The vector vec{a}=-hat{i}+2 hat{j}+hat{k} is | vedclass

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

Question

(A) Given $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$. The vector $\vec{b}$ is obtained by rotating $\vec{a}$ by $90^{\circ}$ such that it passes throug

Vedclass · Accepted Answer

$3 \sqrt{2}$

Vedclass · Answer

(A) Given $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$. The vector $\vec{b}$ is obtained by rotating $\vec{a}$ by $90^{\circ}$ such that it passes through the $y$-axis. This implies $\vec{b}$ lies in the plane of $\vec{a}$ and $\hat{j}$.Thus,$\vec{b} = \lambda(\vec{a} 	imes (\vec{a} 	imes \hat{j}))$.Calculating the triple product: $\vec{a} 	imes \hat{j} = (-\hat{i} + 2\hat{j} + \hat{k}) 	imes \hat{j} = -\hat{k} + \hat{i} = \hat{i} - \hat{k}$.Then $\vec{a} 	imes (\vec{a} 	imes \hat{j}) = (-\hat{i} + 2\hat{j} + \hat{k}) 	imes (\hat{i} - \hat{k}) = \hat{j} + 2\hat{k} + 2\hat{i} + \hat{j} = 2\hat{i} + 2\hat{j} + 2\hat{k}$.Since $|\vec{b}| = |\vec{a}| = \sqrt{(-1)^2 + 2^2 + 1^2} = \sqrt{6}$,we have $\sqrt{6} = |\lambda| \sqrt{2^2 + 2^2 + 2^2} = |\lambda| \sqrt{12} = 2\sqrt{3}|\lambda|$.So,$|\lambda| = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{1}{\sqrt{2}}$.Since $\vec{b}$ passes through the $y$-axis,$\vec{b} \cdot \hat{j} > 0$. Testing $\lambda = -\frac{1}{\sqrt{2}}$,$\vec{b} = -\frac{1}{\sqrt{2}}(2\hat{i} + 2\hat{j} + 2\hat{k}) = -\sqrt{2}\hat{i} - \sqrt{2}\hat{j} - \sqrt{2}\hat{k}$. This gives $\vec{b} \cdot \hat{j} = -\sqrt{2} Testing $\lambda = \frac{1}{\sqrt{2}}$,$\vec{b} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \sqrt{2}\hat{k}$. This gives $\vec{b} \cdot \hat{j} = \sqrt{2} > 0$. Thus $\vec{b} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \sqrt{2}\hat{k}$.Now,$3\vec{a} + \sqrt{2}\vec{b} = 3(-\hat{i} + 2\hat{j} + \hat{k}) + \sqrt{2}(\sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \sqrt{2}\hat{k}) = -3\hat{i} + 6\hat{j} + 3\hat{k} + 2\hat{i} + 2\hat{j} + 2\hat{k} = -\hat{i} + 8\hat{j} + 5\hat{k}$.The projection on $\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$ is $\frac{(- \hat{i} + 8\hat{j} + 5\hat{k}) \cdot (5\hat{i} + 4\hat{j} + 3\hat{k})}{\sqrt{5^2 + 4^2 + 3^2}} = \frac{-5 + 32 + 15}{\sqrt{50}} = \frac{42}{5\sqrt{2}} = \frac{21\sqrt{2}}{5} = 4.2\sqrt{2}$.

The vector $\vec{a}=-\hat{i}+2 \hat{j}+\hat{k}$ is rotated through a right angle,passing through the $y$-axis in its way and the resulting vector is $\vec{b}$. Then the projection of $3 \vec{a}+\sqrt{2} \vec{b}$ on $\vec{c}=5 \hat{i}+4 \hat{j}+3 \hat{k}$ is

Similar Questions

The area of the parallelogram for which the vectors $\hat{i}+\hat{j}+2 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$ are adjacent sides is equal to

If $\overrightarrow{A}=i-2j-3k,\,\overrightarrow{B}=2i+j-k,\,\overrightarrow{C}=i+3j-2k$,then $(\overrightarrow A \times \overrightarrow B ) \times \overrightarrow C $ is

Given $a = i + j - k$,$b = -i + 2j + k$,and $c = -i + 2j - k$. $A$ unit vector perpendicular to both $a + b$ and $b + c$ is

The area of a parallelogram whose adjacent sides are given by the vectors $i + 2j + 3k$ and $-3i - 2j + k$ (in square units) is

Let $\vec \alpha = 3\hat i + \hat j$ and $\vec \beta = 2\hat i - \hat j + 3\hat k.$ If $\vec \beta = \vec \beta _1 - \vec \beta _2,$ where $\vec \beta _1$ is parallel to $\vec \alpha$ and $\vec \beta _2$ is perpendicular to $\vec \alpha,$ then $\vec \beta _1 \times \vec \beta _2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module