| Direction: Let a, b and c be three vectors such that \[|a|\,=\,|b|\,=\,|\,c|\,=4\]and the angle between a and b is \[\pi /3,\] the angle between b and c is \[\pi /3,\] and angle between c and a is \[\pi /3,\]. Then, |
A) \[24\sqrt{2}\]
B) \[24\sqrt{3}\]
C) \[32\sqrt{2}\]
D) \[32\sqrt{3}\]
Correct Answer: C
Solution :
\[\because \] \[[a\,b\,c]\,[u\,v\,w]=\,\left| \begin{matrix} a\cdot u & b\cdot u & c\cdot u \\ a\cdot v & b\cdot v & c\cdot v \\ a\cdot w & b\cdot w & c\cdot w \\ \end{matrix} \right|\] \[\therefore \] \[{{[a\,b\,c]}^{2}}=\,[a\,b\,c]\,[a\,b\,c]\] \[=\,\left| \begin{matrix} a\cdot a & b\cdot a & c\cdot a \\ a\cdot b & b\cdot b & c\cdot b \\ a\cdot c & b\cdot c & c\cdot c \\ \end{matrix} \right|\] Now, \[a\cdot \,a={{a}^{2}}=|a{{|}^{2}}=16\] \[a\cdot b=b\cdot a=|a|\,|b|\,\cos \,\frac{\pi }{3}\] \[=4\times 4\times \frac{1}{2}=8\] \[a\cdot c=c\cdot a\,=|a|\,|c|\,\cos \,\frac{\pi }{3}\] \[=4\times 4\times \frac{1}{2}=8\] \[b\cdot b={{b}^{2}}=\,|b{{|}^{2}}=16\] \[b\cdot c=c\cdot b=|b|\,|c|\,\cos \,\pi /3\] \[=4\times 4\times \frac{1}{2}=8\] \[\therefore \] \[c\cdot c=|c{{|}^{2}}={{4}^{2}}=16\] Now, \[{{[a\,b\,c]}^{2}}=\,\left| \begin{matrix} 16 & 8 & 8 \\ 8 & 16 & 8 \\ 8 & 8 & 16 \\ \end{matrix} \right|\] \[={{8}^{3}}\,\left| \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 1 & 2 \\ \end{matrix} \right|\] \[={{8}^{3}}\times 4=64\times 32\] \[\therefore \] Volume \[=\,[\,a\,b\,c]\,=32\sqrt{2}\]
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