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Question : 29 of 160
Marks:
+1,
-0
Solution:
+3=k 3b+2c=ma Substitute (1) in to (2)
[k−]+2=ma ⇒k−+2=m ⇒[k+2]=[m+1] and
are non-collinear
k+2=0⇒k=−2 m+1=0→m=−1 a+3b+2c =kc+2c k(−2)×c+2c =0
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