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  4. If a.b=0 and a+b makes an angle of 60° with a , then

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Q. If $\overrightarrow{a}.\overrightarrow{b}=0 $ and $\overrightarrow{a}+\overrightarrow{b}$ makes an angle of 60$^\circ$ with $\overrightarrow{a} $ , then

Vector Algebra

Solution:

$cos60^{\circ}=\frac{\left(\vec{a}+\vec{b}\right)\cdot\vec{a}}{\left|\vec{a}+\vec{b}\right|\cdot\frac{\left|\vec{a}\right|}{\left|\vec{a}\right|}}$
$\Rightarrow \frac{1}{2}=\frac{\vec{a}^{2}+\vec{a}\cdot\vec{b}}{\left|\vec{a}+\vec{b}\right|}=\frac{\left|\vec{a}\right|^{2}}{\left|\vec{a}+\vec{b}\right|\cdot\left|\vec{a}\right|^{2}}$
$\Rightarrow \frac{1}{2}=\frac{\left|\vec{a}\right|}{\left|\vec{a}+\vec{b}\right|}$, where $\left|\vec{a}+\vec{b}\right|^{2}=\left|\vec{a}\right|^{2}+\left|\vec{b}\right|^{2}$
$\left[\because \vec{a}\cdot\vec{b}=0\right]$
$\Rightarrow \frac{1}{2}=\frac{\left|\vec{a}\right|}{\sqrt{\left|\vec{a}^{2}+\vec{b}\right|^{2}}}$
$\Rightarrow 4\left|\vec{a}\right|^{2}=\left|\vec{a}\right|^{2}+\left|\vec{b}\right|^{2}$
$\Rightarrow 3\left|\vec{a}\right|^{2}=\left|\vec{b}\right|^{2}$
$\Rightarrow \left|\vec{b}\right|=\sqrt{3}\,\left|\vec{a}\right|$