- #1
mathmari
Gold Member
MHB
- 5,049
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Hey!
We want to show that if $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are on the same plane, then there are $A, B, C$ not all $0$ such that $A \vec a+B \vec b+C \vec c=\vec 0$.
The solution is the following:
If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are on the same plane, then
We want to show that if $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are on the same plane, then there are $A, B, C$ not all $0$ such that $A \vec a+B \vec b+C \vec c=\vec 0$.
The solution is the following:
If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are on the same plane, then
- all are equal to $\overrightarrow{0}$ and then $1 \cdot \overrightarrow{a} + 1 \cdot \overrightarrow{b}+ 1 \cdot \overrightarrow{c}=\overrightarrow{0}$
- all are on the same line and one is not the $\overrightarrow{0}$
for example, $\overrightarrow{a} \neq \overrightarrow{0}$, so $\overrightarrow{b}=m \overrightarrow{a}, \overrightarrow{c}=n\overrightarrow{a}$ and then $-(m+n)\overrightarrow{a}+1\cdot \overrightarrow{b}+1 \overrightarrow{c}=\overrightarrow{0}$ - two are not the $\overrightarrow{0}$ and don't belong to the same line
for example, the $\overrightarrow{b}, \overrightarrow{c}$, then $\overrightarrow{a}=x \overrightarrow{b}+y \overrightarrow{c}$ and then $1 \cdot \overrightarrow{a}+(-x) \overrightarrow{b}+(-y) \overrightarrow{c}=\overrightarrow{0}$.