A) \[r.[b\times c+c\times a+a\times b]\]
B) \[\frac{1}{2}r.\left( a+b+c \right)\]
C) 2r. (a + b + c)
D) 4
Correct Answer: B
Solution :
Now, \[r.a=\alpha \left( a.b\times c \right)+\beta \left( a.c\times a \right)+\gamma \left( a.a\times b \right)\] \[=\alpha \left[ abc \right]+0+0\] Similarly \[\text{r}\text{.b= }\!\!\beta\!\!\text{ }\left[ \text{abc} \right]\text{ and r}\text{.c= }\!\!\gamma\!\!\text{ }\left[ \text{abc} \right]\] \[\therefore \] \[\frac{1}{2}r.\left( a+b+c \right)=\frac{1}{2}\left( r.a+r.b+r.c \right)\] \[=\frac{1}{2}\left( \alpha +\beta +\gamma \right)\left[ abc \right]\] \[=\frac{1}{2}\left( \alpha +\beta +\gamma \right)\times 2=\alpha +\beta +\gamma \]
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