The angles which a vector hat{i} + hat{j} + s | vedclass

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

Question

(C) Let the vector be $\vec{R} = \hat{i} + \hat{j} + \sqrt{2} \hat{k}$.Comparing this with $\vec{R} = R_x \hat{i} + R_y \hat{j} + R_z \hat{k}$,we get

Vedclass · Accepted Answer

$60^\circ, 60^\circ, 45^\circ$

Vedclass · Answer

(C) Let the vector be $\vec{R} = \hat{i} + \hat{j} + \sqrt{2} \hat{k}$.Comparing this with $\vec{R} = R_x \hat{i} + R_y \hat{j} + R_z \hat{k}$,we get $R_x = 1, R_y = 1, R_z = \sqrt{2}$.The magnitude of the vector is $|\vec{R}| = \sqrt{R_x^2 + R_y^2 + R_z^2} = \sqrt{1^2 + 1^2 + (\sqrt{2})^2} = \sqrt{1 + 1 + 2} = \sqrt{4} = 2$.The direction cosines are given by $\cos \alpha = \frac{R_x}{|\vec{R}|}, \cos \beta = \frac{R_y}{|\vec{R}|}, \cos \gamma = \frac{R_z}{|\vec{R}|}$.For the $X$-axis: $\cos \alpha = \frac{1}{2} \Rightarrow \alpha = 60^\circ$.For the $Y$-axis: $\cos \beta = \frac{1}{2} \Rightarrow \beta = 60^\circ$.For the $Z$-axis: $\cos \gamma = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow \gamma = 45^\circ$.Thus,the angles are $60^\circ, 60^\circ, 45^\circ$.

The angles which a vector $\hat{i} + \hat{j} + \sqrt{2} \hat{k}$ makes with $X, Y$ and $Z$ axes respectively are

Similar Questions

Define scalar and vector physical quantities with their examples.

The angle subtended by the vector $\vec{A} = 4 \hat{i} + 3 \hat{j} + 12 \hat{k}$ with the $x$-axis is

Find the unit vector in the direction of $2\hat{i} + 3\hat{j} + 4\hat{k}$.

If $|\vec{P}+\vec{Q}|=|\vec{P}|=|\vec{Q}|$,then the angle between $\vec{P}$ and $\vec{Q}$ is: (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module