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Question

Physics Question on Motion in a plane

Angle (in rad) made by the vector 3i^+j^\sqrt{3\widehat{i}}+\widehat{j} with the X-axis:

A

π6\frac{\pi }{6}

B

π4\frac{\pi }{4}

C

π3\frac{\pi }{3}

D

π2\frac{\pi }{2}

Answer

π6\frac{\pi }{6}

Explanation

Solution

Let A⃗=3i^+j^\vec{A}=\sqrt{3\hat{i}}+\hat{j} and B⃗=i^\vec{B}=\hat{i} (unit vector along X-axis) =A⃗.B⃗=(3i^+j^).i^=\vec{A}.\vec{B}=(\sqrt{3}\hat{i}+\hat{j}).\hat{i} =3+0=3=\sqrt{3}+0=\sqrt{3} (∵  i^.i^=1, j^.i^=0)(\because \,\,\hat{i}.\hat{i}=1,\,\hat{j}.\hat{i}=0) Also, ∣A⃗∣=(3)2+(1)2=3+1=2|\vec{A}|=\sqrt{{{(\sqrt{3})}^{2}}+{{(1)}^{2}}}=\sqrt{3+1}=2 ∣B⃗∣=(1)2=1|\vec{B}|=\sqrt{{{(1)}^{2}}}=1 ∵\because AB cos⁡θ=A⃗.B⃗AB\,\cos \theta =\vec{A}.\vec{B} where θ\theta is the angle between vector A⃗\vec{A} and X - axis. cos⁡θ=A⃗.B⃗AB=32.1=32=cos⁡π6\cos \theta =\frac{\vec{A}.\vec{B}}{AB}=\frac{\sqrt{3}}{2.1}=\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6} θ=π6  rad\theta =\frac{\pi }{6}\,\text{ rad}