Question

If $\overline{a}$ and $\overline{b}$ are non-collinear unit vectors and $|\overline{a}+\overline{b}|=\sqrt{3}$ then $(2\overline{a}+5\overline{b})\cdot (3\overline{a}-\overline{b})=?$

• A. $\frac{15}{4}$
• B. $\frac{15}{2}$
• C. $15$
• D. $16$

Answer 1

The correct option is B $\frac{15}{2}$

$|\overline{a}+\overline{b}|=\sqrt{{\left|\overline{a}\right|}^2+{\left|\overline{b}\right|}^2+2\overline{a}.\overline{b}}=\sqrt{3}$

Since $\left|\overline{a}\right|=\left|\overline{b}\right|=1$

$\overline{a}.\overline{b}=\frac{1}{2}$

So, $(2\overline{a}+5\overline{b}).(3\overline{a}-\overline{b})=6{\left|\overline{a}\right|}^2-2\overline{a}.\overline{b}+15\overline{b}.\overline{a}-5{\left|\overline{b}\right|}^2$

as $\overline{a}.\overline{b}=\overline{b}.\overline{a}$, we can say that

$(2\overline{a}+5\overline{b}).(3\overline{a}-\overline{b})=6+13\left(\frac{1}{2}\right)-5=\frac{15}{2}$