Coplanar Vectors

Coplanar vectors are vectors that lie within the same plane in a three-dimensional space. This means that these vectors are all parallel to that particular plane. In other words, coplanar vectors do not span across different planes but instead remain confined to a single plane. This characteristic is crucial in determining the spatial relationships between vectors, making it an essential concept in fields such as physics, engineering, and computer graphics. Understanding and identifying coplanar vectors helps in analyzing and solving various problems involving vector geometry.

1.0What are Coplanar Vectors

In vector algebra and geometry, vectors are said to be coplanar if they lie within the same plane. This means that any linear combination of these vectors will also lie in that plane. More formally, vectors $\vec{a},\vec{b}and\vec{c}$ are coplanar if there exist scalars x, y, and z, not all zero, such that: 

$x\vec{a}+y\vec{b}+z\vec{c}=0\Rightarrow \left[\begin{array}{lll}\vec{a}& \vec{b}& \vec{c}\end{array}\right]=0$

In simpler terms, vectors are coplanar if they can be expressed as linear combinations of each other and their resultant lies in a single plane. 

For two vectors, $\vec{a}$ and $\vec{b}$ to be coplanar with a third vector, $\vec{c}$ the scalar triple product of the vectors must be zero:

$\overrightarrow{\mathrm{a}}\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=0$

If this condition is met, it indicates that the three vectors do not span a three-dimensional space but rather lie within a two-dimensional plane.

2.0Conditions for Coplanar vectors

• If there are three vectors in 3D space and their scalar triple product is zero, then these vectors are coplanar.
• If there are three vectors in 3D space and they are linearly dependent, then these vectors are coplanar.
• For n vectors, if at most two of them are linearly independent, then all the vectors are coplanar.

3.0How Can Two Vectors Be Coplanar?

Two vectors are inherently coplanar because any two vectors in three-dimensional space always lie on a plane. To understand why this is the case, consider the following:

When you have two vectors, $\vec{a}$ and $\vec{b}$ you can always find a plane that contains both vectors. This is because you can draw both vectors originating from the same point (the origin or any other point). The plane containing these two vectors is defined by the linear combinations of $\vec{a}$ and $\vec{b}$. Mathematically, any vector $\vec{c}$ that lies in this plane can be expressed as:

$\overrightarrow{\mathrm{c}}=x\overrightarrow{\mathrm{a}}+y\overrightarrow{\mathrm{b}}$

where x and y are scalar coefficients. This linear combination shows that $\vec{c}$ , and therefore any point on the plane, is a sum of scaled versions of $\vec{a}$ and $\vec{b}$.

For two vectors to lie in the same plane, no additional conditions are needed beyond their mere existence, as any two vectors will always define a plane. This property is fundamental and makes the concept of coplanarity straightforward when dealing with only two vectors.

4.0Conditions for Coplanar vectors

For vectors to be coplanar, they must lie within the same plane in three-dimensional space. Here are the conditions that need to be met for vectors to be considered coplanar:

1. Two Vectors:

 Any two vectors are always coplanar. This is because two vectors define a plane, and hence, they lie on the same plane by default.

2. Three or More Vectors

For three or more vectors to be coplanar, a specific condition must be satisfied. Three vectors $\vec{a}$ , $\vec{b}$  and $\vec{c}$  are coplanar if the scalar triple product is zero:

$\vec{a}\cdot (\vec{b}\times \vec{c})=0$

The scalar triple product calculates the volume of the parallelepiped formed by the vectors. If this volume is zero, it means the vectors lie in the same plane.

3. Linear Dependence:

Three vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ are coplanar if one of them can be expressed as a linear combination of the other two. For instance:

$\vec{c}=x\vec{a}+y\vec{b}$

Here, x and y are scalars. This relationship indicates that $\vec{c}$ lies in the plane formed by $\vec{a}$ and $\vec{b}$.

4. Determinant Condition:

The vectors $\vec{a}$ ,$\vec{b}$  and $\vec{c}$ can also be tested for coplanarity using the determinant of a matrix composed of these vectors. They are coplanar if the determinant of the matrix is zero:

$\left|\begin{array}{lll}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=0$

_where 

$\vec{a}={\mathrm{a}}_{1\hat{i}}+{\mathrm{a}}_{2\hat{j}}+{\mathrm{a}}_3\hat{k},\vec{b}={\mathrm{b}}_{1\hat{i}}+{\mathrm{b}}_{2\hat{j}}+{\mathrm{b}}_{3\hat{k}}\text{ and }\vec{c}={\mathrm{c}}_{1\hat{i}}+{\mathrm{c}}_2\hat{j}+{\mathrm{c}}_{3\hat{k}}$

This determinant condition provides a clear and calculable method to verify the coplanarity of three vectors.

A given number of vectors are called coplanar if their line of support are all parallel to the same plane. Note that “Two Free Vectors Are Always Coplanar”.

5.0Coplanar Vectors Solved Examples

Example 1: If  $\vec{a}-\hat{i}-\hat{j}+\hat{k},\vec{b}=\hat{i}+2\hat{j}-\hat{k}\text{ and }\vec{c}=3\hat{i}+p\hat{j}+5\hat{k}$  are coplanar then the value of p will be  

(A) –6 (B) –2 (C) 2 (D) 6

Ans. (A)

Solution: 

Since $\vec{a},\vec{b},\vec{c}$ are coplanar vectors 

$\left|\begin{array}{ccc}1& -1& 1\\ 1& 2& -1\\ 3& p& 5\end{array}\right|=0\Rightarrow p=-6$

Example 2: Let $\vec{a},\vec{b},\vec{c}$ be three vectors such that $\vec{a}\cdot \vec{a}=\vec{b}\cdot \vec{b}=\vec{c}\cdot \vec{c}=3and|\vec{a}+\vec{b}-\vec{c}{|}^2+|\vec{b}+\vec{c}-\vec{a}{|}^2+|\vec{c}+\vec{a}-\vec{b}{|}^2=36$, then

(A) $\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}=\frac{-9}{2}$

(B)$\vec{a},\vec{b},\vec{c}$ are coplanar vectors

(C) $\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}=\frac{-27}{2}$

(D) None of these

Ans. (A, B)

Solution:

$|\vec{a}+\vec{b}-\vec{c}{|}^2+|\vec{b}+\vec{c}-\vec{a}{|}^2+|\vec{c}+\vec{a}-\vec{b}{|}^2=36$

⇒ $3\left(|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2\right)-2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=36$

⇒  $\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}=-\frac{9}{2}$

Also $|\vec{a}+\vec{b}+\vec{c}{|}^2=|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=0$

⇒$\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$

⇒ $\vec{c}=-\vec{a}-\vec{b}$

Hence $\vec{a},\vec{b},\vec{c}$ are coplanar and represent sides of a triangle.

Example 3: Show that the vector$\vec{x}=4\hat{i}-6\hat{j}-2\hat{k},\vec{y}=-\hat{i}+4\hat{j}+3\hat{k}\text{ and }\vec{z}=-8\hat{i}-\hat{j}+3\hat{k}$ are coplanar.

Solution: For three vectors,$\overrightarrow{\mathrm{x}},\overrightarrow{\mathrm{y}},\overrightarrow{\mathrm{z}}$ to be coplanar, if

$\left|\begin{array}{ccc}-4& -6& -2\\ -1& 4& 3\\ -8& -1& 3\end{array}\right|=0$

L.H.S = –4(12 + 3) + 6 (–3 + 24) –2 (1 + 32)

= –4(15) + 6(21) –2(33)

= – 60 + 126 – 66 = 0 = RHS.

Example 4: How do you verify coplanarity using determinants?

Solution: You can verify if vectors $\vec{a},\vec{b}and\vec{c}$ are coplanar by checking if the determinant of the matrix formed by these vectors equals zero:

$\left|\begin{array}{lll}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=0$

$where\vec{a}={\mathrm{a}}_{1\hat{i}}+{\mathrm{a}}_{2\hat{j}}+{\mathrm{a}}_3\hat{k},\vec{b}={\mathrm{b}}_{1\hat{i}}+{\mathrm{b}}_{2\hat{j}}+{\mathrm{b}}_{3\hat{k}}\text{ and }\vec{c}={\mathrm{c}}_{1\hat{i}}+{\mathrm{c}}_{2\hat{j}}+{\mathrm{c}}_3\hat{k}\text{. }$

If the determinant is zero, the vectors are coplanar; otherwise, they are not.

6.0Coplanar Vectors Practice Problems

1. Determine whether$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}},\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{z}}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}+1$ are coplanar vectors.
2. If$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}},\overrightarrow{\mathrm{y}}=\hat{\mathrm{i}}+3\hat{\mathrm{j}}+\hat{\mathrm{k}}\text{ and }\vec{z}=2\hat{i}+2\hat{j}+2\hat{\mathrm{k}}$ are three vectors, then prove that they are coplanar.