A) 0
B) \[\frac{1}{3}\]
C) \[-\frac{1}{3}\]
D) \[-\frac{1}{2}\]
Correct Answer: D
Solution :
[d] \[|\overset{\to }{\mathop{p}}\,+\overset{\to }{\mathop{q}}\,|=\sqrt{3}\Rightarrow \overset{\to }{\mathop{{{p}^{2}}}}\,+\overset{\to }{\mathop{{{q}^{2}}}}\,+2\overset{\to }{\mathop{p}}\,\overset{\to }{\mathop{q}}\,=3\] Since \[\overset{\to }{\mathop{p}}\,\] and \[\overset{\to }{\mathop{q}}\,\] are unit vectors So, \[1+1+2pq=3\] \[\Rightarrow 2pq=1\Rightarrow pq=\frac{1}{2}\] \[(2\overset{\to }{\mathop{p}}\,-3\overset{\to }{\mathop{q}}\,)(3\overset{\to }{\mathop{p}}\,+\overset{\to }{\mathop{q}}\,)=6\overset{\to }{\mathop{{{p}^{2}}}}\,+2\overset{\to }{\mathop{p}}\,\overset{\to }{\mathop{q}}\,-9\overset{\to }{\mathop{q}}\,\overset{\to }{\mathop{p}}\,-3\overset{\to }{\mathop{{{q}^{2}}}}\,=\frac{-1}{2}\]
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