Let a and b be two unit vectors. If the vectors c = a + 2b and d = 5a - 4b are perpendicular to each other, then the angle between a and b is: - Mathematics MCQ AIEEE 2012 2026 | ExamTest.live

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Let \overset{a}{^} and \hat{b} be two unit vectors. If the vectors c = \overset{a}{^} + 2 \hat{b} and d = 5 \overset{a}{^} - 4 \hat{b} are perpendicular to each other, then the angle between \overset{a}{^} and \hat{b} is:

Solution & Step-by-step Explanation

Given that c ⊥ d, their dot product must be zero: c \cdot d = 0 (\overset{a}{^} + 2 \hat{b}) \cdot (5 \overset{a}{^} - 4 \hat{b}) = 0 Expanding the dot product: 5 (\overset{a}{^} \cdot \overset{a}{^}) - 4 (\overset{a}{^} \cdot \hat{b}) + 10 (\hat{b} \cdot \overset{a}{^}) - 8 (\hat{b} \cdot \hat{b}) = 0 Since \overset{a}{^} and \hat{b} are unit vectors, ∣ \overset{a}{^} ∣ = 1 and ∣ \hat{b} ∣ = 1 . Let the angle between them be θ, then \overset{a}{^} \cdot \hat{b} = cos θ . 5 (1)^{2} + 6 (\overset{a}{^} \cdot \hat{b}) - 8 (1)^{2} = 0 - 3 + 6 cos θ = 0 6 cos θ = 3 ⟹ cos θ = \frac{1}{2} θ = \frac{π}{3}

Try it yourself before checking the explanation above.

Let \overset{a}{^} and \hat{b} be two unit vectors. If the vectors c = \overset{a}{^} + 2 \hat{b} and d = 5 \overset{a}{^} - 4 \hat{b} are perpendicular to each other, then the angle between \overset{a}{^} and \hat{b} is:

\frac{π}{6}

\frac{π}{2}

\frac{π}{3}

\frac{π}{4}

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