Statement I: Two forces (overrightarrow{P}+ov | vedclass

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

Question

(B) Let $\overrightarrow{A} = \overrightarrow{P} + \overrightarrow{Q}$ and $\overrightarrow{B} = \overrightarrow{P} - \overrightarrow{Q}$. Since $\ove

Vedclass · Accepted Answer

Both Statement-$I$ and Statement-$II$ are true.

Vedclass · Answer

(B) Let $\overrightarrow{A} = \overrightarrow{P} + \overrightarrow{Q}$ and $\overrightarrow{B} = \overrightarrow{P} - \overrightarrow{Q}$. Since $\overrightarrow{P} \perp \overrightarrow{Q}$,the magnitudes are $|\overrightarrow{A}| = |\overrightarrow{B}| = \sqrt{P^{2} + Q^{2}}$.The resultant magnitude is $R = \sqrt{|\overrightarrow{A}|^{2} + |\overrightarrow{B}|^{2} + 2|\overrightarrow{A}||\overrightarrow{B}| \cos 	heta} = \sqrt{2(P^{2} + Q^{2}) + 2(P^{2} + Q^{2}) \cos 	heta} = \sqrt{2(P^{2} + Q^{2})(1 + \cos 	heta)}$.For $	heta_{1}$,$R_{1} = \sqrt{3(P^{2} + Q^{2})} \implies 2(1 + \cos 	heta_{1}) = 3 \implies \cos 	heta_{1} = 0.5 \implies 	heta_{1} = 60^{\circ}$.For $	heta_{2}$,$R_{2} = \sqrt{2(P^{2} + Q^{2})} \implies 2(1 + \cos 	heta_{2}) = 2 \implies \cos 	heta_{2} = 0 \implies 	heta_{2} = 90^{\circ}$.Since $60^{\circ} < 90^{\circ}$,$	heta_{1} < 	heta_{2}$ is true. Thus,both statements are true.

Column $-I$	Column $-II$
$(a) \vec a + \vec b = \vec c$	$(i)$ Vector $\vec a$ is along $+Y$,$\vec c$ is along $+X$,and $\vec b$ connects the origin to the tip of $\vec c$
$(b) \vec a - \vec c = \vec b$	$(ii)$ Vector $\vec a$ is along $+X$,$\vec b$ is along $+Y$,and $\vec c$ connects the origin to the tip of $\vec b$
$(c) \vec b - \vec a = \vec c$	$(iii)$ Vector $\vec c$ is along $+X$,$\vec a$ is along $+Y$,and $\vec b$ connects the tip of $\vec c$ to the tip of $\vec a$
$(d) \vec a + \vec b + \vec c = 0$	$(iv)$ Vector $\vec a$ is along $-X$,$\vec b$ is along $-Y$,and $\vec c$ connects the origin to the tip of $\vec b$

Similar Questions

If $\vec{a}$ and $\vec{b}$ are two unit vectors inclined at an angle of $60^{\circ}$ to each other,then:

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

When vector $\overrightarrow{A} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ is subtracted from vector $\vec{B}$,it gives a vector equal to $2\hat{j}$. Then the magnitude of vector $\vec{B}$ will be:

The angle made by the resultant vector of two vectors $2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}-7 \hat{j}-4 \hat{k}$ with the $x$-axis is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module