What is non-collinearity? What is the equation of a plane that passes through three non collinear points? The equation of such a plane can be found in Vector form and in Cartesian form. Using the position vectors and the Cartesian product of the vector perpendicular to the plane, the equation of the plane can be found. Refer to the solved example to understand how to perform calculations.

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## Plane Equation Passing Through Three Non Collinear Points

As the name suggests, **non collinear points **refer to those points that do not all lie on the same line. From our knowledge from previous lessons, we know that an infinite number of planes can pass through a given vector that is perpendicular to it but there will always be one and only one plane that is perpendicular to the vector and passes through a given point.

In this lesson, we shall focus on the equation of a plane passing through three non collinear points. We shall look at both – the Vector form and the Cartesian form of equations.

**Browse more Topics Under Three Dimensional Geometry**

- Angle Between a Line and a Plane
- Angle Between Two Lines
- Coplanarity of Two Lines
- Angle Between Two Planes
- Direction Cosines and Direction Ratios of a Line
- Distance Between Parallel Lines
- The Distance Between Two Skew Lines
- Distance of a Point from a Plane
- Equation of a Plane in Normal Form
- Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point
- The Equation of Line for Space
- Equation of Plane Passing Through Three Non Collinear Points
- Intercept Form of the Equation of a Plane
- Plane Passing Through the Intersection of Two Given Planes

### Vector Form

Let us consider three non collinear points – R, S, T. Let the position vectors of these points be \( \vec{a}, \vec{b} \; and \; \vec{c}. \) We know that the product of the vectors \( \vec{RS} \; and \; \vec{RT} \) must be perpendicular to the plane containing the points R, S and T.

*Source: examfear*

Let us also assume that the position vector of the point P is \( \vec{r} \) and that of point R is \(\vec{a}. \) Again, we know that the equation of the plane perpendicular to \( \vec{RS} \times \vec{RT} \) and passing through point P must be

$$ ( \vec{r} – \vec{a} ) . ( \vec{RS} \times \vec{RT} ) = 0 $$

From the figure we can write,

$$ ( \vec{r} – \vec{a} ) . [ ( \vec{b} – \vec{a}) \times ( \vec{c} – \vec{a} ) ] = 0 $$

Note that here \( \vec{b} \) and \( \vec{c}\) are the position vectors of the points S and T respectively. Thus, the above equation can be taken to represent the equation of a plane passing through three non collinear points in Vector form.

We shall now move on to the Cartesian equation.

### Cartesian Form

Let us now consider the coordinates of the three non collinear points as under –

$$ ( R (x_1, y_1, z_1), S (x_2, y_2, z_2), T (x_3, y_3, z_3) $$

Let P be a point on the plane that must contain these points. The position vector of the point P be \( \vec{r} \) and its coordinates are (x, y, z). Now, we can write the vectors as –

$$ \vec{RP} = (x – x_1) \hat{i} + (y – y_1) \hat{j} + (z – z_1) \hat{k} $$

$$ \vec{RS} = (x_2 – x_1) \hat{i} + (y_2 – y_1) \hat{j} + (z_2 – z_1) \hat{k} $$

Also, $$ \vec{RP} = (x_3 – x_1) \hat{i} + (y_3 – y_1) \hat{j} + (z_3 – z_1) \hat{k} $$

We can now use the Vector form to substitute the above vectors. Doing so would give us the advantage of presenting the equation in determinant form as under –

$$ \begin{vmatrix}(x – x_1) & (y – y_1) & (z – z_1) \\ (x_2 – x_1) & (y_2 – y_1) & (z_2 – z_1) \\ (x_3 – x_1) & (y_3 – y_1) & (z_3 – z_1) \end{vmatrix} = 0 $$

Thus the determinant form gives us nothing but the Cartesian equation of a plane passing through three non collinear points.

## Solved Example for You

**Question 1: **Find the equation of the plane in Vector form that passes through the points (1, 1, 0), (1, 2, 1) and (-2, 2, -1).

**Answer: ** We shall first check the determinant of the three points to check for collinearity of the points.

$$ \begin{bmatrix}1 & 1 & 0 \\ 1 & 2 & 1 \\ -2 & 2 & -1 \end{bmatrix} = -8 $$

Since the value of the determinant is not zero, it implies that the points are non collinear. Now, we shall find the equation of the plane in Cartesian from as under –

$$ \begin{vmatrix}(x – 1) & (y – 1) & (z – 0) \\ (1 – 1) & (2 – 1) & (1 – 0) \\ (-2 – 1) & (2 – 1) & (-1 – 0) \end{vmatrix} = 0 $$

$$ \begin{vmatrix}(x – 1) & (y – 1) & z \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} = 0 $$

On solving, we shall get :

(x – 1) (-2) – (y – 1) (-3) + z (-3) = 0

-2x + 3y -3z -1 = 0

2x – 3y + 3z = -1 is the required equation of the plane.

**Question 2: What is the difference between collinear and non-collinear points?**

**Answer: **A different line consists of points T, O, and M, thus those three points will be collinear, however they will not be collinear to points A, B, and C. Moreover, when points are not collinear, we refer to them as non-collinear. So, for instance, points A, T, and O will be non-collinear since no line will be able to pass through the three of them together.

**Question 3: What are the three non-collinear points?**

**Answer:** If any three non-collinear points are defining the circumference of one circle, and one circle only. If the three points are collinear i.e. lying on a straight line, after that the circle that passes through all three will be having a radius of infinity. Thus, there will be no practical circle that may pass through three collinear points.

**Question 4: What will be the locus of three collinear points?**

**Answer:** The locus of a point halfway from three non-collinear points A, B and C refers to the circumcentre of the triangle. Further, the locus of the points will be the point of intersection of the perpendicular bisectors of each of the two points

**Question 5: What are coplanar points?**

**Answer:** Coplanar points refer to the three or more points that are lying in the same plane. Remember that a plane is a flat surface that extends without end in all directions. Moreover, we generally show it in math textbooks as a 4-sided figure