Let $\vec{a}$ and $\vec{b}$ be unit vectors such that $|\vec{a}+\vec{b}|=\sqrt{3}$, then the value of $(2 \vec{a}+5 \vec{b}) . (3 \vec{a}+\vec{b}+\vec{a} \times \vec{b})$ is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option 3: $\frac{39}{2}$
$(2 \vec{a}+5 \vec{b}) .(3 \vec{a}+\vec{b}+\vec{a} \times \vec{b})$
$=6 \vec{a} . \vec{a}+17 \vec{a} . \vec{b}+5 \vec{b} . \vec{b}$
$=11+17 \vec{a} . \vec{b}$
Now, $|\vec{a}+\vec{b}|=\sqrt{3}=|\vec{a}+\vec{b}|^2=3$
$\Rightarrow|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} . \vec{b}=3$
$\Rightarrow \vec{a} . \vec{b}=\frac{1}{2}$
$\Rightarrow(2 \vec{a}+5 \vec{b}) .(3 \vec{a}+\vec{b}+\vec{a} \times \vec{b})$
$=11+\frac{17}{2}=\frac{39}{2}$
Hence (3) is correct answer.