A) 6
B) 10
C) 8
D) 4
Correct Answer: D
Solution :
| We know,\[{{\left[ a\,b\,c \right]}^{2}}=\left[ \begin{matrix} a.a & a.b & a.c \\ b.a & b.b & b.c \\ c.a & c.b & c.c \\ \end{matrix} \right]\] |
| \[{{\left[ a\,b\,c \right]}^{2}}=\left[ \begin{matrix} 1 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 \\ \end{matrix} \right]\] |
| \[{{\left[ a\,b\,c \right]}^{2}}=\frac{5}{4}-\frac{3}{4}=\frac{1}{2}\] |
| \[\left[ a\,b\,c \right]=\frac{1}{\sqrt{2}}\] |
| As given \[a\times b+b\times c=pa+qb+rc\]Take dot product with a is |
| \[a.\left( a\times b \right)+a.\left( b\times c \right)\]\[=p{{\left( a \right)}^{2}}+qa..b+r\left( a.c \right)\] |
| \[0+\frac{1}{\sqrt{2}}=p+\frac{q}{2}+\frac{r}{2}\] ?? (i) |
| Similarly take dot product with b and c |
| \[0=\frac{p}{2}+q+\frac{r}{2}\] ?.. (ii) |
| \[\frac{1}{\sqrt{2}}=\frac{p}{q}+\frac{q}{2}+r\] ?.. (iii) |
| From Eqs. (i) and (iii), we get \[P=r\]and \[p+q=0\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\frac{{{p}^{2}}+2{{q}^{2}}+{{r}^{2}}}{{{q}^{2}}}=\frac{{{p}^{2}}+2{{p}^{2}}+{{p}^{2}}}{{{p}^{4}}}=4\] |
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