InterviewSolution
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InterviewSolution
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on of two vectors is also a unit vector then the angle between tbstractivo vectors is |
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Answer» Let the two vectors be (ai^+bj^+ck^) & (pi^+qj^+rk^)Then,(ai^+bj^+ck^) - (pi^+qj^+rk^)= (a-p)i^ + (b-q)j^ + (c-k)k^For this to be an unit vector, EITHER two of (a-p), (b-q) or (c-r) have to be zero. For the sake of simplicity, I take (b-q) and (c-r) to be zero.Therefore, b=q and c=r, giving us vectors ai^+bj^+ck^ and pi^+bj^+ck^,Therefore, angle between them is,∆ = arc(cos[(ap + b^2 + c^2)/{√(a^2 + b^2 + c^2)√(p^2 + b^2 + c^2)}]) If b=c=q=r=0,Angle ∆ = arc(cos(ap/ap))= arc(cos 1)= 0° |
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