If theta is the angle between the unit vector | vedclass

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

Question

(B) Given that $\mathbf{a}$ and $\mathbf{b}$ are unit vectors,so $|\mathbf{a}| = 1$ and $|\mathbf{b}| = 1$. The dot product $\mathbf{a} \cdot \mathbf{

Vedclass · Accepted Answer

$\frac{1}{2} |\mathbf{a} + \mathbf{b}|$

Vedclass · Answer

(B) Given that $\mathbf{a}$ and $\mathbf{b}$ are unit vectors,so $|\mathbf{a}| = 1$ and $|\mathbf{b}| = 1$. The dot product $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos 	heta = \cos 	heta$. Consider the magnitude squared of the sum of the vectors: $|\mathbf{a} + \mathbf{b}|^2 = (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2(\mathbf{a} \cdot \mathbf{b})$. Substituting the values: $|\mathbf{a} + \mathbf{b}|^2 = 1^2 + 1^2 + 2 \cos 	heta = 2 + 2 \cos 	heta$. Using the trigonometric identity $1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$: $|\mathbf{a} + \mathbf{b}|^2 = 2(1 + \cos 	heta) = 2(2 \cos^2 \frac{	heta}{2}) = 4 \cos^2 \frac{	heta}{2}$. Taking the square root on both sides: $|\mathbf{a} + \mathbf{b}| = 2 \cos \frac{	heta}{2}$. Therefore,$\cos \frac{	heta}{2} = \frac{1}{2} |\mathbf{a} + \mathbf{b}|$.

If $\theta$ is the angle between the unit vectors $\mathbf{a}$ and $\mathbf{b}$,then $\cos \frac{\theta}{2} = $

Similar Questions

$A$ particle acted on by two forces $3i + 2j - 3k$ and $2i + 4j + 2k$ is displaced from the point $i + 2j + k$ to $5i + 4j + 2k.$ The total work done by the forces is equal to ............ $unit$.

If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of a geometric progression are $a$,$b$,and $c$ respectively,then find the angle between the vectors $\vec{u} = (\log a)\hat{i} + (\log b)\hat{j} + (\log c)\hat{k}$ and $\vec{v} = (q - r)\hat{i} + (r - p)\hat{j} + (p - q)\hat{k}$.

If $\vec{a}$ and $\vec{b}$ are two vectors such that $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} < 0$ and $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$,then the angle between the vectors $\vec{a}$ and $\vec{b}$ is

The value of $\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module