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

(D) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$. Consider the expression $|a - b|^2$: $|a - b|^2 = |a|^2 + |b|^2 - 2(a \cdot b)

Vedclass · Accepted Answer

$\frac{1}{2}|a-b|$

Vedclass · Answer

(D) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$. Consider the expression $|a - b|^2$: $|a - b|^2 = |a|^2 + |b|^2 - 2(a \cdot b)$ Since $a \cdot b = |a||b| \cos 	heta = 1 	imes 1 	imes \cos 	heta = \cos 	heta$,we have: $|a - b|^2 = 1^2 + 1^2 - 2 \cos 	heta$ $|a - b|^2 = 2 - 2 \cos 	heta$ $|a - b|^2 = 2(1 - \cos 	heta)$ Using the trigonometric identity $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$: $|a - b|^2 = 2(2 \sin^2 \frac{	heta}{2})$ $|a - b|^2 = 4 \sin^2 \frac{	heta}{2}$ Taking the square root on both sides: $|a - b| = 2 \sin \frac{	heta}{2}$ Therefore,$\sin \frac{	heta}{2} = \frac{1}{2} |a - b|$.

If $\theta$ is the angle between the unit vectors $a$ and $b$,then $\sin \frac{\theta}{2}$ is equal to

Similar Questions

Let $ABC$ be a triangle. Let $u = \overrightarrow{AB}$ and $v = \overrightarrow{AC}$. If $D$ is the midpoint of $BC$,then the length of the median of $\triangle ABD$ through the vertex $B$ is:

If $a, b, c$ are coplanar vectors,then

$A$ particle is acted upon by constant forces $4\hat{i} + \hat{j} - 3\hat{k}$ and $3\hat{i} + \hat{j} - \hat{k}$. The displacement of the particle from the point $\hat{i} + 2\hat{j} + 3\hat{k}$ to the point $5\hat{i} + 4\hat{j} + \hat{k}$ is given. Find the total work done by the forces in units.

If $3\vec{a} - 5\vec{b}$ and $2\vec{a} + \vec{b}$ are perpendicular to each other,and $\vec{a} + 4\vec{b}$ and $-\vec{a} + \vec{b}$ are also perpendicular to each other,and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,then find $\cos \theta$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module