A) \[x=3,y=1\]
B) \[x=0,y=3\]
C) \[~x=1,\text{ }y=3\]
D) \[x=0,\text{ }y=0\]
Correct Answer: D
Solution :
[b] \[\because \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0.\] \[\Rightarrow \overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}\,\Rightarrow {{(\overrightarrow{a}+\overrightarrow{b})}^{2}}={{(-\overrightarrow{c})}^{2}}\] \[\Rightarrow {{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}+2\overrightarrow{a}.\overrightarrow{b}={{\left| \overrightarrow{c} \right|}^{2}}~\Rightarrow {{\left( 3 \right)}^{2}}+{{\left( 5 \right)}^{2}}+2\left| \overrightarrow{a} \right|.\left| \overrightarrow{b} \right|.\cos \theta =49\]\[\Rightarrow 9+25+2\times 3\times 5.cos\theta =49\Rightarrow 30.cos\theta =49-34=15\] \[cos\theta =\frac{15}{30}=\frac{1}{2}~~~~~~~~~~~\Rightarrow cos\theta =cos\frac{\pi }{3}\] \[\Rightarrow \theta =\frac{\pi }{3}\] Hence, option [b] is correct.
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