Vector Algebra Question 64
Question: If $ \vec{p} $ and $ \vec{q} $ are non-collinear unit vectors and $ | \vec{p}+\vec{q} |=\sqrt{3} $ , then $ (2\vec{p}-3\vec{q})\cdot (3\vec{p}+\vec{q}) $ is equal to
Options:
A) 0
B) $ \frac{1}{3} $
C) $ -\frac{1}{3} $
D) $ -\frac{1}{2} $
Show Answer
Answer:
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} $
BETA
