The line L1 is parallel to the vector a = -3i + 2j + 4k and passes through the point (7, 6, 2) and the line L2 is parallel to the vector b = 2i + j + 3k and passes through the point (5, 3, 4). The shortest distance between the lines L1 and L2 is :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q15 – JEE-MAIN 2025

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Home Maths Vector AlgebraQuestion #517

The line

L_{1}

is parallel to the vector

\vec{a} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}

and passes through the point

(7, 6, 2)

and the line

L_{2}

is parallel to the vector

\vec{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 :

Clue 1 of 3: Conceptual Principle

The shortest distance between two skew lines is the length of the common perpendicular between them.

🧠 Full Solution Path

Step 1: Identify Vectors and Calculate Connecting Vector✦ Active

The given information provides the direction vectors and points for two lines $L_{1}$ and $L_{2}$ . For $L_{1}$ , the point is $P_{1} (7, 6, 2)$ and the direction vector is $\vec{a} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ . For $L_{2}$ , the point is $P_{2} (5, 3, 4)$ and the direction vector is $\vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ . We first find the vector connecting the two points:

\vec{p_{1}} = 7 \hat{i} + 6 \hat{j} + 2 \hat{k}

\vec{p_{2}} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}

\vec{p_{2}} - \vec{p_{1}} = (5 - 7) \hat{i} + (3 - 6) \hat{j} + (4 - 2) \hat{k} = - 2 \hat{i} - 3 \hat{j} + 2 \hat{k}

Step 2: Calculate Cross Product of Direction Vectors○ Expand

Next, we calculate the cross product of the direction vectors $\vec{a}$ and $\vec{b}$ :

\vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 3 & 2 & 4 \\ 2 & 1 & 3 \end{matrix} |

= \hat{i} (2 \cdot 3 - 4 \cdot 1) - \hat{j} (- 3 \cdot 3 - 4 \cdot 2) + \hat{k} (- 3 \cdot 1 - 2 \cdot 2)

= \hat{i} (6 - 4) - \hat{j} (- 9 - 8) + \hat{k} (- 3 - 4)

= 2 \hat{i} + 17 \hat{j} - 7 \hat{k}

💡 Teacher's Secret Hint

Ensure careful calculation of the determinant to avoid sign errors.

Step 3: Apply Shortest Distance Formula○ Expand

The shortest distance $D$ between two skew lines is given by the formula $D = \frac{| (\vec{p_{2}} - \vec{p_{1}}) \cdot (\vec{a} \times \vec{b}) |}{| ‖ \vec{a} \times \vec{b} ‖ |}$ . We calculate the dot product and the magnitude:

(\vec{p_{2}} - \vec{p_{1}}) \cdot (\vec{a} \times \vec{b}) = (- 2 \hat{i} - 3 \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} + 17 \hat{j} - 7 \hat{k})

= (- 2) (2) + (- 3) (17) + (2) (- 7) = - 4 - 51 - 14 = - 69

| ‖ \vec{a} \times \vec{b} ‖ | = ‖ ‖ 2 \hat{i} + 17 \hat{j} - 7 \hat{k} ‖ | = \sqrt{2^{2} + 17^{2} + (- 7)^{2}}

= \sqrt{4 + 289 + 49} = \sqrt{342}

D = \frac{| - 69 |}{\sqrt{342}} = \frac{69}{\sqrt{342}}

\text{To simplify the denominator, we factorize 342:}

342 = 9 \times 38

\sqrt{342} = \sqrt{9 \times 38} = 3 \sqrt{38}

\text{Substitute this back into the distance formula:}

D = \frac{69}{3 \sqrt{38}} = \frac{23}{\sqrt{38}}

\text{This matches option 4.}

💡 Teacher's Secret Hint

Remember to take the absolute value of the scalar triple product in the numerator, as distance must be positive. Simplify the radical in the denominator if possible.

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Maths Question 15 – JEE-MAIN 2025

Clue 1 of 3: Conceptual Principle

🧠 Full Solution Path