If |vec{a} times vec{b}|^{2}+|vec{a} cdot vec | vedclass

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

Question

(C) Given that,$|\vec{a} 	imes \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144$ and $|\vec{a}|=4$. We know the Lagrange's identity for vectors: $|\vec{a

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given that,$|\vec{a} 	imes \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144$ and $|\vec{a}|=4$. We know the Lagrange's identity for vectors: $|\vec{a} 	imes \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} |\vec{b}|^{2} \sin^{2} 	heta + |\vec{a}|^{2} |\vec{b}|^{2} \cos^{2} 	heta$. Factoring out $|\vec{a}|^{2} |\vec{b}|^{2}$,we get: $|\vec{a}|^{2} |\vec{b}|^{2} (\sin^{2} 	heta + \cos^{2} 	heta) = |\vec{a}|^{2} |\vec{b}|^{2}$. Therefore,$|\vec{a} 	imes \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} |\vec{b}|^{2}$. Substituting the given values: $144 = (4)^{2} |\vec{b}|^{2}$. $144 = 16 |\vec{b}|^{2}$. $|\vec{b}|^{2} = \frac{144}{16} = 9$. Taking the square root,we get $|\vec{b}| = 3$.

If $|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144$ and $|\vec{a}|=4$,then the value of $|\vec{b}|$ is

Similar Questions

If $a, b, c$ are unit vectors satisfying the relation $a+b+\sqrt{3} c=0$,then the angle between $a$ and $b$ is

If $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$,and $\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}$ are given vectors. If $\vec{a}$ is perpendicular to $\lambda \vec{b} + \vec{c}$,then $\lambda = . . . . . .$.

Find a unit vector in the $xy$-plane which makes an angle of $45^{\circ}$ with the vector $i + j$ and an angle of $60^{\circ}$ with the vector $3i - 4j$.

Given $ABC$ is an equilateral triangle of side length $1$ unit and $P$ be any arbitrary point on the circumcircle of triangle $ABC,$ then $|\vec{PA}|^2+|\vec{PB}|^2+|\vec{PC}|^2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module