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 $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.Since $a - \sqrt{2}b$ is a unit vector,its magnitude is $1$,so $|a - \sqrt{2}b|

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.Since $a - \sqrt{2}b$ is a unit vector,its magnitude is $1$,so $|a - \sqrt{2}b| = 1$.Squaring both sides,we get $|a - \sqrt{2}b|^2 = 1^2$.Using the property $|u - v|^2 = |u|^2 + |v|^2 - 2(u \cdot v)$,we have:$|a|^2 + |\sqrt{2}b|^2 - 2(a \cdot \sqrt{2}b) = 1$.Substituting the values $|a| = 1$ and $|b| = 1$:$1^2 + 2|b|^2 - 2\sqrt{2}(a \cdot b) = 1$.$1 + 2(1) - 2\sqrt{2}(a \cdot b) = 1$.$3 - 2\sqrt{2}(a \cdot b) = 1$.$2\sqrt{2}(a \cdot b) = 2$.$a \cdot b = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.Since $a \cdot b = |a||b|\cos	heta$,we have $1 \cdot 1 \cdot \cos	heta = \frac{1}{\sqrt{2}}$.$\cos	heta = \frac{1}{\sqrt{2}}$,which implies $	heta = \frac{\pi}{4}$.

If $\theta$ is the angle between the unit vectors $a$ and $b$,then $a - \sqrt{2}b$ will be a unit vector if $\theta = $

Similar Questions

$a, b, c$ are three vectors such that $|a|=3, |b|=5, |c|=7$. If $a, b, c$ are perpendicular to the vectors $b+c, c+a, a+b$ respectively,then $\sqrt{(a+b+c)^2-2}=$

The angle between the vectors $\bar{a} = 6 \hat{i} + 2 \hat{j} - 8 \hat{k}$ and $\bar{b} = 4 \hat{i} - 4 \hat{j} + 2 \hat{k}$ is . . . . . . .

$a = 4 \hat{i} + 3 \hat{j}$ and $b$ are two vectors in the $XOY$ plane, and $a$ is perpendicular to $b$. $A$ vector $c$ lying in the same plane and having projections $1$ and $2$ on $a$ and $b$ respectively is:

It is given that $a, b, c$ are vectors of lengths $6, 8, 10$ respectively. If $a$ is perpendicular to $(b+c)$, $b$ is perpendicular to $(c+a)$, and $c$ is perpendicular to $(a+b)$, then the length of the vector $a+b+c$ is (in $\sqrt{2}$)

If $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$, $\vec{a} \cdot \vec{b} < 0$ and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then the value of $\sin \theta + \tan \theta$ is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module