Three Dimensional Geometry Part 4

Section: Mathematics

Question : 9 of 73

Marks: +1, -0

Let P(α,β,γ) be the point on the line ‌

x−1

=‌

y+1

−3

=z at a distance 4√14 from the point (1,−1,0) and nearer to the origin. Then the shortest distance, between the lines ‌

x−α

=‌

y−β

=‌

z−γ

and ‌

x+5

=‌

y−10

=‌

z−3

, is equal to

[22 Jan 2026 Shift 1]

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