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) \[\frac{1}{3}(2\overset{\to }{\mathop{p}}\,+\overset{\to }{\mathop{q}}\,-\overset{\to }{\mathop{r}}\,)\]
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 |
| \[{{k}^{2}}(\overset{\to }{\mathop{x}}\,-\overset{\to }{\mathop{q}}\,-{{a}_{1}}\hat{p}+\overset{\to }{\mathop{x}}\,-\overset{\to }{\mathop{r}}\,-{{a}_{2}}\hat{q}+\overset{\to }{\mathop{x}}\,-\overset{\to }{\mathop{p}}\,-{{a}_{3}}\hat{r})=0\] |
| \[\Rightarrow {{k}^{2}}\{3\overset{\to }{\mathop{x}}\,-(\overset{\to }{\mathop{p}}\,+\overset{\to }{\mathop{q}}\,+\overset{\to }{\mathop{r}}\,)-({{a}_{1}}\hat{p}+{{a}_{2}}\hat{q}+{{a}_{3}}\hat{r})\}=0\] |
| \[\Rightarrow {{k}^{2}}[2\overset{\to }{\mathop{x}}\,-(\overset{\to }{\mathop{p}}\,+\overset{\to }{\mathop{q}}\,+\overset{\to }{\mathop{r}}\,)]=\overset{\to }{\mathop{0}}\,\] |
| \[\Rightarrow \overset{\to }{\mathop{x}}\,=\frac{1}{2}(\overset{\to }{\mathop{p}}\,+\overset{\to }{\mathop{q}}\,+\overset{\to }{\mathop{r}}\,)[\because \,k\ne 0]\] |
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