Angle Between Vectors

Question

Two vectors a and b have magnitudes 3 and 5 respectively while their dot product equals 15/2; find the angle between vectors a and b.

RankGuru · Accepted Answer

The angle between two vectors follows from rearranging the geometric form of the dot product, a . b = |a||b|cos(theta), into cos(theta) = (a . b)/(|a||b|). Isolating the cosine and then inverting is the standard JEE Advanced route to recover an angle from numerical data. Substituting the given values, cos(theta) = (15/2)/((3)(5)) = (15/2)/15 = 1/2. The angle whose cosine equals 1/2 in the standard range is theta = 60 degrees. Option 30 degrees corresponds to cos(theta) = sqrt(3)/2, which would require a dot product of (3)(5)(sqrt 3/2), larger than given. Option 45 degrees needs cos(theta) = 1/sqrt(2), again inconsistent with 1/2. Option 90 degrees would demand a zero dot product, contradicting the positive value 15/2. Because the cosine is positive but less than one, the angle must be acute and strictly between 0 and 90 degrees, which 60 degrees satisfies. Plausibility check: the dot product 15/2 = 7.5 is positive and below the maximum |a||b| = 15, so cos(theta) lies in (0,1), confirming a valid acute angle.

Angle Between Vectors

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