A unit vector perpendicular to each of the ve | vedclass

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

Question

(A) vector perpendicular to both $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ is given by their cross product: $(\vec{a} + \vec{b}) 	imes (\vec{a}

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} - \frac{1}{\sqrt{6}}\hat{k}$

Vedclass · Answer

(A) vector perpendicular to both $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ is given by their cross product: $(\vec{a} + \vec{b}) 	imes (\vec{a} - \vec{b})$.Expanding this, we get: $\vec{a} 	imes \vec{a} - \vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{a} - \vec{b} 	imes \vec{b} = 0 - (\vec{a} 	imes \vec{b}) - (\vec{a} 	imes \vec{b}) - 0 = -2(\vec{a} 	imes \vec{b})$.First, calculate $\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1) = \hat{i} - 2\hat{j} + \hat{k}$.Thus, the vector perpendicular to both is $-2(\hat{i} - 2\hat{j} + \hat{k}) = -2\hat{i} + 4\hat{j} - 2\hat{k}$.The unit vector is $\pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{\sqrt{(-2)^2 + 4^2 + (-2)^2}} = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{\sqrt{4 + 16 + 4}} = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{\sqrt{24}} = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \pm \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}}$.This simplifies to $\pm (-\frac{1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} - \frac{1}{\sqrt{6}}\hat{k})$. Option $A$ matches this result.

$A$ unit vector perpendicular to each of the vectors $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ is . . . . . . where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.

Similar Questions

The adjacent sides of a parallelogram are $\bar{a} = \hat{j} + 2\hat{k}$ and $\bar{b} = \hat{i} + 2\hat{j}$. Find its area.

Let $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+2 \hat{k}$. Then the number of vectors $\vec{b}$ such that $\vec{b} \times \vec{c}=\vec{a}$ and $|\vec{b}| \in\{1, 2, \ldots, 10\}$ is

If $a = i + j + k$,$b = i + 3j + 5k$,and $c = 7i + 9j + 11k$,then the area of the parallelogram having diagonals $a + b$ and $b + c$ is

$A$ unit vector which is perpendicular to the vector $2\hat{i} - \hat{j} + 2\hat{k}$ and is coplanar with the vectors $\hat{i} + \hat{j} - \hat{k}$ and $2\hat{i} + 2\hat{j} - \hat{k}$ is

The position vectors of the points $A, B$ and $C$ are $i + j, j + k$ and $k + i$ respectively. The vector area of the $\Delta ABC = \pm \frac{1}{2} \vec{\alpha}$ where $\vec{\alpha} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module