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

$2 \vec{ V }+\vec{ V } \times(\hat{ i }+2 \hat{ j })=(2 \hat{ i }+\hat{ k })$
(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 } $
$\therefore m =6$