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Let a⃗\vec{a} and b⃗\vec{b} be two unit vectors. If the vectors c⃗=5a⃗−4b⃗\vec{c} = 5\vec{a} - 4\vec{b} and d⃗=a⃗+2b⃗\vec{d} = \vec{a} + 2\vec{b} are perpendicular to each other, then the angle between a⃗\vec{a} and b⃗\vec{b} is :

Solution

✅ Correct Option: 2

c⃗⋅d⃗=5∣a⃗∣2+6a⃗⋅b⃗−8∣b⃗∣2=0\vec{c} \cdot \vec{d} = 5|\vec{a}|^2 + 6\vec{a}\cdot\vec{b} - 8|\vec{b}|^2 = 0.

5+6a⃗⋅b⃗−8=0⇒a⃗⋅b⃗=125 + 6\vec{a}\cdot\vec{b} - 8 = 0 \Rightarrow \vec{a}\cdot\vec{b} = \frac{1}{2}.

So cos⁡θ=12⇒θ=π3\cos\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}.

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