Question

Vectors $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$are inclined at angle θ = 120°. If $\left|\overrightarrow{a}\right|=1,\left|\overrightarrow{b}\right|=2,$then ${\left[\left(\overrightarrow{a}+3\overrightarrow{b}\right)\times \left(3\overrightarrow{a}-\overrightarrow{b}\right)\right]}^2$ is equal to
(a) 300
(b) 325
(c) 275
(d) 225

Answer 1

(a) 300

$$\begin{gathered} \left(\overrightarrow{a}+3\overrightarrow{b}\right)\times \left(3\overrightarrow{a}-\overrightarrow{b}\right) \\ =3\left(\overrightarrow{a}\times \overrightarrow{a}\right)-\overrightarrow{a}\times \overrightarrow{b}+9\left(\overrightarrow{b}\times \overrightarrow{a}\right)-3\left(\overrightarrow{b}\times \overrightarrow{b}\right) \\ =3\left(0\right)-\overrightarrow{a}\times \overrightarrow{b}-9\left(\overrightarrow{a}\times \overrightarrow{b}\right)-3\left(0\right) \\ =-10\left(\overrightarrow{a}\times \overrightarrow{b}\right) \\ \text{Now,} \\ \left|\left(\overrightarrow{a}\times 3\overrightarrow{b}\right)\times \left(3\overrightarrow{a}-\overrightarrow{b}\right)\right|{}^2 \\ ={\left|-10\left(\overrightarrow{a}\times \overrightarrow{b}\right)\right|}^2 \\ =100{\left|\left(\overrightarrow{a}\times \overrightarrow{b}\right)\right|}^2 \\ =100{\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2{\sin }^{2}120 \\ =100{\left(1\right)}^2{\left(2\right)}^2{\left(\frac{\sqrt{3}}{2}\right)}^2 \\ =400\times \frac{3}{4} \\ =300 \end{gathered}$$