If P + Q = R and |P| = |Q| = sqrt{3} and |R| | vedclass

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

Question

(C) The magnitude of the resultant vector $R$ is given by the formula: $|R| = \sqrt{|P|^2 + |Q|^2 + 2|P||Q| \cos 	heta}$,where $	heta$ is the angle

Vedclass · Accepted Answer

$\pi / 3$

Vedclass · Answer

(C) The magnitude of the resultant vector $R$ is given by the formula: $|R| = \sqrt{|P|^2 + |Q|^2 + 2|P||Q| \cos 	heta}$,where $	heta$ is the angle between vectors $P$ and $Q$.Given $|P| = |Q| = \sqrt{3}$ and $|R| = 3$.Substituting the values into the formula:$3 = \sqrt{(\sqrt{3})^2 + (\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{3}) \cos 	heta}$$3 = \sqrt{3 + 3 + 6 \cos 	heta}$Squaring both sides:$9 = 6 + 6 \cos 	heta$$3 = 6 \cos 	heta$$\cos 	heta = 3 / 6 = 1 / 2$Since $\cos 	heta = 1 / 2$,the angle $	heta = 60^{\circ}$ or $\pi / 3$ radians.

If $P + Q = R$ and $|P| = |Q| = \sqrt{3}$ and $|R| = 3$,then the angle between $P$ and $Q$ is

Similar Questions

If $\overrightarrow A = 4\hat i - 3\hat j$ and $\overrightarrow B = 6\hat i + 8\hat j$,find the magnitude and direction of $\overrightarrow A + \overrightarrow B$.

Two forces of magnitude $F$ have a resultant of the same magnitude $F$. The angle between the two forces is ........ $^o$

The vectors $5\hat{i} + 8\hat{j}$ and $2\hat{i} + 7\hat{j}$ are added. The magnitude of the sum of these vectors is

Two forces are such that the sum of their magnitudes is $18 \,N$ and their resultant is perpendicular to the smaller force and the magnitude of the resultant is $12 \,N$. Then the magnitudes of the forces are:

Two vectors $\vec A$ and $\vec B$ have equal magnitudes. The magnitude of $(\vec A + \vec B)$ is $n$ times the magnitude of $(\vec A - \vec B)$. The angle between $\vec A$ and $\vec B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module