Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three vectors such that each is inclined at an angle $\frac{\pi }{3}$ with the other two and $|\overrightarrow{a}|=1,|\overrightarrow{b}|=2,|\overrightarrow{c}|=3$, then the scalar product of the vectors $2\overrightarrow{a}+3\overrightarrow{b}-5\overrightarrow{c}$ and $4\overrightarrow{a}-6\overrightarrow{b}+10\overrightarrow{c}$ is

• A. $188$
• B. $-334$
• C. $-522$
• D. $-514$

Answer 1

The correct option is B $-334$

We have

$\overline{a}.\overline{b}=|a||b|\cos \theta =1$
$\overline{c}.\overline{b}=|c||b|\cos \theta =3$
$\overline{a}.\overline{c}=|a||c|\cos \theta =1.5$
$\therefore$ the scalar product is $(2\overline{a}+3\overline{b}-5\overline{c}).(4\overline{a}-6\overline{b}+10\overline{c})$

$=8|a{|}^2-18|b{|}^2-50|c{|}^2+60\overline{c}.\overline{b}$

$=8-72-450+180$

$=-334$