 # Coplanar Vectors

Coplanar vectors are the vectors which lie on the same plane, in a three-dimensional space. These are vectors which are parallel to the same plane. We can always find in a plane any two random vectors, which are coplanar. Also learn, coplanarity of two lines in a three dimensional space, represented in vector form.

## Conditions for Coplanar vectors

• If there are three vectors in a 3d-space and their scalar triple product is zero, then these three vectors are coplanar.
• If there are three vectors in a 3d-space and they are linearly independent, then these three vectors are coplanar.
• In case of n vectors, if no more than two vectors are linearly independent, then all vectors are coplanar.

A linear combination of vectors v1, …, vn with coefficients a1, …, an is a vector, such that;

a1v1 + … + anvn

A linear combination a1v1 + … + anvn is called trivial if all the coefficients a1, …, an is zero and if at least one of the coefficients is not zero, then it is known as non-trivial.

What are Linearly independent vectors?

The vectors, v1,……vn are linearly independent if no non-trivial combination of these vectors is equal to the zero vector. That means a1v1 + … + anvn = 0 and the coefficients a1= 0 …, an=0.

What are Linearly dependent vectors?

The vectors, v1,……vn are linearly dependent if there exist at least one non-trivial combination of these vectors equal to zero vector.

### Solved Examples

Question 1: Determine whether x = {1; 2; 3}, y = {1; 1; 1}, z = {1; 2; 1} are coplanar vectors.

Solution: To check whether the three vectors x, y and z are coplanar or not, we have to calculate the scalar triple product:

x . [y × z] = (1)·(1)·(1) + (1)·(1)·(2) + (1)·(2)·(3) – (1)·(1)·(3) – (1)·(1)·(2) – (1)·(1)·(2)

= 1 + 2 + 6 – 3 – 2 – 2

= 2

As we can see, the scalar triple product is not equal to zero, hence, vectors x, y and z are not coplanar.

Question 2. If x = {1; 1; 1}, y = {1; 3; 1} and z = {2; 2; 2} are three vectors, then prove that they are coplanar.

Solution:

Calculate a scalar triple product of vectors

x · [y × z] = (1)·(2)·(3) + (1)·(1)·(2) + (1)·(1)·(2) – (1)·(2)·(3) – (1)·(1)·(2) – (1)·(1)·(2)

= 6 + 2 + 2 – 6 – 2 – 2

= 0

As we can see, the scalar triple product is equal to zero, hence vectors x, y and z are coplanar.

1. Tabinda wani

let a=i+j+k and b=i-j+2k and c =xi+(x-2)j-k.if the vector c lies in the plane of a and b,then x equals [AIEEE 2007]
a). 0
b). 1
c). _4
d). _2

i have heard that
,◼️ if a vector c is in the plane of a and b ,then c=a+constant times b
,◼️also their STP=0,
But when ist method is used x equals 0,and when second method is used x equals -2 ???