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.
Also, read:
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.
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 ???
please help me sorting this confusion.
Use the below formula:
u,v,w are coplanar if there exist two real numbers a and b
w=au+bv
The answer will be x = -2