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3D Geometry · 2020 · 8 Jan · Shift 1

1 Mathematics questions from this JEE Main paper

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JEE Main
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2020 · 8 Jan · Shift 1
Mathematics
3D Geometry

3D Geometry · 2020 · 8 Jan · Shift 1

1 Mathematics questions from this JEE Main paper

Q24.The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is :
Mathematics · 3D Geometry · MCQ