Let |vec{a}| = |vec{b}| = 1 and |vec{a} + vec | vedclass

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(D) Given that $|\vec{a} + \vec{b}| = \sqrt{3}$.Squaring both sides,we get $|\vec{a} + \vec{b}|^2 = 3$.$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \v

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(D) Given that $|\vec{a} + \vec{b}| = \sqrt{3}$.Squaring both sides,we get $|\vec{a} + \vec{b}|^2 = 3$.$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 3$.Since $|\vec{a}| = 1$ and $|\vec{b}| = 1$,we have $1 + 1 + 2(\vec{a} \cdot \vec{b}) = 3$.$2(\vec{a} \cdot \vec{b}) = 1$,so $\vec{a} \cdot \vec{b} = 1/2$  $(1)$.Now,consider the given equation $\vec{c} - \vec{a} - 2\vec{b} = 3(\vec{a} 	imes \vec{b})$.Taking the dot product with $\vec{b}$ on both sides:$(\vec{c} - \vec{a} - 2\vec{b}) \cdot \vec{b} = 3(\vec{a} 	imes \vec{b}) \cdot \vec{b}$.$\vec{c} \cdot \vec{b} - \vec{a} \cdot \vec{b} - 2(\vec{b} \cdot \vec{b}) = 3[\vec{a} \vec{b} \vec{b}]$.Since the scalar triple product $[\vec{a} \vec{b} \vec{b}] = 0$ (as two vectors are identical),we have:$\vec{c} \cdot \vec{b} - 1/2 - 2(1) = 0$.$\vec{c} \cdot \vec{b} = 1/2 + 2 = 5/2$.

Let $|\vec{a}| = |\vec{b}| = 1$ and $|\vec{a} + \vec{b}| = \sqrt{3}$. If $\vec{c}$ is a vector satisfying the condition $\vec{c} - \vec{a} - 2\vec{b} = 3(\vec{a} \times \vec{b})$,then $\vec{c} \cdot \vec{b} = \dots$ (in $/2$)

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