MCQ
Let $\vec{v}=\alpha \hat{i}+2 \hat{j}-3 \hat{k}, \vec{w}=2 \alpha \hat{i}+\hat{j}-\hat{k}$, and $\overrightarrow{ u }$ 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 $|\vec{u} . \hat{i}|^2=\frac{m}{n}$ where $m$ and $n$ are coprime natural numbers, then $m + n$ is equal to $.........$.
  • A
    $3502$
  • B
    $3503$
  • $3501$
  • D
    $3504$

Answer

Correct option: C.
$3501$
c
${[\vec{u} \vec{v} \vec{w}]=\vec{u} \cdot(\vec{v} \times \vec{w})}$

$\min .(|u||\vec{v} \times \vec{w}| \cos \theta)=-\alpha \sqrt{3401}$

$\cos \theta=-1$

$|u|=\alpha \text { (Given) }$

$|\vec{v} \times \vec{w}|=\sqrt{3401}$

$\vec{v} \times \vec{w}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -3 \\ 2 \alpha & 1 & -1\end{array}\right|$

$\vec{v} \times \vec{w}=\hat{i}-5 \alpha \hat{j}-3 \alpha \hat{k}$

$|\vec{v} \times \vec{w}|=\sqrt{1+25 \alpha^2+9 \alpha^2}=\sqrt{3401}$

$34 \alpha^2=3400$

$\alpha^2=100$

$\alpha=10 \quad(\text { as } \alpha > 0)$

So $\vec{u} =\lambda(\hat{i}-5 \alpha \hat{j}-3 \alpha \hat{k})$

$\vec{u} =\sqrt{\lambda^2+25 \alpha^2 \lambda^2+9 \alpha^2 \lambda}$

$\alpha^2 =\lambda^2\left(1+25 \alpha^2+9 \alpha^2\right)$

$100 =\lambda^2(1+34 \times 100)$

$\lambda^2 =\frac{100}{3401}=\frac{m}{n}$

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