Vector Algebra Question 17
Question: If $\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vectors such that $\vec{a}+\vec{b}+\vec{c}=\alpha \vec{d}$ and $\vec{b}+\vec{c}+\vec{d}=\beta \vec{a}$, then $\vec{a}+\vec{b}+\vec{c}+\vec{d}$ is equal to
Options:
A) $\vec{0}$
B) $\alpha \vec{a}$
C) $\beta \vec{b}$
D) $(\alpha+\beta) \vec{c}$
Show Answer
Answer:
Correct Answer: A
Solution:
- We have $ \mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha \mathbf{d} $ and $ \mathbf{b}+\mathbf{c}+\mathbf{d}=\beta \mathbf{a} $
$ \therefore \mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=(\alpha +1)\mathbf{d} $ and
$ \mathbf{a+\mathbf{b}+\mathbf{c}+\mathbf{d}=(\beta +1)\mathbf{a}$
$ \Rightarrow (\alpha +1)\mathbf{d}=(\beta +1)\mathbf{a} $ If $ \alpha \ne -1, $ then $ (\alpha +1)\mathbf{d}=(\beta +1)\mathbf{a}\Rightarrow \mathbf{d}=\frac{\beta +1}{\alpha +1}\mathbf{a} $
$ \Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha \mathbf{d}\Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha ( \frac{\beta +1}{\alpha +1} )\mathbf{a} $
$ \Rightarrow ( 1-\frac{\alpha (\beta +1)}{\alpha +1} )\mathbf{a}+\mathbf{b}+\mathbf{c}=0 $
$ \Rightarrow \mathbf{a},\mathbf{b},\mathbf{c} $ are coplanar which is contradiction to the given condition,
$ \therefore \alpha =-1 $ and so $ \mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=0. $
BETA
