JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ , and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ is $a$ , then $28a^2$ is equal to:

Choose the correct answer:

Explanation

Solving

1. Lines identifying parameters:

Line 1 ( $L_1$ ):
- $\vec{a_1} = (1, 2, 3)$
- $\vec{b_1} = (2-1, 3-2, 4-3) = (1, 1, 1)$
Line 2 ( $L_2$ ):
- $\vec{a_2} = (1, -1, 2)$
- $\vec{b_2} = (2, -1, 0)$

2. Calculating $(\vec{a_2} - \vec{a_1})$ :

\vec{a_2} - \vec{a_1} = (1-1, -1-2, 2-3) = (0, -3, -1)

3. Calculating $(\vec{b_1} \times \vec{b_2})$ :

\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix} = \hat{i}(1) - \hat{j}(-2) + \hat{k}(-3) = (1, 2, -3)

4. Shortest Distance ( $a$ ):

a = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}

a = \frac{|(0)(1) + (-3)(2) + (-1)(-3)|}{\sqrt{1^2 + 2^2 + (-3)^2}}

a = \frac{|0 - 6 + 3|}{\sqrt{1 + 4 + 9}} = \frac{3}{\sqrt{14}}

5. Final Calculation:

a^2 = \frac{9}{14}

28a^2 = 28 \times \frac{9}{14}

28a^2 = 18

Answer: 18

Choose the correct answer:

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