ and $\hat{w}=\frac{1}{\sqrt{6}}({}\hat{i}+\hat{j}+2\hat{k})$. Given that there exists vector  $\overrightarrow{v}$  in  $R^{3}$ , such  that  $|\hat{u}+\vec{v}|=1$ and  $\hat{w}.(\hat{u}+\vec{v})=1$   which of the following statement(s) is/are correct?

Answer:

Explanation:

Let θ be the angle between  $\hat{u}$ and $\vec{v}$

$\therefore|\hat{u} \times \overrightarrow{v}|=1\Rightarrow|\hat{u}|\overrightarrow{v}|\sin \theta=1$

$\therefore |\overrightarrow{v}| \sin \theta=1$ $[\therefore |\hat{u}|=1]$  .......(i)

Clearly, there may be infinite vectors

 $\overrightarrow{OP}=\overrightarrow{v}$ , such that  P is always 1 unit distance from  $\hat{u}$.

$\therefore$    Option (b) is correct.

Again, let Φ be the angle between $\hat{w}$ and  $\hat{u}\times\overrightarrow{v}.$

$\therefore \hat{w}.(\hat{u}\times\overrightarrow{v})=1\Rightarrow |\hat{w}||\hat{u}\times\overrightarrow{v}|\cos \phi =1$

$\Rightarrow \cos \phi=1\Rightarrow\phi=0$

Thus,    $\hat{w}=\hat{u}\times \overrightarrow{v}$

Now, if  $\hat{u}$ lies in XY-plane, then

$\begin{vmatrix}\hat{i} & \hat{j}& \hat{k} \\u_{1} & u_{2}&0 \\ v_{1}&v_{2}&v_{3} \end{vmatrix}$

 or      $\hat{u}=u\hat{i}+u_{2}\hat{j}$

     $\therefore\hat{w}=(u_{2}v_{3})\hat{i}-(u_{1}v_{3})\hat{j}+(u_{1}v_{2}-v_{1}u_{2})\hat{k}$

 $\therefore\hat{w}=\frac{1}{\sqrt{6}}(\hat{i}+\hat{j}+2\hat{k})$

$\therefore$     $u_{2}v_{3}=\frac{1}{\sqrt{6}}\Rightarrow u_{1}v_{3}=\frac{-1}{\sqrt{6}}$

$\Rightarrow  \frac{u_{2}v_{3}}{u_{1}v_{3}}=-1$ or   $|u_{1}|=|u_{2}|$

$\therefore$ Option (c) is correct.

Now, if  $\hat{u}$ lies un XZ-plane, then  $\hat{u}= u_{1}\hat{i}+u_{3}\hat{k}$

    $\therefore$       

$\begin{vmatrix}\hat{i} & \hat{j}& \hat{k} \\u_{1} & 0&u_{3} \\ v_{1}&v_{2}&v_{3} \end{vmatrix}$

$\Rightarrow w=(-v_{2}u_{3})\hat{i}-(u_{1}v_{3}-u_{3}v_{1})\hat{j}+(u_{1}v_{2})\hat{k}$

$\Rightarrow \hat{w}=\frac{1}{\sqrt{6}}(\hat{i}+\hat{j}+2\hat{k})$

$= -v_{2}u_{3}=\frac{1}{\sqrt{6}}$  and  $u_{1}v_{2}=\frac{2}{\sqrt{6}}$

$=|u_{2}|=2|u_{3}|$

  $\therefore$  Option (d) is incorrect