# Coplanar Vectors – Definition and Example

Here you will learn definition of coplanar vectors with example and test of coplanarity of four points.

Let’s begin –

## Coplanar Vectors

A system of vectors is said to be coplanar, if their supports are parallel to the same plane.

Note : Two vectors are coplanar.

Theorem 1 (Test of Coplanarity of Three vectors)

Let $$\vec{a}$$ and $$\vec{b}$$ be two given non-zero non-collinear vectors. Then, any vector $$\vec{r}$$ coplanar with $$\vec{a}$$ and $$\vec{b}$$ can be uniquely expressed as $$\vec{r}$$ = $$x\vec{r}$$ + $$y\vec{b}$$, for some scalars x and y.

Theorem 2

The necessary and sufficient condition for three vectors $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ to be coplanar is that there exist scalars l, m, n not all zero simultaneously such that $$l\vec{a} + m\vec{b} + n\vec{c}$$ = $$\vec{0}$$.

Theorem 3

If $$\vec{a}$$, $$\vec{b}$$ and $$\vec{c}$$ are three non-zero non-coplanar vector and x, y, z are three scalars, then $$x\vec{a} + y\vec{b} + z\vec{c}$$ = $$\vec{0}$$ $$\implies$$ x = y = z = 0.

Example : Show that the vectors $$\vec{a} – 2\vec{b} + 3\vec{c}$$, $$\vec{a} – 3\vec{b} + 5\vec{c}$$ and $$-2\vec{a} + 3\vec{b} – 4\vec{c}$$ are coplanar, where $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ are non-coplanar vector.

Solution : From Theorem 1,

Three vectors are co-planar if one of the given vectors are expressible as a linear combination of the other two. Let

$$\vec{a} – 2\vec{b} + 3\vec{c}$$ = x$$\vec{a} – 3\vec{b} + 5\vec{c}$$ + y$$-2\vec{a} + 3\vec{b} – 4\vec{c}$$ for some scalars x and y.

$$-2\vec{a} + 3\vec{b} – 4\vec{c}$$ = (x – 2y)$$\vec{a}$$ + (-3x + 3y)$$\vec{b}$$ + (5x – 4y)$$\vec{c}$$

$$\implies$$ 1 = x – 2y, -2 = -3x + 3y and 3 = 5x – 4y

Solving first two of these equation, we get

x = 1/3, y = -1/3.

Clearly, these values of x and y satisfy the third equation.

Hence, the given vectors are co-planar.

## Test of Coplanarity of Four Points

Four points with position vectors $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ and $$\vec{d}$$ are co-planar iff there exist scalars x, y z, u not allzero such that $$x\vec{a} + y\vec{b} + z\vec{c} + u\vec{d}$$ = $$\vec{0}$$, where x + y + z + u =0.