The modulus of the vector product of two vect | vedclass

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

Question

(A) Let the two vectors be $\vec{A}$ and $\vec{B}$ and the angle between them be $	heta$.According to the problem,the magnitude of the vector product

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) Let the two vectors be $\vec{A}$ and $\vec{B}$ and the angle between them be $	heta$.According to the problem,the magnitude of the vector product is $\frac{1}{\sqrt{3}}$ times the scalar product:$|\vec{A} 	imes \vec{B}| = \frac{1}{\sqrt{3}} (\vec{A} \cdot \vec{B})$$AB \sin 	heta = \frac{1}{\sqrt{3}} AB \cos 	heta$Dividing both sides by $AB \cos 	heta$ (assuming $A, B 
eq 0$ and $	heta 
eq 90^{\circ}$):$	an 	heta = \frac{1}{\sqrt{3}}$Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $	heta = 30^{\circ}$.Converting degrees to radians: $	heta = 30^{\circ} 	imes \frac{\pi}{180^{\circ}} = \frac{\pi}{6}$.

The modulus of the vector product of two vectors is $\frac{1}{\sqrt{3}}$ times their scalar product. The angle between the vectors is

Similar Questions

Find the angle in $^o$ between two vectors $\overrightarrow{A} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ and $\overrightarrow{B} = 4\hat{i} + 2\hat{j} - 4\hat{k}$.

The diagonals of a parallelogram are $2\,\hat{i}$ and $2\,\hat{j}$. What is the area of the parallelogram in square units?

The area of the parallelogram formed by vectors $\vec{P} = 2\hat{i} - \hat{j} + 5\hat{k}$ and $\vec{Q} = 3\hat{i} - 2\hat{j} + 4\hat{k}$ is

$\vec{A}$,$\vec{B}$,and $\vec{C}$ are three non-collinear,non-coplanar vectors. What can you say about the direction of $\vec{A} \times (\vec{B} \times \vec{C})$?

Two unit vectors $\hat{a}_{1}$ and $\hat{a}_{2}$ are inclined to each other at an angle $\theta$. If $|\hat{a}_{1}-\hat{a}_{2}|=\sqrt{3}$,then the value of $(\hat{a}_{1}-\hat{a}_{2}) \cdot (2\hat{a}_{1}-\hat{a}_{2})$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module