JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $\vec{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$ , $\vec{b} = \hat{i} + \hat{k}$ and $\vec{c} = \hat{i} + 2\hat{j} - 3\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$ and $\vec{r} \cdot \vec{b} = 0$ , then $|\vec{r}|$ is equal to:

Choose the correct answer:

Explanation

Solution

Step 1: Use the cross product relation

Given the equation:

\vec{r} \times \vec{a} = \vec{c} \times \vec{a}

Rearranging the terms:

\vec{r} \times \vec{a} - \vec{c} \times \vec{a} = 0

(\vec{r} - \vec{c}) \times \vec{a} = 0

Since the cross product of two vectors is zero, they must be parallel. Therefore:

\vec{r} - \vec{c} = \lambda\vec{a}

\vec{r} = \vec{c} + \lambda\vec{a}

Step 2: Express $\vec{r}$ in terms of $\lambda$

Substitute the values of $\vec{c}$ and $\vec{a}$ :

\vec{r} = (\hat{i} + 2\hat{j} - 3\hat{k}) + \lambda(2\hat{i} - 7\hat{j} + 5\hat{k})

\vec{r} = (1 + 2\lambda)\hat{i} + (2 - 7\lambda)\hat{j} + (-3 + 5\lambda)\hat{k}

Step 3: Apply the condition $\vec{r} \cdot \vec{b} = 0$

Given $\vec{b} = \hat{i} + 0\hat{j} + \hat{k}$ :

\vec{r} \cdot \vec{b} = [(1 + 2\lambda)(1)] + [(2 - 7\lambda)(0)] + [(-3 + 5\lambda)(1)] = 0

1 + 2\lambda - 3 + 5\lambda = 0

7\lambda - 2 = 0 \implies \lambda = \frac{2}{7}

Step 4: Find the vector $\vec{r}$

Substitute $\lambda = \frac{2}{7}$ back into the equation for $\vec{r}$ :

\vec{r} = \left(1 + \frac{4}{7}\right)\hat{i} + \left(2 - \frac{14}{7}\right)\hat{j} + \left(-3 + \frac{10}{7}\right)\hat{k}

\vec{r} = \frac{11}{7}\hat{i} + 0\hat{j} - \frac{11}{7}\hat{k}

\vec{r} = \frac{11}{7}(\hat{i} - \hat{k})

Step 5: Calculate the magnitude $|\vec{r}|$

|\vec{r}| = \frac{11}{7} \sqrt{1^2 + (-1)^2}

|\vec{r}| = \frac{11}{7}\sqrt{2}

Correct Option: (3)

Choose the correct answer:

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