Mathematics in Physics

JEE Main level, single correct, medium

18 of 121

Question 18 of 121

Which of the following relations is true for two unit vector A^\hat{A} and B^\hat{B} making an angle θ\theta to each other?

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Answer: (B) ∣A^−B^∣=∣A^+B^∣tan⁡θ2\left|\hat{A}-\hat{B}\right|=\left|\hat{A}+\hat{B}\right|\dfrac{\tan \theta }{2}

We know,
1−cos⁡θ=2sin⁡2(θ2)1+cos⁡θ=2cos⁡2(θ2)1-\cos \theta =2{\sin }^{2}\left(\dfrac{\theta }{2}\right) 1+\cos \theta =2{\cos }^{2}\left(\dfrac{\theta }{2}\right)
Now,(a⃗+b⃗)⋅(a⃗+b⃗)=∣a⃗∣2+2a⃗⋅b⃗+∣b⃗∣2⇒(a⃗+b⃗)⋅(a⃗+b⃗)=(1)2+2×1×1×cos⁡θ+(1)2⇒(a⃗+b⃗)⋅(a⃗+b⃗)=2(1+cos⁡θ)⇒∣a⃗+b⃗∣2=4cos⁡2(θ2)⇒∣a⃗+b⃗∣=2cos⁡(θ2)...(1)\left(\vec{a}+\vec{b}\right)\cdot \left(\vec{a}+\vec{b}\right)={\left|\vec{a}\right|}^{2}+2\vec{a}\cdot \vec{b}+{\left|\vec{b}\right|}^{2} \Rightarrow \left(\vec{a}+\vec{b}\right)\cdot \left(\vec{a}+\vec{b}\right)={\left(1\right)}^{2}+2\times 1\times 1\times \cos \theta +{\left(1\right)}^{2} \Rightarrow \left(\vec{a}+\vec{b}\right)\cdot \left(\vec{a}+\vec{b}\right)=2\left(1+\cos \theta \right) \Rightarrow {\left|\vec{a}+\vec{b}\right|}^{2}=4{\cos }^{2}\left(\dfrac{\theta }{2}\right) \Rightarrow \left|\vec{a}+\vec{b}\right|=2\cos \left(\dfrac{\theta }{2}\right)...(1)
Similarly,
(a⃗−b⃗)⋅(a⃗−b⃗)=∣a⃗∣2−2a⃗⋅b⃗+∣b⃗∣2⇒(a⃗−b⃗)⋅(a⃗−b⃗)=(1)2−2×1×1×cos⁡θ+(1)2⇒(a⃗−b⃗)⋅(a⃗−b⃗)=2(1−cos⁡θ)⇒∣a⃗−b⃗∣2=4sin⁡2(θ2)⇒∣a⃗−b⃗∣=2sin⁡(θ2)...(2)\left(\vec{a}-\vec{b}\right)\cdot \left(\vec{a}-\vec{b}\right)={\left|\vec{a}\right|}^{2}-2\vec{a}\cdot \vec{b}+{\left|\vec{b}\right|}^{2} \Rightarrow \left(\vec{a}-\vec{b}\right)\cdot \left(\vec{a}-\vec{b}\right)={\left(1\right)}^{2}-2\times 1\times 1\times \cos \theta +{\left(1\right)}^{2} \Rightarrow \left(\vec{a}-\vec{b}\right)\cdot \left(\vec{a}-\vec{b}\right)=2\left(1-\cos \theta \right) \Rightarrow {\left|\vec{a}-\vec{b}\right|}^{2}=4{\sin }^{2}\left(\dfrac{\theta }{2}\right) \Rightarrow \left|\vec{a}-\vec{b}\right|=2\sin \left(\dfrac{\theta }{2}\right)...(2)
Therefore,
∣A^−B^∣=∣A^+B^∣tan⁡θ2\left|\hat{A}-\hat{B}\right|=\left|\hat{A}+\hat{B}\right|\dfrac{\tan \theta }{2}

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