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Question : 34 of 160
Marks:
+1,
-0
Solution:
Given:
r=−4i−k+t(3i−2j−2k) r=6i+2j+2k+s(i−2j+2k) a1=−4i−k b1=3i−2j−2k a2=6i+2j+2k b2=i−2j+2k Now,
a2−a1=10i+2j+3k b1×b2=|| |b1×b2| =√(−8)2+(−8)2+(−4)2 =√64+64+16 =12 Now,
(a2−a1).(b1×b2)=(10i+2j+3k) .(−8i−8j−4k) =−80−16−12 =−108 The required distance is,
=|| (a2−a1).(b1×b2) |
| |b1×b2| |
| =|| =9
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