If $\vec{a}, \vec{b}$ are two unit vectors with $(\vec{a}, \vec{b})=\theta$ and $|\vec{a}-\vec{b}|=1$, then…

If $\vec{a}, \vec{b}$ are two unit vectors with $(\vec{a}, \vec{b})=\theta$ and $|\vec{a}-\vec{b}|=1$, then $2|\vec{a}+\vec{b}| \cos \frac{\theta}{2}=$
  1. 3
  2. 1
  3. $\sqrt{3}$
  4. 9

Solution

$ \begin{aligned} & \text { } \because \quad \vec{a}, \vec{b} \text { are unit vectors. So, }|\vec{a}|=1 \&|\vec{b}|=1 \\ & \text { Also, }|\vec{a}-\vec{b}|^2=1 \Rightarrow|\vec{a}-\vec{b}|^2=1 \\ & \Rightarrow|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b}=1 \\ & \Rightarrow 1+1-2 \vec{a} \cdot \vec{b}=1 \Rightarrow \vec{a} \cdot \vec{b}=\frac{1}{2} \\ & |\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=1+1+2 \times \frac{1}{2}=3 \\ & \Rightarrow|\vec{a}+\vec{b}|=\sqrt{3} \\ & \because \vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cos \theta=\frac{1}{2} \Rightarrow \cos \theta=\frac{1}{2} \\ & \Rightarrow 2 \cos ^2 \frac{\theta}{2}-1=\frac{1}{2} \Rightarrow 2 \cos ^2 \frac{\theta}{2}=\frac{3}{2} \Rightarrow \cos ^2 \frac{\theta}{2}=\frac{3}{4} \\ & \Rightarrow \cos \frac{\theta}{2}=\frac{\sqrt{3}}{2} \end{aligned} $ Now, $2|\vec{a}+\vec{b}| \cos \frac{\theta}{2}=2 \times \sqrt{3} \times \frac{\sqrt{3}}{2}=3$

Asked in: AP EAMCET 2023 (18 May Shift 2)

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