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Question : 32 of 160
Marks:
+1,
-0
Solution:
The distance is,
d=d1i+d2j+d3k a.d=0 (i+j).(d1i+d2j+d3k)=0 d1+d2=0 d2=−d1 The
d is the unit vector.
b.(c×d)=0 [b c d]=0 ||=0 −d3+d1+d2=0 d3=0 From above calculation,
d12+d12=1 2d12=1 d1=±(−)
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