Question

Vectors $ \overrightarrow{a} $ and $ \overrightarrow{b} $ are inclined at an angle $ \theta =120{}^\circ $ .If $ |\overrightarrow{a}|=1,|\overrightarrow{b}|=2, $ then $ {{[(\overrightarrow{a}+3\overrightarrow{b})\times (3\overrightarrow{a}+\overrightarrow{b})]}^{2}} $ is equal to

• A. 190
• B. 275
• C. 300
• D. 320
• E. 192

Answer 1

Answer: (E)

Given, $ |\overrightarrow{a}|=1,|\overrightarrow{b}|=2 $
$ \therefore $ $ {{[(\overrightarrow{a}+3\overrightarrow{b})\times (3\overrightarrow{a}+\overrightarrow{b})]}^{2}} $
$={{[0+\overrightarrow{a}\times \overrightarrow{b}+9\overrightarrow{b}\times \overrightarrow{a}+0]}^{2}} $
$={{[-8\overrightarrow{a}\times \overrightarrow{b}]}^{2}} $
$=64[|\overrightarrow{a}{{|}^{2}}|\overrightarrow{b}{{|}^{2}}{{\sin }^{2}}\theta ] $
$=64[1\times 4\times {{\sin }^{2}}120{}^\circ ] $
$=64\times 4\times \frac{3}{4} $
$=192 $