Question

Let a three- dimensional vector $\vec{V}$ satisfy the condition. $2\vec{V}+\vec{V}\times (\hat{i}+2\hat{j})=2\hat{i}+\hat{k}$. If $3|\vec{V}|=\sqrt{m}$ then find the value of $m$.

Answer 1

Answer: 6

$2\vec{V}+\vec{V}\times (\hat{i}+2\hat{j})=(2\hat{i}+\hat{k})$
(i)
or $2\vec{V}\cdot (\hat{i}+2\hat{j})=2$
or $|\vec{V}\cdot (\hat{i}+2\hat{j}){|}^2=1$
or $|\vec{V}{|}^2\cdot |\hat{i}+2\hat{j}{|}^2{\cos }^2\theta =1$
$(\theta$ is the angle between $\vec{V}$ and $\hat{i}+2\hat{j})$
or $|\vec{V}{|}^{2}5\left(1-{\sin }^2\theta \right)=1$
or $|\vec{V}{|}^{2}5{\sin }^2\theta =5|\vec{V}{|}^2-1$
(ii)
From Eq. (i) we have
$|2\vec{V}+\vec{V}\times (\hat{i}+2\hat{j}){|}^2=|2\hat{i}+\hat{k}{|}^2$
or $4|\vec{V}{|}^2+|\vec{V}\times (\hat{i}+2\hat{j}){|}^2=5$
or $4|\vec{V}{|}^2+|\vec{V}{|}^2|\hat{i}+2\hat{j}{|}^2{\sin }^2\theta =5$
or $4|\vec{V}{|}^2+5|\vec{V}{|}^2{\sin }^2\theta =5$
or $4|\vec{V}{|}^2+5|\vec{V}{|}^2-1=5$
or $9|\vec{V}{|}^2=6$
or $3|\vec{V}|=\sqrt{6}$
or $\sqrt{6}=\sqrt{m}$
$\therefore m=6$