The shortest distance between the lines r = (1/3i + 2j + 8/3k) + λ(2i - 5j + 6k) and r = (-2/3i - 1/3k) + μ(j - k), λ, μ ∈ R, is:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q13 – JEE-MAIN 2026

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Home Maths Vector AlgebraQuestion #508

The shortest distance between the lines

\vec{r} = (\frac{1}{3} \hat{i} + 2 \hat{j} + \frac{8}{3} \hat{k}) + λ (2 \hat{i} - 5 \hat{j} + 6 \hat{k})

and

\vec{r} = (- \frac{2}{3} \hat{i} - \frac{1}{3} \hat{k}) + μ (\hat{j} - \hat{k}), λ, μ \in R, is:

Clue 1 of 3: Identify Line Parameters

Recognize the given equations as vector forms of lines, $\vec{r} = \vec{a} + λ \vec{b}$ , and extract the position vectors ( $\vec{a_{1}}, \vec{a_{2}}$ ) and direction vectors ( $\vec{b_{1}}, \vec{b_{2}}$ ) for each line.

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Step 1: Identify Position and Direction Vectors✦ Active

The given lines are in the form $\vec{r} = \vec{a} + λ \vec{b}$ . From the equations, we identify the position vectors and direction vectors:

\vec{a_{1}} = \frac{1}{3} \hat{i} + 2 \hat{j} + \frac{8}{3} \hat{k}, \vec{b_{1}} = 2 \hat{i} - 5 \hat{j} + 6 \hat{k}

\vec{a_{2}} = - \frac{2}{3} \hat{i} - \frac{1}{3} \hat{k}, \vec{b_{2}} = \hat{j} - \hat{k}

Now, calculate the difference vector $\vec{a_{2}} - \vec{a_{1}}$ :

\vec{a_{2}} - \vec{a_{1}} = (- \frac{2}{3} - \frac{1}{3}) \hat{i} + (0 - 2) \hat{j} + (- \frac{1}{3} - \frac{8}{3}) \hat{k} = - \hat{i} - 2 \hat{j} - 3 \hat{k}

Step 2: Calculate Cross Product and Scalar Triple Product○ Expand

Next, calculate the cross product of the direction vectors $\vec{b_{1}} \times \vec{b_{2}}$ :

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

Then, compute the scalar triple product $(\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}})$ :

(\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}}) = (- \hat{i} - 2 \hat{j} - 3 \hat{k}) \cdot (- \hat{i} + 2 \hat{j} + 2 \hat{k}) = (- 1) (- 1) + (- 2) (2) + (- 3) (2) = 1 - 4 - 6 = - 9

Step 3: Compute the Shortest Distance○ Expand

Calculate the magnitude of the cross product:

| \vec{b_{1}} \times \vec{b_{2}} | = | - \hat{i} + 2 \hat{j} + 2 \hat{k} | = \sqrt{(- 1)^{2} + 2^{2} + 2^{2}} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3

Finally, use the formula for the shortest distance between two skew lines:

D = \frac{| (\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}}) |}{| \vec{b_{1}} \times \vec{b_{2}} |} = \frac{| - 9 |}{3} = \frac{9}{3} = 3

The shortest distance between the lines is $3$ units.

💡 Teacher's Secret Hint

Ensure all vector operations (subtraction, cross product, dot product, magnitude) are performed carefully to avoid calculation errors.

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Maths Question 13 – JEE-MAIN 2026

Clue 1 of 3: Identify Line Parameters

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