The shortest distance between the lines whose vector equations are r = (6i+2j+2k) + (i-2j+2k) and r = (-4i-k) + (3i-2j-2k) Option 1: 109/12 units Option 2: 12 units Option 3: 25/3 units. Option 4: 9 units

The answer is '9 units'. View full solution:

CUET Mathematics - The shortest distance between the lines whose vector equations are r = (6i+2j+2k) + (i-2j+2k) and r = (-4i-k) + (3i-2j-2k) | PYQs + Solutions

CUET Mathematics 2026 19th May Shift 1

Geometry

Medium

core

$\frac{109}{12}$ units

12 units

$\frac{25}{3}$ units.

9 units

✅ Correct Option: 4

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Consider the equation of the line $\vec{r} = -\hat{i} + 2\hat{k} + \mu(4\hat{i} - \hat{j} + 2\hat{k})$ .

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List-I	List-II
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The shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$ and $\vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 8\hat{k})$ is equal to

CUET Mathematics 2026 19th May Shift 1 PYQs

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