Vector Algebra Question 48
Question: If $ \vec{r}\cdot \vec{a}=\vec{r}\cdot b=\vec{r}\cdot \vec{c}=\frac{1}{2} $ for some non-zero vector $ \vec{r} $ , then the area of the triangle whose vertices are $ A(\vec{a}),B(\vec{b}) $ and $ C( {\vec{c}} ) $ is ( $ \vec{a},\vec{b},\vec{c} $ are non-coplanar)
Options:
A) $ | [\vec{a},\vec{b},\vec{c}] | $
B) $ | {\vec{r}} | $
C) $ | [\vec{a},\vec{b},\vec{c}]\vec{r} | $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Any vector $ \vec{r} $ can be represented in terms of three non-coplanar vectors $ \vec{a},,\vec{b} $ and $ \vec{c} $ as $ \vec{r}=x(\vec{a}\times \vec{b})+y(\vec{b}\times \vec{c})+z(\vec{c}+\vec{a}) $ ?(i) Taking dot product with $ \vec{a},\vec{b} $ and $ \vec{c} $ , respectively we have $ x=\frac{\vec{r}\cdot \vec{c}}{[\vec{a},\vec{b},\vec{c}]},y=\frac{\vec{r}\cdot \vec{a}}{[\vec{a},\vec{b},\vec{c}]} $ and $ z=\frac{\vec{r}\cdot \vec{b}}{[\vec{a},\vec{b},\vec{c}]} $ From (i), we have $ [\vec{a}\vec{b}\vec{c}]\vec{r}=\frac{1}{2}(\vec{a}\times \vec{b}\times \vec{c}+\vec{c}\times \vec{a}) $
$ \therefore $ Area of $ \Delta ABC=\frac{1}{2}|\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}| $ $ =|[\vec{a}\vec{b},\vec{c}]\vec{r}| $
BETA
