Vector Algebra Question 62
Question: Let $ \overset{\to }{\mathop{p}},,\overset{\to }{\mathop{q}},,\overset{\to }{\mathop{r}}, $ be three mutually perpendicular vectors of the same magnitude. If a vector $ \vec{x} $ satisfies the equation $ \overset{\to }{\mathop{p}},\times {(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{q}},)\times \overset{\to }{\mathop{p}},}+\overset{\to }{\mathop{q}},\times {(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{r}},))\times \overset{\to }{\mathop{q}},} $ $ +\overset{\to }{\mathop{r}},\times {(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{p}},)\times \overset{\to }{\mathop{r}},}=\overset{\to }{\mathop{0}}, $ then $ \overset{\to }{\mathop{x}}, $ is given by
Options:
A) $ \frac{1}{2}(\overset{\to }{\mathop{p}},+\overset{\to }{\mathop{q}},-2\overset{\to }{\mathop{r}},) $
B) $ \frac{1}{2}(\overset{\to }{\mathop{p}},+\overset{\to }{\mathop{q}},+\overset{\to }{\mathop{r}},) $
C) $ \frac{1}{3}(\overset{\to }{\mathop{p}},+\overset{\to }{\mathop{q}},+\overset{\to }{\mathop{r}},) $
D) $ \frac{1}{3}(2\overset{\to }{\mathop{p}},+\overset{\to }{\mathop{q}},-\overset{\to }{\mathop{r}},) $
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Answer:
Correct Answer: B
Solution:
- [b] Let $ |\overset{\to }{\mathop{p}},|=|\overset{\to }{\mathop{q}},|=|\overset{\to }{\mathop{r}},|=k $
Let $ \hat{p},\hat{q},\hat{r} $ be unit vectors along $ \overset{\to }{\mathop{p}},,\overset{\to }{\mathop{q}},,\overset{\to }{\mathop{r}}, $ respectively. Clearly $ \hat{p},\hat{q},\hat{r} $ are mutually perpendicular vectors, so any vector $ \overset{\to }{\mathop{x}}, $ can be written as $ a_1\hat{p}+a_2\hat{q}+a_3\hat{r} $ .
$ \therefore \overset{\to }{\mathop{p}},\times {(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{q}},)\times \overset{\to }{\mathop{p}},}=(\overset{\to }{\mathop{p}},.\overset{\to }{\mathop{p}},)(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{q}},)-{\overset{\to }{\mathop{p}},.(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{q}},)}\overset{\to }{\mathop{p}}, $
$ =k^{2}(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{q}},)-(\overset{\to }{\mathop{p}},.\overset{\to }{\mathop{x}},)\overset{\to }{\mathop{p}},[\because ,\overset{\to }{\mathop{p}},.\overset{\to }{\mathop{q}},=0] $
$ =k^{2}(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{q}},)-k\hat{p}.(a_1\hat{p}+a_2\hat{q}+a_3\hat{r})k\hat{p} $
$ =k^{2}(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{q}},-a_1\hat{p}) $
Similarly, $ \overset{\to }{\mathop{q}},\times {(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{r}},)\times \overset{\to }{\mathop{q}},}=k^{2}(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{r}},-a_2\hat{q}) $
and $ \overset{\to }{\mathop{r}},\times {(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{p}},)\times \overset{\to }{\mathop{r}},}=k^{2}(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{p}},-a_3\hat{r}) $
According to the given condition
$ k^{2}(\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{q}},-a_1\hat{p}+\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{r}},-a_2\hat{q}+\overset{\to }{\mathop{x}},-\overset{\to }{\mathop{p}},-a_3\hat{r})=0 $
$ \Rightarrow k^{2}{3\overset{\to }{\mathop{x}},-(\overset{\to }{\mathop{p}},+\overset{\to }{\mathop{q}},+\overset{\to }{\mathop{r}},)-(a_1\hat{p}+a_2\hat{q}+a_3\hat{r})}=0 $
$ \Rightarrow k^{2}[2\overset{\to }{\mathop{x}},-(\overset{\to }{\mathop{p}},+\overset{\to }{\mathop{q}},+\overset{\to }{\mathop{r}},)]=\overset{\to }{\mathop{0}}, $
$ \Rightarrow \overset{\to }{\mathop{x}},=\frac{1}{2}(\overset{\to }{\mathop{p}},+\overset{\to }{\mathop{q}},+\overset{\to }{\mathop{r}},)[\because ,k\ne 0] $
BETA
