Let bar{a} and bar{b} be two non-collinear un | vedclass

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

Question

(A) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,so $|\bar{a}| = 1$ and $|\bar{b}| = 1$. Let $	heta$ be the angle between $\bar{a}$ and $\bar{

Vedclass · Accepted Answer

$|\bar{u}|$

Vedclass · Answer

(A) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,so $|\bar{a}| = 1$ and $|\bar{b}| = 1$. Let $	heta$ be the angle between $\bar{a}$ and $\bar{b}$. Then $\bar{a} \cdot \bar{b} = |\bar{a}||\bar{b}| \cos 	heta = \cos 	heta$. Also,$|\bar{v}| = |\bar{a} 	imes \bar{b}| = |\bar{a}||\bar{b}| \sin 	heta = \sin 	heta$. Now,consider $\bar{u} = \bar{a} - (\bar{a} \cdot \bar{b}) \bar{b} = \bar{a} - (\cos 	heta) \bar{b}$. Then $|\bar{u}|^2 = |\bar{a} - (\cos 	heta) \bar{b}|^2 = |\bar{a}|^2 + \cos^2 	heta |\bar{b}|^2 - 2 \cos 	heta (\bar{a} \cdot \bar{b})$. $|\bar{u}|^2 = 1 + \cos^2 	heta - 2 \cos^2 	heta = 1 - \cos^2 	heta = \sin^2 	heta$. Thus,$|\bar{u}| = \sin 	heta$. Since $|\bar{v}| = \sin 	heta$ and $|\bar{u}| = \sin 	heta$,we have $|\bar{v}| = |\bar{u}|$.

Let $\bar{a}$ and $\bar{b}$ be two non-collinear unit vectors. If $\bar{u}=\bar{a}-(\bar{a} \cdot \bar{b}) \bar{b}$ and $\bar{v}=\bar{a} \times \bar{b}$,then $|\bar{v}|=$

Similar Questions

The cosine of the angle between any two diagonals of a cube is

If $a = x^2 \hat{i} + x \hat{j} + 3 \hat{k}$ and $b = x \hat{i} - 4 \hat{j} + 2 \hat{k}$ and $a \cdot b > 6$,then:

Let $a = i + 2j + k$,$b = i - j + k$,and $c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has a projection of $\frac{1}{\sqrt{3}}$ on $c$. Then,one such vector is:

The magnitude of the projection of the vector $\vec{a} = 4\hat{i} - 3\hat{j} + 2\hat{k}$ on the line which makes equal angles with the coordinate axes is

$3 \hat{i}-2 \hat{j}-\hat{k}, -2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle ABC$ respectively. If $H$ is its orthocenter,then $\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module