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.

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