If $\vec{A} = 2 \hat{i} + \hat{j} + \hat{k} and \vec{B} = \hat{i} + \hat{j} + \hat{k}$ are two vectors, then the unit vector is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$Parallel to \vec{A} is \frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}$

$Perpendicular to \vec{A} is \frac{- \hat{j} + \hat{k}}{\sqrt{2}}$

$Perpendicular to \vec{B} is (\frac{- \hat{j} + \hat{k}}{\sqrt{2}})$

$Parallel to \vec{A} is \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \vec{A} \times \vec{B} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}| \\ = \hat{i} (1 - 1) - \hat{j} (2 - 1) + \hat{k} (2 - 1) = - \hat{j} + \hat{k} \end{array}$

The unit vector perpendicular to $\vec{A} and \vec{B} is (\frac{- \hat{j} + \hat{k}}{\sqrt{2}})$ . So choices (a) and (c) are correct.
Any vector whose magnitude is K (constant) times $(2 \hat{i} + \hat{j} + \hat{k})$ is parallel to $\vec{A}$ .
So, unit vector $\frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}$ is parallel to $\vec{A}$ .
So, choice (b) is correct.