The shortest distance between the lines x-4/1 = y-3/2 = z-2/-3 and x+2/2 = y-6/4 = z-5/-5 is:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q14 – JEE-MAIN 2026

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Home Maths 3D GeometryQuestion #13

The shortest distance between the lines

\frac{x - 4}{1} = \frac{y - 3}{2} = \frac{z - 2}{- 3}

and

\frac{x + 2}{2} = \frac{y - 6}{4} = \frac{z - 5}{- 5}

is:

Clue 1 of 3: Identify Line Parameters

Extract a point and direction vector for each line from their symmetric equations.

🧠 Full Solution Path

Step 1: Extract Points and Direction Vectors✦ Active

For the first line, $\frac{x - 4}{1} = \frac{y - 3}{2} = \frac{z - 2}{- 3}$ , a point on the line is $A_{1} = (4, 3, 2)$ and its direction vector is $\vec{b_{1}} = (1, 2, - 3)$ . For the second line, $\frac{x + 2}{2} = \frac{y - 6}{4} = \frac{z - 5}{- 5}$ , a point on the line is $A_{2} = (- 2, 6, 5)$ and its direction vector is $\vec{b_{2}} = (2, 4, - 5)$ . The vector connecting the two points is $\vec{a_{2}} - \vec{a_{1}} = (- 2 - 4, 6 - 3, 5 - 2) = (- 6, 3, 3)$ .

Step 2: Calculate Cross Product and its Magnitude○ Expand

Calculate the cross product of the direction vectors $\vec{b_{1}}$ and $\vec{b_{2}}$ :

\vec{b_{1}} \times \vec{b_{2}} = | \begin{matrix} i & j & k \\ 1 & 2 & - 3 \\ 2 & 4 & - 5 \end{matrix} | = i (- 10 - (- 12)) - j (- 5 - (- 6)) + k (4 - 4) = 2 i - j + 0 k = (2, - 1, 0)

The magnitude of this cross product is:

| \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{2^{2} + (- 1)^{2} + 0^{2}} = \sqrt{4 + 1 + 0} = \sqrt{5}

💡 Teacher's Secret Hint

Ensure correct sign conventions when calculating the determinant for the cross product.

Step 3: Calculate Shortest Distance○ Expand

Now, calculate the dot product of $(\vec{a_{2}} - \vec{a_{1}})$ and $(\vec{b_{1}} \times \vec{b_{2}})$ :

(\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}}) = (- 6, 3, 3) \cdot (2, - 1, 0) = (- 6) (2) + (3) (- 1) + (3) (0) = - 12 - 3 + 0 = - 15

Finally, apply the shortest distance formula:

d = \frac{| (\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}}) |}{| \vec{b_{1}} \times \vec{b_{2}} |} = \frac{| - 15 |}{\sqrt{5}} = \frac{15}{\sqrt{5}}

Rationalize the denominator:

d = \frac{15}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{15 \sqrt{5}}{5} = 3 \sqrt{5}

This matches option 3.

💡 Teacher's Secret Hint

Remember to take the absolute value of the scalar triple product in the numerator.

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Maths Question 14 – JEE-MAIN 2026

Clue 1 of 3: Identify Line Parameters

🧠 Full Solution Path