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If the position vectors of the points A,B and C be `hati+hatj,hati-hatj` and `ahati+bhatj+chatk` respectively, then the points A,B and C are collinear, ifA. a=b=c=1B. a=1,b and c are arbitrary scalarsC. ab=c=0D. c=0,a=1 and b is arbitrary scalars |
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Answer» Correct Answer - D Here, `AB=-2hatj,BC=(a-1)hati+(b+1)hatj+chatk` The points are collinear, then `AB=k(BC)` `-2hatj=k{(a-1)hati+(b+1)hatj+chatk}` On comparing, `k(a-1)=0,k(b+1)=-2,kc=0` Hence, c=0,a=1 and b is arbitrary scalar. |
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