Three forces of magnitudes 1, 2, 3 text{ dyne | vedclass

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

Question

(C) Let the cube be placed in the coordinate system such that its edges lie along the $x, y,$ and $z$ axes. The diagonals of the three adjacent faces

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let the cube be placed in the coordinate system such that its edges lie along the $x, y,$ and $z$ axes. The diagonals of the three adjacent faces meeting at the origin are represented by the vectors $\vec{a} = \hat{i} + \hat{j}$,$\vec{b} = \hat{j} + \hat{k}$,and $\vec{c} = \hat{k} + \hat{i}$.The unit vectors along these directions are $\hat{u}_1 = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$,$\hat{u}_2 = \frac{\hat{j} + \hat{k}}{\sqrt{2}}$,and $\hat{u}_3 = \frac{\hat{k} + \hat{i}}{\sqrt{2}}$.The forces are given as $\vec{F}_1 = 1 \cdot \hat{u}_1$,$\vec{F}_2 = 2 \cdot \hat{u}_2$,and $\vec{F}_3 = 3 \cdot \hat{u}_3$.The resultant force $\vec{R} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \frac{1(\hat{i} + \hat{j}) + 2(\hat{j} + \hat{k}) + 3(\hat{k} + \hat{i})}{\sqrt{2}} = \frac{(1+3)\hat{i} + (1+2)\hat{j} + (2+3)\hat{k}}{\sqrt{2}} = \frac{4\hat{i} + 3\hat{j} + 5\hat{k}}{\sqrt{2}}$.The magnitude of the resultant force is $|\vec{R}| = \frac{\sqrt{4^2 + 3^2 + 5^2}}{\sqrt{2}} = \frac{\sqrt{16 + 9 + 25}}{\sqrt{2}} = \frac{\sqrt{50}}{\sqrt{2}} = \sqrt{25} = 5 	ext{ dynes}$.

Three forces of magnitudes $1, 2, 3 \text{ dynes}$ meet at a point and act along the diagonals of three adjacent faces of a cube. The resultant force is ............ $\text{dyne}$.

Similar Questions

If $G$ is the centroid of the $\triangle ABC$,then $\vec{GA} + \vec{GB} + \vec{GC}$ is equal to

Answer the following as true or false.
Two collinear vectors having the same magnitude are equal.

If $\bar{a}$ and $\bar{b}$ are vectors such that $|\bar{a}|=3$,$|\bar{b}|=2$ and $\bar{a} \cdot \bar{b}=5$,then $|\bar{a}-\bar{b}|=$

The position vector of a point that divides the line segment joining $P \equiv(1,2,-1)$ and $Q \equiv(-1,1,1)$ externally in the ratio $1: 2$ is:

$\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors. If $\alpha \vec{d}=\vec{a}+\vec{b}+\vec{c}$ and $\beta \vec{a}=\vec{b}+\vec{c}+\vec{d}$,then $|\vec{a}+\vec{b}+\vec{c}+\vec{d}|=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module