Question
Prove that a necessary and sufficient condition for three vectors $\vec{\text{a}},\ \vec{\text{b}}$ and $\vec{\text{c}}$ to be coplanar is that there exist scalars l, m, n not all zero simultaneously such that $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=\vec0$.

Answer

Necessary Condition: Let $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are three coplanar vectors. Then one of them can be expressed as the linear combination of other two vectors. Let, $\vec{\text{c}}=\text{x}\vec{\text{a}}+\text{y}\vec{\text{b}}$ $\text{x}\vec{\text{a}}+\text{y}\vec{\text{b}}-\vec{\text{c}}=0$ Put $\text{x}=1,\ \text{y}=\text{m},\ (-1)=\text{n}$$\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=0$
Thus, if $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors, then there exist scalars l, m, n $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=0$ Such that l, m, n are not all zero simultaneously. Sufficient Condition: Let $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ be three vectors such that there exist scalars l, m, n not all zero simultaneously satisfying $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=0$ $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=0$ $\text{n}\vec{\text{c}}=-\text{l}\vec{\text{a}}-\text{m}\vec{\text{b}}$ Dividing by n, both the sides $\frac{\text{n}\vec{\text{c}}}{\text{n}}=\frac{-\text{l}\vec{\text{a}}}{\text{n}}-\frac{\text{m}\vec{\text{b}}}{\text{n}}$ $\vec{\text{c}}=\Big(-\frac{\text{l}}{\text{n}}\Big)\vec{\text{a}}+\Big(-\frac{\text{m}}{\text{n}}\Big)\vec{\text{b}}$ $\vec{\text{c}}$ is a linear combination of $\vec{\text{a}} \text{ and }\vec{\text{b}}$ Hence, $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors.

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