A) 190
B) 275
C) 300
D) 320
E) 192
Correct Answer: E
Solution :
Given, \[|\overrightarrow{a}|=1,|\overrightarrow{b}|=2\] \[\therefore \] \[{{[(\overrightarrow{a}+3\overrightarrow{b})\times (3\overrightarrow{a}+\overrightarrow{b})]}^{2}}\] \[={{[0+\overrightarrow{a}\times \overrightarrow{b}+9\overrightarrow{b}\times \overrightarrow{a}+0]}^{2}}\] \[={{[-8\overrightarrow{a}\times \overrightarrow{b}]}^{2}}\] \[=64[|\overrightarrow{a}{{|}^{2}}|\overrightarrow{b}{{|}^{2}}{{\sin }^{2}}\theta ]\] \[=64[1\times 4\times {{\sin }^{2}}120{}^\circ ]\] \[=60\times 4\times \frac{3}{4}\] \[=192\]
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