If bar{a} and bar{b} are unit vectors such th | vedclass

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

Question

(B) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,we have $|\bar{a}| = 1$ and $|\bar{b}| = 1$. We are given the equation $|\bar{a}+\bar{b}| = \s

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(B) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,we have $|\bar{a}| = 1$ and $|\bar{b}| = 1$. We are given the equation $|\bar{a}+\bar{b}| = \sqrt{3}$. Squaring both sides,we get $|\bar{a}+\bar{b}|^2 = 3$. Using the property $|\bar{x}|^2 = \bar{x} \cdot \bar{x}$,we have $(\bar{a}+\bar{b}) \cdot (\bar{a}+\bar{b}) = 3$. Expanding the dot product,we get $|\bar{a}|^2 + |\bar{b}|^2 + 2(\bar{a} \cdot \bar{b}) = 3$. Substituting the values $|\bar{a}| = 1$ and $|\bar{b}| = 1$,we get $1^2 + 1^2 + 2(\bar{a} \cdot \bar{b}) = 3$. This simplifies to $2 + 2(\bar{a} \cdot \bar{b}) = 3$,which means $2(\bar{a} \cdot \bar{b}) = 1$,so $\bar{a} \cdot \bar{b} = \frac{1}{2}$. We know that $\bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos 	heta$,where $	heta$ is the angle between the vectors. Substituting the known values,$\frac{1}{2} = (1)(1) \cos 	heta$,so $\cos 	heta = \frac{1}{2}$. Therefore,$	heta = \frac{\pi}{3}$.

If $\bar{a}$ and $\bar{b}$ are unit vectors such that $|\bar{a}+\bar{b}|=\sqrt{3}$,then the angle between $\bar{a}$ and $\bar{b}$ is

Similar Questions

If $a = i + 2j - 3k$ and $b = 3i - j + 2k$,find the angle between the vectors $a + b$ and $a - b$ in degrees.

If $\vec{a} = 4 \hat{i} + 5 \hat{j} - 3 \hat{k}$ and $\vec{b} = 6 \hat{i} - 2 \hat{j} - 2 \hat{k}$ are two vectors,then the magnitude of the component of $\vec{b}$ parallel to $\vec{a}$ is: (in $\sqrt{2}$)

If the vectors $\bar{a}, \bar{b}, \bar{c}$ satisfy the condition $|\bar{a}-\bar{c}|=|\bar{b}-\bar{c}|$,then $(\bar{b}-\bar{a}) \cdot \left(\bar{c}-\frac{\bar{a}+\bar{b}}{2}\right) = $

The angle between the vectors $3\,i + j + 2\,k$ and $2\,i - 2\,j + 4\,k$ is

If $|\bar{a}|=2, |\bar{b}|=3$ and $\bar{a}, \bar{b}$ are mutually perpendicular vectors,then the area of the triangle whose vertices are $0, \bar{a}+2\bar{b}, \bar{a}-2\bar{b}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module