A) -112.5
B) -106.5
C) -118.5
D) -99.5
Correct Answer: C
Solution :
\[\left| {{{\vec{A}}}_{1}} \right|=3\,\,\,\,\,\,\,\,\,\,\left| {{{\vec{A}}}_{2}} \right|=5\,\,\,\,\,\,\,\,\,\,\,\left| {{{\vec{A}}}_{1}}+{{{\vec{A}}}_{2}} \right|=5\] \[\,\left| {{{\vec{A}}}_{1}}+{{{\vec{A}}}_{2}} \right|=\sqrt{\,{{\left| {{{\vec{A}}}_{1}} \right|}^{2}}+{{\left| {{{\vec{A}}}_{2}} \right|}^{2}}+2\left| {{{\vec{A}}}_{1}} \right|\left| {{{\vec{A}}}_{2}} \right|\cos \theta }\] \[5=\sqrt{9+25+2\times 3\times 5\cos \theta }\] \[\cos \theta =-\frac{9}{2\times 3\times 5}=-\frac{3}{10}\] \[\left( 2{{{\vec{A}}}_{1}}+3{{{\vec{A}}}_{2}} \right).\left( 3{{{\vec{A}}}_{1}}-2{{{\vec{A}}}_{2}} \right)\] \[=6{{\left| {{{\vec{A}}}_{1}} \right|}^{2}}+9{{\vec{A}}_{1}}.{{\vec{A}}_{2}}-4{{\vec{A}}_{1}}{{\vec{A}}_{2}}-6{{\left| {{{\vec{A}}}_{2}} \right|}^{2}}\] \[54+5\times 3\times 5\left( -\frac{3}{10} \right)-6\times 25\] \[=54-150-\frac{45}{2}=-118.5\]
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