JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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:

Choose the correct answer:

Explanation

Solution

1. Calculate $\vec{v} \times \vec{w}$ :

\vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -3 \\ 2\alpha & 1 & -1 \end{vmatrix} = \hat{i}(-2 + 3) - \hat{j}(-\alpha + 6\alpha) + \hat{k}(\alpha - 4\alpha)

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

2. Minimum Scalar Triple Product: The scalar triple product $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$ .

The minimum value occurs when $\vec{u}$ is in the opposite direction of $(\vec{v} \times \vec{w})$ .

$[\vec{u} \vec{v} \vec{w}]_{min} = -|\vec{u}| |\vec{v} \times \vec{w}| = -\alpha \sqrt{1^2 + (-5\alpha)^2 + (-3\alpha)^2}$

-\alpha\sqrt{1 + 34\alpha^2} = -\alpha\sqrt{3401}

1 + 34\alpha^2 = 3401 \implies 34\alpha^2 = 3400 \implies \alpha^2 = 100 \implies \alpha = 10

3. Find $|\vec{u} \cdot \hat{i}|^2$ : For the minimum value, $\vec{u} = -\alpha \frac{(\vec{v} \times \vec{w})}{|\vec{v} \times \vec{w}|} = -10 \frac{(\hat{i} - 50\hat{j} - 30\hat{k})}{\sqrt{3401}}$ .

$\vec{u} \cdot \hat{i} = \frac{-10}{\sqrt{3401}}$ .

$|\vec{u} \cdot \hat{i}|^2 = \frac{100}{3401}$ .

Here, $m = 100$ and $n = 3401$ . They are coprime.

4. Result: $m + n = 100 + 3401 = 3501$ .

Answer: 3501

Choose the correct answer:

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