Vector-Algebra Question 402
Question: Let $ \overset{\to }{\mathop{a}},,\overset{\to }{\mathop{b}},,\overset{\to }{\mathop{c}}, $ be non-coplanar vectors and $ \overset{\to }{\mathop{p}},=\frac{\overset{\to }{\mathop{b}},\times \overset{\to }{\mathop{c}},}{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]},,\overset{\to }{\mathop{q}},=\frac{\overset{\to }{\mathop{c}},\times \overset{\to }{\mathop{a}},}{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]},,\overset{\to }{\mathop{r}},=\frac{\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}},}{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]}. $ What is the value of $ (\vec{a}-\vec{b}-\vec{c}).\vec{p}+(\vec{b}-\vec{c}-\vec{a}).\vec{q}+(\vec{c}-\vec{a}-\vec{b}).\vec{r} $ ?
Options:
A) 0
B) -3
C) 3
D) -9
Correct Answer: C $ \therefore (\overset{\to }{\mathop{a}},-\overset{\to }{\mathop{b}},-\overset{\to }{\mathop{c}},).\overset{\to }{\mathop{p}},+(\overset{\to }{\mathop{b}},-\overset{\to }{\mathop{c}},-\overset{\to }{\mathop{a}},).\overset{\to }{\mathop{q}},+(\overset{\to }{\mathop{c}},-\overset{\to }{\mathop{a}},-\overset{\to }{\mathop{b}},).\overset{\to }{\mathop{r}}, $Show Answer
Answer:
Solution:
$ =\frac{a.(\overset{\to }{\mathop{b}},\times \overset{\to }{\mathop{c}},)}{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]}+\frac{\overset{\to }{\mathop{b}},.(\overset{\to }{\mathop{c}},\times \overset{\to }{\mathop{a}},)}{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]}+\frac{\overset{\to }{\mathop{c}},.(\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}},)}{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]} $
[Since $ \overset{\to }{\mathop{b}},.(\overset{\to }{\mathop{b}},\times \overset{\to }{\mathop{c}},)=0, $
$ \overset{\to }{\mathop{c}},.(\overset{\to }{\mathop{b}},\times \overset{\to }{\mathop{c}},),=0,\overset{\to }{\mathop{c}},.(\overset{\to }{\mathop{c}},\times \overset{\to }{\mathop{a}},)=0\overset{\to }{\mathop{a}},.(\overset{\to }{\mathop{c}},\times \overset{\to }{\mathop{a}},)] $
$ =0,\overset{\to }{\mathop{a}},.(\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}},)=0 $ and $ \overset{\to }{\mathop{b}},.(\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}},)=0 $
$ =\frac{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]}{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]}+\frac{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{c}{\mathop{c}},]}{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]}+\frac{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]}{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]}=3 $
BETA
