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Question : 32 of 160
Marks:
+1,
-0
Solution:
‌ ‌r=(+5+5)+t(4−4+5‌) ‌r=(2+4+5)+s(8−3+) Let's find the shortest distance between them projection of
(2−1)+(4−5)+(5−5) on
‌ ‌‌‌n1×n2=|| ,,;4,−4,5;8,−3,1|=11+36+20 |
‌ Perpendicular distance ‌=(−)⋅‌| (11+36+20‌) |
| √112+362+202 |
Clearly, this is not zero.
∴ Both the lines are skew lines.
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