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If $\mathbf{a}$ and $\mathbf{b}$ are wit vectors and $\alpha$ is the angle between them, then $\mathbf{a}+\mathbf{b}$ is a wnit vector when $\cos \alpha=$
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Verified Answer
The correct answer is:
$-\frac{1}{2}$
We have, $|\mathbf{a}|=|\mathbf{b}|=1$ and $\alpha$ is the angle between $\mathbf{a}$ and $\mathbf{b}$.
$\begin{aligned}
& \text {Clearly, } \cos \alpha=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \\
& \Rightarrow \quad \cos \alpha=\mathbf{a} \cdot \mathbf{b}
\end{aligned}$
Now, let $\mathbf{a}+\mathbf{b}$ is a unit vector, then
$\begin{aligned}
& |\mathbf{a}+\mathbf{b}|=1 \\
& \Rightarrow \quad|\mathbf{a}+\mathbf{b}|^2=1 \\
& \Rightarrow \quad\langle\mathbf{a}+\mathbf{b}\rangle \cdot\langle\mathbf{a}+\mathbf{b}\rangle=1 \\
& \Rightarrow \mathbf{a} \cdot \mathbf{a}+\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{a}+\mathbf{b} \cdot \mathbf{b}=1 \\
& \Rightarrow \quad|\mathbf{a}|^2+2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2=1 \quad[\because \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{a}] \\
& \Rightarrow \quad 1+2 \cos \alpha+1=1 \quad \text { [using Eq. (i)] } \\
& \Rightarrow \quad 1+2 \cos \alpha=0 \\
& \Rightarrow \quad \cos \alpha=-\frac{1}{2}
\end{aligned}$
$\begin{aligned}
& \text {Clearly, } \cos \alpha=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \\
& \Rightarrow \quad \cos \alpha=\mathbf{a} \cdot \mathbf{b}
\end{aligned}$
Now, let $\mathbf{a}+\mathbf{b}$ is a unit vector, then
$\begin{aligned}
& |\mathbf{a}+\mathbf{b}|=1 \\
& \Rightarrow \quad|\mathbf{a}+\mathbf{b}|^2=1 \\
& \Rightarrow \quad\langle\mathbf{a}+\mathbf{b}\rangle \cdot\langle\mathbf{a}+\mathbf{b}\rangle=1 \\
& \Rightarrow \mathbf{a} \cdot \mathbf{a}+\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{a}+\mathbf{b} \cdot \mathbf{b}=1 \\
& \Rightarrow \quad|\mathbf{a}|^2+2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2=1 \quad[\because \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{a}] \\
& \Rightarrow \quad 1+2 \cos \alpha+1=1 \quad \text { [using Eq. (i)] } \\
& \Rightarrow \quad 1+2 \cos \alpha=0 \\
& \Rightarrow \quad \cos \alpha=-\frac{1}{2}
\end{aligned}$
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