The unit vector perpendicular to the vectors $6 \hat{i} + 2 \hat{j} + 3 \hat{k} and 3 \hat{i} - 6 \hat{j} - 2 \hat{k}$ is

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

$\frac{2 \hat{i} - 3 \hat{j} - 6 \hat{k}}{7}$

$\frac{2 \hat{i} + 3 \hat{j} + 6 \hat{k}}{7}$

$\frac{2 \hat{i} - 3 \hat{j} + 6 \hat{k}}{7}$

$\frac{2 \hat{i} + 3 \hat{j} - 6 \hat{k}}{7}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\vec{P} = 0.866 \hat{i} + 0.500 \hat{j} + 0 \hat{k}, | \vec{P} | = 1$
$\vec{Q} = 0.259 \hat{i} + 0.966 \hat{j} + 0 \hat{k}, | \vec{Q} | = 1$
$Angle between two vectors \vec{P} and \vec{Q},$
$\cos θ = \frac{\vec{P} \cdot \vec{Q}}{| \vec{P} | | \vec{Q} |}$

$\begin{array}{l} \cos θ = (0.866, 0.500, 0) \cdot (0.259, 0.966, 0) \\ \cos θ = 0.224 + 0.483 \\ \cos θ = 0.707 \\ ∴ θ = 45^{\circ} \end{array}$

Unit vector perpendicular to both the given vectors is

$\frac{(6 \hat{i} + 2 \hat{j} + 3 \hat{k}) \times (3 \hat{i} - 6 \hat{j} - 2 \hat{k})}{| (6 \hat{i} + 2 \hat{j} + 3 \hat{k}) \times (3 \hat{i} - 6 \hat{j} - 2 \hat{k}) |} = \frac{2 \hat{i} + 3 \hat{j} - 6 \hat{k}}{7}$