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Question : 34 of 160
Marks:
+1,
-0
Solution:
Given, lines
r=(−+3)+t(2+3+6) and
‌‌r=(3+−)+s(−+2) Shortest distance between lines is
((3+1)+−(+3))−[(+3+6)SD=‌| ×(2i−j+2k)] |
| |(2+3+6)|×|(2−+2)| |
SD=‌| (4+−4)⋅(12+8−8) |
| |12+8−8| |
SD=‌=‌=‌
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