Vector Algebra Question 85
Question: Let $ \overset{\to }{\mathop{a}},,\overset{\to }{\mathop{b}}, $ and $ \overset{\to }{\mathop{c}}, $ be three non-coplanar vectors, and let $ \overset{\to }{\mathop{p}},,\overset{\to }{\mathop{q}}, $ and $ \overset{\to }{\mathop{r}}, $ be the vectors defined by the relations $ \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}},]} $ and $ \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}},]}. $ Then the value of the expression $ (\overset{\to }{\mathop{a}},+\overset{\to }{\mathop{b}},).\overset{\to }{\mathop{p}},+(\overset{\to }{\mathop{b}},+\overset{\to }{\mathop{c}},).\overset{\to }{\mathop{q}},+(\overset{\to }{\mathop{c}},+\overset{\to }{\mathop{a}},).\overset{\to }{\mathop{r}}, $ is equal to
Options:
A) 0
B) 1
C) 2
D) 3
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] $ \overset{\to }{\mathop{a}},.\overset{\to }{\mathop{p}},=\frac{\overset{\to }{\mathop{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{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]}{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]}=1=\overset{\to }{\mathop{b}},.\overset{\to }{\mathop{q}},=\overset{\to }{\mathop{c}},.\overset{\to }{\mathop{r}}, $ $ \overset{\to }{\mathop{b}},.\overset{\to }{\mathop{p}},=\frac{\overset{\to }{\mathop{b}},.(\overset{\to }{\mathop{b}},\times \overset{\to }{\mathop{c}},)}{[\overset{\to }{\mathop{a}}\overset{\to }{\mathop{b}}\overset{\to }{\mathop{c}},]}=\frac{0}{[\overset{\to }{\mathop{a}},\overset{\to }{\mathop{b}},\overset{\to }{\mathop{c}},]}=0=\overset{\to }{\mathop{c}},.\overset{\to }{\mathop{p}},=\overset{\to }{\mathop{a}},.\overset{\to }{\mathop{r}}, $ Therefore, the given expression is equal to $ 1+0+1+0+1+0=3 $ .
BETA
