Question

Let $\vec{v}=\alpha \hat{i}+2\hat{j}-3\hat{k},\vec{w}=2\alpha \hat{i}+\hat{j}-\hat{k}$ and $\vec{u}$ be a vector such that $|\vec{u}|=\alpha >0$. If the minimum value of the scalar triple product $[\vec{u}\vec{v}\vec{w}]$ is $-\alpha \sqrt{3401}$, and $|\vec{u}\cdot \hat{i}{|}^2=\frac{m}{n}$ where $m$ and $n$ are coprime natural numbers, then $m+n$ is equal to _____

Answer 1

Answer: 3501

$[\vec{u}\vec{v}\vec{w}]=\vec{u}\cdot (\vec{v}\times \vec{w})$
$\min .(|u||\vec{v}\times \vec{w}|\cos \theta )=-\alpha \sqrt{3401}$
$\Rightarrow \cos \theta =-1$
$|u|=\alpha \text{ (Given) }$
$|\vec{v}\times \vec{w}|=\sqrt{3401}$
$\vec{v}\times \vec{w}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \alpha & 2& -3\\ 2\alpha & 1& -1\end{array}\right|$
$\vec{v}\times \vec{w}=\hat{i}-5\alpha \hat{j}-3\alpha \hat{k}$
$|\vec{v}\times \vec{w}|=\sqrt{1+25{\alpha }^2+9{\alpha }^2}=\sqrt{3401}$
$34{\alpha }^2=3400$
${\alpha }^2=100$
$\alpha =10(\text{ as }\alpha >0)$
So $\vec{u}=\lambda (\hat{i}-5\alpha \hat{j}-3\alpha \hat{k})$
$\vec{u}=\sqrt{{\lambda }^2+25{\alpha }^2{\lambda }^2+9{\alpha }^2\lambda }$
${\alpha }^2={\lambda }^2\left(1+25{\alpha }^2+9{\alpha }^2\right)$
$100={\lambda }^2(1+34\times 100)$
${\lambda }^2=\frac{100}{3401}=\frac{m}{n}$