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Question : 34 of 160
Marks:
+1,
-0
Solution:
Given,
r1=(3−5+2)+t(4+3−)So,
a1=(3−5+2),
b1=(4+3−)And
r2=(+2−4)+s(6+3−2)So,
a2=(+2−4),b2=(6+3−2)Since, shortest distance,
d=‌| |(a2−a1)⋅(b1×b2)| |
| |b1×b2| |
Now,
a2−a1=⟨1,2,−4⟩−⟨3,−5,2⟩=⟨1−3,2−(−5),−4−2⟩=<−2,7,−6>b1×b2=||=(−6−(−3))−(−8−(−6))+(12−18)=−3+2−6|b1×b2|=√(−3)2+22+(−6)2=√9+4+36=√49=7So,
‌‌d=‌| |⟨−2,7,−6⟩⋅⟨−3,2,−6⟩| |
| 7 |
=‌=‌=8
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