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(D) Given,$a \cdot b = 0$ and $(a + b)$ makes a $60^{\circ}$ angle with $a$. Using the formula for the angle between two vectors: $\cos 60^{\circ} = \

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$\sqrt{3}|a| = |b|$

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(D) Given,$a \cdot b = 0$ and $(a + b)$ makes a $60^{\circ}$ angle with $a$. Using the formula for the angle between two vectors: $\cos 60^{\circ} = \frac{(a + b) \cdot a}{|a + b||a|}$ $\Rightarrow \frac{1}{2} = \frac{a \cdot a + b \cdot a}{|a + b||a|}$ Since $a \cdot b = 0$,we have $b \cdot a = 0$. $\Rightarrow \frac{1}{2} = \frac{|a|^2}{|a + b||a|} = \frac{|a|}{|a + b|}$ $\Rightarrow |a + b| = 2|a|$ Squaring both sides: $|a + b|^2 = 4|a|^2$ $|a|^2 + |b|^2 + 2(a \cdot b) = 4|a|^2$ Since $a \cdot b = 0$: $|a|^2 + |b|^2 = 4|a|^2$ $|b|^2 = 3|a|^2$ $|b| = \sqrt{3}|a|$

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

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