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If $\mathbf{a} \cdot \mathbf{b}=0$ and $\mathbf{a}+\mathbf{b}$ makes an angle $60^{\circ}$ with a, then
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Verified Answer
The correct answer is:
$\sqrt{3}|\mathbf{a}|=|\mathbf{b}|$
Given, $\mathbf{a} \cdot \mathbf{b}=0$ and $(\mathbf{a}+\mathbf{b})$ makes $60^{\circ}$ angle with a.
$\cos 60^{\circ}=\frac{(\mathbf{a}+\mathbf{b}) \cdot \mathbf{a}}{|\mathbf{a}+\mathbf{b}||\mathbf{a}|}$
$\Rightarrow \quad \frac{1}{2}=\frac{|\mathbf{a}|^{2}+\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}+\mathbf{b}||\mathbf{a}|}$
$\Rightarrow \quad|\mathbf{a}+\mathbf{b}|=2|\mathbf{a}|$
$\Rightarrow \quad|\mathbf{a}+\mathbf{b}|^{2}=4|\mathbf{a}|^{2}$
$\Rightarrow \quad(\mathbf{a}+\mathbf{b})(\mathbf{a}+\mathbf{b})=4|\mathbf{a}|^{2}$
$\Rightarrow \quad|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+2 \mathbf{a} \cdot \mathbf{b}=4|\mathbf{a}|^{2}$
$\Rightarrow \quad \mathbf{b}^{2}=3|\mathbf{a}|^{2}$
$\Rightarrow \quad|\mathbf{b}|=\sqrt{3}|\mathbf{a}|$
$\cos 60^{\circ}=\frac{(\mathbf{a}+\mathbf{b}) \cdot \mathbf{a}}{|\mathbf{a}+\mathbf{b}||\mathbf{a}|}$
$\Rightarrow \quad \frac{1}{2}=\frac{|\mathbf{a}|^{2}+\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}+\mathbf{b}||\mathbf{a}|}$
$\Rightarrow \quad|\mathbf{a}+\mathbf{b}|=2|\mathbf{a}|$
$\Rightarrow \quad|\mathbf{a}+\mathbf{b}|^{2}=4|\mathbf{a}|^{2}$
$\Rightarrow \quad(\mathbf{a}+\mathbf{b})(\mathbf{a}+\mathbf{b})=4|\mathbf{a}|^{2}$
$\Rightarrow \quad|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+2 \mathbf{a} \cdot \mathbf{b}=4|\mathbf{a}|^{2}$
$\Rightarrow \quad \mathbf{b}^{2}=3|\mathbf{a}|^{2}$
$\Rightarrow \quad|\mathbf{b}|=\sqrt{3}|\mathbf{a}|$
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