JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}$ , $\vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}$ and $\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}$ . If a vector $\vec{d}$ satisfies $\vec{d} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{d} \cdot \vec{a} = 24$ , then $|\vec{d}|^2$ is equal to:

Choose the correct answer:

Explanation

Solution

Given:

\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}, \vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k} \text{ and } \vec{c} = 2\hat{i} - \hat{j} + 4\hat{k} \text{ and } \vec{d} \times \vec{b} = \vec{c} \times \vec{b}, \vec{d} \cdot \vec{a} = 24 \text{}

Step 1: Simplify the cross product equation

\Rightarrow (\vec{d} - \vec{c}) \times \vec{b} = 0 \text{}

Since the cross product is zero, $\vec{d} - \vec{c}$ and $\vec{b}$ are parallel.

\text{So } \vec{d} - \vec{c} = \lambda \vec{b} \text{ or } \vec{d} = \vec{c} + \lambda \vec{b} \text{}

Step 2: Solve for } $\lambda$ using the dot product

\vec{d} \cdot \vec{a} = 24 \Rightarrow (\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 24 \text{}

\Rightarrow \vec{c} \cdot \vec{a} + \lambda \vec{a} \cdot \vec{b} = 24 \text{}

Substitute the component values:

(2 - 4 + 8) + \lambda (3 - 8 + 14) = 24 \text{}

6 + \lambda (9) = 24 \text{ or } \lambda = 2 \text{}

**Step 3: Find vector } $\vec{d}$ and } $|\vec{d}|^2$

\Rightarrow \vec{d} = \vec{c} + 2\vec{b} \text{}

\vec{d} = 2\hat{i} - \hat{j} + 4\hat{k} + 6\hat{i} - 4\hat{j} + 14\hat{k} \text{}

\Rightarrow \vec{d} = 8\hat{i} - 5\hat{j} + 18\hat{k} \text{}

Finally, calculate the squared magnitude:

|\vec{d}|^2 = 64 + 25 + 324 = 413 \text{}

Correct Option: (3) 413

Choose the correct answer:

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