Vector Algebra Question 20
Question 20 - 01 February - Shift 1
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 $|\overrightarrow{{}u} . \hat{i}|^{2}=\frac{m}{n}$ where $m$ and $n$ are coprime natural numbers, then $m+n$ is equal to
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Answer: (3501)
Solution:
Formula: Scalar triple product, Scalar product of two vectors
$[\vec{u} \vec{v} w]=\vec{u} \cdot(\vec{v} \times \vec{w})$
$\min .(|u||\vec{v} \times \vec{w}| \cos \theta)=-\alpha \sqrt{3401}$
$\Rightarrow \cos \theta=-1$
$|u|=\alpha$ (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} $
$\vec{v} \times \vec{w}=\hat{i}-5 \alpha \hat{j}-3 \alpha \hat{k}$
$|\vec{v} \times \vec{w}|=\sqrt{1+25 \alpha^{2}+9 \alpha^{2}}=\sqrt{3401}$
$34 \alpha^{2}=3400$
$\alpha^{2}=100$
$\alpha=10$
(as $\alpha>0$ )
So $\quad \vec{u}=\lambda(\hat{i}-5 \alpha \hat{j}-3 \alpha \hat{k})$
$\vec{u}=\sqrt{\lambda^{2}+25 \alpha^{2} \lambda^{2}+9 \alpha^{2} \lambda}$
$\alpha^{2}=\lambda^{2}(1+25 \alpha^{2}+9 \alpha^{2})$
$100=\lambda^{2}(1+34 \times 100) $
$\Rightarrow \lambda^{2}= \frac{100}{3401} = \frac{m}{n}$
$\therefore m+n = 100+3401 = 3501$
