A) \[\sqrt{54}/4\]
B) \[\sqrt{55}/8\]
C) \[\sqrt{3}/16\]
D) \[\sqrt{247}/16\]
Correct Answer: B
Solution :
\[(3\vec{p}+\vec{q}).(5\vec{p}-3\vec{q})=0\] Or \[15{{\vec{p}}^{2}}-3{{\vec{p}}^{2}}=4\vec{p}.\vec{q}\] ? \[(2\vec{p}+\vec{q}).(4\vec{p}-2\vec{q})=0\] Or \[8{{\vec{p}}^{2}}=2{{\vec{q}}^{2}}\Rightarrow \,{{\vec{q}}^{2}}=4{{\vec{p}}^{2}}\]? Now \[\cos \theta =\frac{\vec{p}.\vec{q}}{|\vec{p}||\vec{q}|};\] substituting \[{{\vec{q}}^{2}}=4{{\vec{p}}^{2}}\] in , \[\therefore \,\,3{{\vec{p}}^{2}}=4\vec{p}.\vec{q}\] \[\cos \theta =\frac{3}{4}.\,\frac{{{{\vec{p}}}^{2}}}{|\vec{p}|\,2|\vec{p}|}\,=\frac{3}{8}\] \[\Rightarrow \,\,\sin \theta \,=\frac{\sqrt{55}}{8}\]
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