If vec{a} and vec{b} are unit vectors,then th | vedclass

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

Question

(C) Let $	heta$ be the angle between the unit vectors $\vec{a}$ and $\vec{b}$. Since they are unit vectors,$|\vec{a}| = 1$ and $|\vec{b}| = 1$.We kno

Vedclass · Accepted Answer

$2\sqrt{2}$

Vedclass · Answer

(C) Let $	heta$ be the angle between the unit vectors $\vec{a}$ and $\vec{b}$. Since they are unit vectors,$|\vec{a}| = 1$ and $|\vec{b}| = 1$.We know that $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 1 + 1 + 2\cos	heta = 2(1 + \cos	heta) = 4\cos^2(\frac{	heta}{2})$.Thus,$|\vec{a} + \vec{b}| = 2\cos(\frac{	heta}{2})$ (assuming $\cos(\frac{	heta}{2}) \ge 0$ for $0 \le 	heta \le \pi$).Similarly,$|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) = 1 + 1 - 2\cos	heta = 2(1 - \cos	heta) = 4\sin^2(\frac{	heta}{2})$.Thus,$|\vec{a} - \vec{b}| = 2\sin(\frac{	heta}{2})$.Let $f(	heta) = |\vec{a} + \vec{b}| + |\vec{a} - \vec{b}| = 2(\cos(\frac{	heta}{2}) + \sin(\frac{	heta}{2}))$.Using the identity $\cos x + \sin x = \sqrt{2}\sin(x + \frac{\pi}{4})$,we get $f(	heta) = 2\sqrt{2}\sin(\frac{	heta}{2} + \frac{\pi}{4})$.The maximum value of $\sin(\frac{	heta}{2} + \frac{\pi}{4})$ is $1$ (when $\frac{	heta}{2} + \frac{\pi}{4} = \frac{\pi}{2}$,i.e.,$	heta = \frac{\pi}{2}$).Therefore,the maximum value is $2\sqrt{2} 	imes 1 = 2\sqrt{2}$.

If $\vec{a}$ and $\vec{b}$ are unit vectors,then the maximum value of $|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|$ is:

Similar Questions

The vector $a$ coplanar with the vectors $i$ and $j$, perpendicular to the vector $b = 4i - 3j + 5k$ such that $|a| = |b|$ is

If $\vec{a} = \hat{i} + 3\hat{j} - 2\hat{k}$ and $\vec{b} = 4\hat{i} - 2\hat{j} + 4\hat{k}$,then $(2\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b}) = \dots$

If $ABCDEF$ is a regular hexagon,the length of whose side is $a$,then $\overrightarrow{AB} \cdot \overrightarrow{AF} + \frac{1}{2} \overrightarrow{BC}^2 = $

The angle between two diagonals of a cube is:

If $\overrightarrow{a}=\hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}-3 \hat{j}+\hat{k}$ and the orthogonal projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\frac{4}{3}(\hat{i}-\hat{j}-\hat{k})$,then $\lambda$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module