If vec{A} and vec{B} are two non-zero vectors | vedclass

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

Question

(C) Given: $|\vec{A} + \vec{B}| = \frac{1}{2} |\vec{A} - \vec{B}|$ and $|\vec{A}| = 2|\vec{B}|$.Squaring both sides of the first equation:$|\vec{A} +

Vedclass · Accepted Answer

$\cos^{-1}(-3/4)$

Vedclass · Answer

(C) Given: $|\vec{A} + \vec{B}| = \frac{1}{2} |\vec{A} - \vec{B}|$ and $|\vec{A}| = 2|\vec{B}|$.Squaring both sides of the first equation:$|\vec{A} + \vec{B}|^2 = \frac{1}{4} |\vec{A} - \vec{B}|^2$$A^2 + B^2 + 2AB \cos 	heta = \frac{1}{4} (A^2 + B^2 - 2AB \cos 	heta)$$4(A^2 + B^2 + 2AB \cos 	heta) = A^2 + B^2 - 2AB \cos 	heta$$4A^2 + 4B^2 + 8AB \cos 	heta = A^2 + B^2 - 2AB \cos 	heta$$3A^2 + 3B^2 + 10AB \cos 	heta = 0$Substitute $A = 2B$ into the equation:$3(2B)^2 + 3B^2 + 10(2B)(B) \cos 	heta = 0$$3(4B^2) + 3B^2 + 20B^2 \cos 	heta = 0$$12B^2 + 3B^2 + 20B^2 \cos 	heta = 0$$15B^2 + 20B^2 \cos 	heta = 0$$20B^2 \cos 	heta = -15B^2$$\cos 	heta = -\frac{15}{20} = -\frac{3}{4}$$	heta = \cos^{-1}\left(-\frac{3}{4}ight)$

If $\vec{A}$ and $\vec{B}$ are two non-zero vectors such that $|\vec{A} + \vec{B}| = \frac{|\vec{A} - \vec{B}|}{2}$ and $|\vec{A}| = 2|\vec{B}|$,then the angle between $\vec{A}$ and $\vec{B}$ is

Similar Questions

$ABCDEF$ is a regular hexagon and forces represented in magnitude and direction by $\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}, \overrightarrow{AE}$ and $\overrightarrow{AF}$ act at point $A$. Their resultant is:

What should be the angle $\theta$ between two vectors $\vec{A}$ and $\vec{B}$ so that the magnitude of the resultant vector $\vec{R}$ is maximum (in $^{\circ}$)?

The angle between vectors $\vec a$ and $\vec b$ is $\frac{\pi }{6}$. The angle between vectors $-3\vec a$ and $2\vec b$ is

If $\overrightarrow A = 2\hat i + 4\hat j - 5\hat k$,the direction cosines of the vector $\overrightarrow A$ are:

Define scalar and vector physical quantities with their examples.

Vedclass Test Series

Exam Paper Generator

Online Exam Module