Medium MCQ +4 / -1 PYQ · JEE Mains 2025

The line $\mathrm{L}_1$ is parallel to the vector $\overrightarrow{\mathrm{a}}=-3 \hat{i}+2 \hat{j}+4 \hat{k}$ and passes through the point $(7,6,2)$ and the line $\mathrm{L}_2$ is parallel to the vector $\overrightarrow{\mathrm{b}}=2 \hat{i}+\hat{j}+3 \hat{k}$ and passes through the point $(5,3,4)$. The shortest distance between the lines $L_1$ and $L_2$ is :

  1. A $\frac{23}{\sqrt{38}}$ Correct answer
  2. B $\frac{21}{\sqrt{38}}$
  3. C $\frac{23}{\sqrt{57}}$
  4. D $\frac{21}{\sqrt{57}}$

Solution

<p>The line $ L_1 $ is parallel to the vector $ \overrightarrow{\mathrm{a}} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k} $ and passes through the point $ (7, 6, 2) $. The line $ L_2 $ is parallel to the vector $ \overrightarrow{\mathrm{b}} = 2 \hat{i} + \hat{j} + 3 \hat{k} $ and passes through the point $ (5, 3, 4) $.</p> <p><strong>Equations of the Lines:</strong></p> <p><p>Equation of $ L_1 $: </p> <p>$ \mathbf{r} = 7 \hat{i} + 6 \hat{j} + 2 \hat{k} + \lambda(-3 \hat{i} + 2 \hat{j} + 4 \hat{k}) $</p></p> <p><p>Equation of $ L_2 $: </p> <p>$ \mathbf{r} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k} + \mu(2 \hat{i} + \hat{j} + 3 \hat{k}) $</p></p> <p><strong>Shortest Distance Between $ L_1 $ and $ L_2 $:</strong></p> <p>The formula for the shortest distance between two skew lines is given by:</p> <p>$ \text{Distance} = \left|\frac{\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right| $</p> <p><strong>Calculations:</strong></p> <p><p>Vector between the points:</p> <p>$ \vec{a}_2 - \vec{a}_1 = (5 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (7 \hat{i} + 6 \hat{j} + 2 \hat{k}) = -2 \hat{i} - 3 \hat{j} + 2 \hat{k} $</p></p> <p><p>Cross product of direction vectors:</p> <p>$ \vec{b}_1 = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}, \quad \vec{b}_2 = 2 \hat{i} + \hat{j} + 3 \hat{k} $</p> <p>$ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 4 \\ 2 & 1 & 3 \end{vmatrix} = 2 \hat{i} + 17 \hat{j} - 7 \hat{k} $</p></p> <p><p>Magnitude of the cross product:</p> <p>$ \left|\vec{b}_1 \times \vec{b}_2\right| = \sqrt{2^2 + 17^2 + (-7)^2} = \sqrt{342} $</p></p> <p><p>Dot product:</p> <p>$ \left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right) = (-2 \hat{i} - 3 \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} + 17 \hat{j} - 7 \hat{k}) $</p> <p>$ = (-2)(2) + (-3)(17) + (2)(-7) = -4 - 51 - 14 = -69 $</p></p> <p><strong>Shortest Distance:</strong></p> <p>$ \text{Distance} = \left|\frac{-69}{\sqrt{342}}\right| = \frac{23}{\sqrt{38}} $</p>

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Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios

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