The angle made by a vector vec{B} = 3 hat{i} | vedclass

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

Question

(D) Given vector $\vec{B} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$.The unit vector along the $y$-axis is $\hat{j} = 0 \hat{i} + 1 \hat{j} + 0 \hat{k}$.The

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{2}{\sqrt{29}}\right)$

Vedclass · Answer

(D) Given vector $\vec{B} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$.The unit vector along the $y$-axis is $\hat{j} = 0 \hat{i} + 1 \hat{j} + 0 \hat{k}$.The angle $	heta$ between a vector $\vec{B}$ and the $y$-axis is given by the formula $\cos 	heta = \frac{\vec{B} \cdot \hat{j}}{|\vec{B}| |\hat{j}|}$.First,calculate the magnitude of $\vec{B}$: $|\vec{B}| = \sqrt{3^2 + 2^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29}$.The magnitude of $\hat{j}$ is $1$.The dot product $\vec{B} \cdot \hat{j} = (3 \hat{i} + 2 \hat{j} + 4 \hat{k}) \cdot (0 \hat{i} + 1 \hat{j} + 0 \hat{k}) = 2$.Therefore,$\cos 	heta = \frac{2}{\sqrt{29} 	imes 1} = \frac{2}{\sqrt{29}}$.Thus,$	heta = \cos^{-1}\left(\frac{2}{\sqrt{29}}ight)$.

The angle made by a vector $\vec{B} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ with the $y$-axis is:

Similar Questions

If the angle between $\hat{a}$ and $\hat{b}$ is $60^{\circ}$,then which of the following vector$(s)$ have magnitude one?

$\vec{A}$ is a vector of magnitude $2.7$ units due east. What is the magnitude and direction of vector $4 \vec{A}$?

If a vector $\overrightarrow{P}$ makes angles $\alpha, \beta,$ and $\gamma$ with the $X, Y,$ and $Z$ axes respectively,then $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = $

$A$ physical quantity that has a direction is called...

$A$ vector has a magnitude the same as that of $\overrightarrow{A} = 3 \hat{i} + 4 \hat{j}$ and is parallel to $\overrightarrow{B} = 4 \hat{i} + 3 \hat{j}$. The $x$ and $y$ components of this vector in the first quadrant are $X$ and $3$ respectively,where $X = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module