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

$ {[\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}$
$\vec{v} \times \vec{w}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -3 \\ 2 \alpha & 1 & -1\end{vmatrix}$
$ \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}$