Coplanar Vectors: Condition, Examples, Solved Problems

Coplanar vectors lie on the same plane. In this page , we learn about coplanar vectors, conditions to be coplanar along with some solved and practice problems.

Coplanar Vectors

Definition: A set of vectors are said to be coplanar if they lie on the same plane. In other words, any linear combination of this set of vectors will also lie on that plane, that is, this set is linearly dependent. More precisely, if $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar then there are scalars x₁, x₂, x₃ (not all zero) such that

$x_1\vec{a}+x_2\vec{b}+x_3\vec{c}=\vec{0}$.

Condition to be Coplanar Vectors

For three vectors to be coplanar, their scalar triple product must be zero. This means that if three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar, then we must have $[\vec{a} \vec{b} \vec{c}]$ $=\vec{a} \cdot (\vec{b} \times \vec{c})=0$.

Conditions for Coplanarity: $\vec{a} \cdot (\vec{b} \times \vec{c})=0$.

So the three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar if either of the following happens

1. the scalar triple product is 0, that is, $\vec{a} \cdot (\vec{b} \times \vec{c})=0$.
2. the three vectors are linearly dependent, that is, we can express $\vec{c}$ as a linear combination of $\vec{a}$ and $\vec{b}$. So $\vec{c}=x\vec{a}+y\vec{b}$ for scalars x and y, not zero together.

Solved Problems

Question 1: Find the value of $m$ for which the vectors $\vec{a}=2\hat{i}+\hat{j}+3\hat{k}$, $\vec{b}=4\hat{i}+m\hat{j}+\hat{k}$ and $\vec{c}=2\hat{i}-\hat{j}+2\hat{k}$ are coplanar.

Answer:

As the three vectors are coplanar, so we must have that $\vec{a} \cdot (\vec{b} \times \vec{c})=0$. In other words,

$\begin{vmatrix} 2 & 1 & 3 \\ 4 & m & 1 \\ 2 & -1 & 2 \end{vmatrix} = 0$

Expanding along the first row, we obtain that

2(2m + 1) – 1(8 – 2) + 3(-4 – 2m) = 0

⇒ 4m+2 -6 -12-6m = 0

⇒ 2p = 2-6-12 = -16

⇒ m = -16/2 = -8.

So the value of m is -8.

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