A) \[a=b=c=1\]
B) \[a=1,\,\,b\] and \[c\] are arbitrary scalars
C) \[a=b=c=0\]
D) \[c=0,\,\,a=1\] and b is arbitrary scalars
Correct Answer: D
Solution :
Here \[\overrightarrow{AB}=-2\mathbf{j},\] \[\overrightarrow{BC}=(a-1)\mathbf{i}+(b+1)\mathbf{j}+c\mathbf{k}\] The points are collinear, then \[\overrightarrow{AB}=k\,(\overrightarrow{BC})\] \[-2\mathbf{j}=k\{(a-1)\mathbf{i}+(b+1)\,\mathbf{j}+c\mathbf{k}\}\] On comparing, \[k\,(a-1)=0\], \[k(b+1)=-2,\] \[kc=0\]. Hence \[c=0,\] \[a=1\] and \[b\]is arbitrary scalar.You need to login to perform this action.
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