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If θ\theta is an acute angle and the vector a⃗=(sin⁡θ)i⃗+(cos⁡θ)j⃗\vec{a} = (\sin \theta)\vec{i} + (\cos\theta)\vec{j} is perpendicular to the vector b⃗=i−3j\vec{b} = i - \sqrt{3}j then θ\theta is equal to

Solution

✅ Correct Option: 2

Since a⃗\vec{a} is perpendicular to b⃗\vec{b}, their dot product is zero:

a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0

Given:

a⃗=(sin⁡θ)i^+(cos⁡θ)j^\vec{a} = (\sin \theta)\hat{i} + (\cos \theta)\hat{j}

b⃗=i^−3j^\vec{b} = \hat{i} - \sqrt{3}\hat{j}

Taking the dot product:

(sin⁡θ)(1)+(cos⁡θ)(−3)=0(\sin \theta)(1) + (\cos \theta)(-\sqrt{3}) = 0

sin⁡θ−3cos⁡θ=0\sin \theta - \sqrt{3}\cos \theta = 0

sin⁡θ=3cos⁡θ\sin \theta = \sqrt{3}\cos \theta

tan⁡θ=3\tan \theta = \sqrt{3}

Since θ\theta is an acute angle:

θ=π3\theta = \frac{\pi}{3}

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