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 $|\vec{u}|=\alpha>$ 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

Correct Answer: 3501

The correct answer is 3501

\([\vec{u}\vec{v}\vec{w}]=\vec{u}.(\vec{v}\times\vec{w})\)

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

\(\Rightarrow cos\theta=-1\)

\(|u|=\alpha(Given)\)

\(|\vec{v}\times\vec{w}|=\sqrt{3401}\)

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

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

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

\(34\alpha ^{2}=3400 \)

\(\alpha ^{2}=100\)

\(\alpha =10(as \: \alpha>0)\)

\(so\, \: \vec{u}=\lambda (\hat{i}-5\alpha\hat{j}-3\alpha\hat{k})\)

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

\( \alpha ^{2}=\lambda ^{2}(1+25\alpha ^{2}+9\alpha ^{2 })\)

\( 100=\lambda ^{2}(1+34\times100)\)

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

Answer 2

Computing the scalar triple product: \[ [\vec{u} \quad \vec{v} \quad \vec{w}] = \hat{u} \cdot (\vec{v} \times \vec{w}) \] Given that the minimum value is \( -\alpha \sqrt{3401} \), \[ \cos \theta = -1 \] \[ |\hat{u}| = \alpha \] Computing cross product: 

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

 Expanding the determinant: \[ \vec{v} \times \vec{w} = \hat{i} (-2 + 3) - \hat{j} (\alpha + 6\alpha) + \hat{k} (\alpha - 4\alpha) \] \[ = \hat{i} - 5\alpha \hat{j} - 3\alpha \hat{k} \] Computing magnitude: 

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

Equating for \( \alpha^2 \), \[ 34\alpha^2 = 3400 \] \[ \alpha^2 = 100 \Rightarrow \alpha = 10 \] Thus, \[ \hat{u} = \lambda ( \hat{i} - 5\alpha \hat{j} - 3\alpha \hat{k} ) \] Computing: \[ |\hat{u} \cdot \hat{i}|^2 = \frac{100}{3401} = \frac{m}{n} \] Since \( m = 100, n = 3401 \), \[ m + n = 3501 \]