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Question : 34 of 160
Marks:
+1,
-0
Solution:
The given vectors are,
r=(−2i+j−k)+k(2i+3j−k) r=(i−j+2k)+k(−i+2j+4k) Then
a2−a1=3i−2j+3k This implies,
b1×b2‌‌=|| ‌‌=14i−7j+7k The shortest distance is given by,
‌| (a2−a1)⋅(b1×b2) |
| |b1×a2| |
‌‌=‌| (3i−2j+3k)⋅(14i−7j+7k) |
| 7√6 |
‌‌=‌ ‌‌=‌
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