3.find the value(s) of x for which the angle between the vectors a=xi-3j-k and b=2xi+xj-k is obtuse

If the angle between the vectors a=xi-3j-k b=2xi+xj-k is obtuse then cosĪ˜ < 0 or a.b < 0 thus, the dot product would be a.b = 2x.2 -3.x +1 < 0 (x-1)(2x-1) < 0 thus, the values of x are x > 1/2 or x < 1

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