The unit vector perpendicular to both vectors | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To find the unit vector $\hat{n}$ perpendicular to both vectors $A$ and $B$,we use the formula $\hat{n} = \pm \frac{A 	imes B}{|A 	imes B|}$.Fir

Vedclass · Accepted Answer

$\frac{3 \hat{j} - 2 \hat{k}}{\sqrt{13}}$

Vedclass · Answer

(A) To find the unit vector $\hat{n}$ perpendicular to both vectors $A$ and $B$,we use the formula $\hat{n} = \pm \frac{A 	imes B}{|A 	imes B|}$.First,calculate the cross product $A 	imes B$:$A 	imes B = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & -2 & -3 \ 2 & 4 & 6 \end{vmatrix}$$= \hat{i}(-12 - (-12)) - \hat{j}(-18 - (-6)) + \hat{k}(-12 - (-4))$$= \hat{i}(0) - \hat{j}(-12) + \hat{k}(-8) = 12 \hat{j} - 8 \hat{k}$.Next,calculate the magnitude $|A 	imes B|$:$|A 	imes B| = \sqrt{12^2 + (-8)^2} = \sqrt{144 + 64} = \sqrt{208} = \sqrt{16 	imes 13} = 4\sqrt{13}$.Now,find the unit vector:$\hat{n} = \frac{12 \hat{j} - 8 \hat{k}}{4\sqrt{13}} = \frac{3 \hat{j} - 2 \hat{k}}{\sqrt{13}}$.Thus,the correct option is $A$.

The unit vector perpendicular to both vectors $A = -3 \hat{i} - 2 \hat{j} - 3 \hat{k}$ and $B = 2 \hat{i} + 4 \hat{j} + 6 \hat{k}$ is:

Similar Questions

In a right-handed Cartesian coordinate system (often referred to as a clockwise system when viewed from the positive axis),which of the following vector relations is correct?

Define the vector product of two vectors.

If $\vec{A} \times \vec{B} = \vec{B} \times \vec{C} = \vec{C} \times \vec{A}$,then $\vec{A} + \vec{B} + \vec{C}$ is equal to:

If the projection of $2 \hat{i} + 4 \hat{j} - 2 \hat{k}$ on $\hat{i} + 2 \hat{j} + \alpha \hat{k}$ is zero,then the value of $\alpha$ will be.

$\vec A$ and $\vec B$ are two vectors and $\theta$ is the angle between them. If $|\vec A \times \vec B| = \sqrt{3}(\vec A \cdot \vec B)$,the value of $\theta$ is ......... $^\circ$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module