At the intersection of planes, another plane passing through the line of intersection of these two planes can be expressed through the three-dimensional geometry. The equation of such a plane can be found in Vector form or Cartesian form using additional information such as which point this required plane passes through.

This lesson explains how the equation of the required plane for the intersection of planes can be found. Reference to a solved example will also help you understand how to approach problems on this topic.

**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

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## Intersection of Planes

Two planes can intersect in the three-dimensional space. Imagine two adjacent pages of a book. These two pages are nothing but an intersection of planes, intersecting each other and the line between them is called the line of intersection. A new plane i.e. a third plane can be given to be passing through this line of intersection of planes. We are to find out the equation of this plane.

Let us assume that the equation of the first plane is \( \pi_1\) and that of the second is \(\pi_2\). The equation of our required plane is \(\pi\) and we are to find out this equation itself. This equation is given by â€“

$$ \pi_1 + \lambda \pi_2 = 0Â Â Â Â Â Â ………..Â (1) $$

Now, the equation of the planes can be given in vector form or Cartesian form. We shall explore both these forms in the following sections and see how the equation of the required plane can be found using the given information.

**Vector form**

If the equation of the two planes is given in Vector form â€“

$$ ( \vec{r} – \vec{a_1} ) . \vec{n_1} = 0 $$ and

$$ ( \vec{r} – \vec{a_2} ) . \vec{n_2} = 0 $$

These two equations can alternatively be written as â€“

$$ \vec{r} . \vec{n_1} – \vec{d_1} = 0 $$ and

$$ \vec{r} . \vec{n_2} – \vec{d_2} = 0 $$

So the equation of the required plane by using (1) can be written as â€“

$$ ( \vec{r} . \vec{n_1} – \vec{d_1} ) + \lambda (\vec{r} . \vec{n_2} – \vec{d_2} ) = 0Â Â Â ………..Â (2) $$

i.e. $$ \vec{r} ( \vec{n_1} + \lambda \vec{n_2} ) â€“ ( \vec{d_1} + \lambda \vec{d_2} )Â $$

By substitution, we have â€“

$$ ( x \hat{i} + y \hat{j} + z\hat{k} ) . (\vec{n_1} + \lambda \vec{n_2} ) â€“ ( \vec{d_1} + \lambda \vec{d_2} )Â $$

Usually a point on the third plane will be given to you. You must then substitute the coordinates of the point for x, y and z to find the value of \( \lambda \). Then use this value of \( \lambda \) to get the equation of the plane from (2). This equation will be nothing but the equation of the required plane that is passing through the line of intersection of two planes and a given point.

**Cartesian form**

The Cartesian equations of two planes can also be given when you must find the equation of the third. To be able to solve such problems, let us look at how the equation of the given planes in Cartesian form would look like.

The two equations will be of the form â€“

$$ a x_1 + b y_1 + c z_1 + d_1 = 0 $$

$$ a x_2 + b y_2 + c z_2 + d_2 = 0 $$

Thus the equation of the required plane becomes â€“

$$ (a x_1 + b y_1 + c z_1 + d_1 ) + \lambda (a x_2 + b y_2 + c z_2 + d_2) = 0 $$

If a point on the required plane is given, you must substitute the coordinates for the values of x, y and z to obtain the value of \( \lambda \) in order to then replace it in the above equation. That will give you the required equation. A simple example will make the process clear.

**Solved Example for You**

**Question 1:Â Find the equation of the plane that passes through the intersection of the planes \( \vec{r} . ( \hat{i} + \vec{j} + \vec{k} ) = 6\) and \(\vec{r} . ( 2 \hat{i} + 3 \vec{j} + 4 \vec{k} ) = 5\) and passing through the point (1,1,1).**

**Answer:** You must note that the vector equations of the two planes are given so the equation of the required (third) plane will also be in vector form.

#### Given:

$$ \pi_1 = \vec{r} . ( \hat{i} + \vec{j} + \vec{k} ) = 6 $$

$$ \pi_2 = \vec{r} . ( 2 \hat{i} + 3 \vec{j} + 4 \vec{k} ) = 5 $$

Now, we already know that the equation of the required plane is \( \pi_1 + \lambda \pi_2 = 0 \) i.e.

$$ [ \vec{r} . ( \hat{i} + \vec{j} + \vec{k} ) â€“ 6 ] + \lambda [ \vec{r} . ( 2 \hat{i} + 3 \vec{j} + 4 \vec{k} â€“ 5 ] = 0 $$

$$ \vec{r}. [ ( \hat{i} + \hat{j} + \hat{k} ) + \lambda (2 \hat{i} + 3 \vec{j} + 4 \vec{k} ) â€“ 6 + 5 \lambda = 0 $$

$$ ( x \hat{i} + y \hat{j} + z\hat{k} ) . [ ( \hat{i} + \hat{j} + \hat{k} ) + \lambda (2 \hat{i} + 3 \vec{j} + 4 \vec{k} ) â€“ 6 + 5 \lambda = 0 $$

Now since the required plane passes through (1,1,1), the point must satisfy the equation of the plane. Putting x =1 , y = 1, z = 1 we have â€“

$$ ( \hat{i} + \hat{j} + \hat{k} ) . [ ( \hat{i} + \hat{j} + \hat{k} ) + \lambda (2 \hat{i} + 3 \vec{j} + 4 \vec{k} ) â€“ 6 + 5 \lambda = 0 $$

$$ ( \hat{i} + \hat{j} + \hat{k} ) . [ ( 1 + 2 \lambda ) \hat{i} + ( 1 + 3 \lambda ) \hat{j} + ( 1 + 4 \lambda ) \hat{k} ] â€“ 6 + 5 \lambda = 0 $$

#### Taking the dot product, we have:

$$ 1 + 2 \lambda + 1 + 3 \lambda + 1 + 4 \lambda â€“ 6 + 5 \lambda = 0 $$

$$ \lambda = \frac{3}{14} $$

Replacing the value of $$\lambda$$ in the equation of the required plane gives us â€“

$$ \vec{r}. [ ( \hat{i} + \hat{j} + \hat{k} ) + \frac{3}{14} (2 \hat{i} + 3 \vec{j} + 4 \vec{k} ) â€“ 6 + 5 \frac{3}{14} = 0 $$

$$ \vec{r}. [ (1 + 2. \frac{3}{14} ) \hat{i} + ( 1 + 3 \frac{3}{14} ) \hat{j} + ( 1 + 4 \frac{3}{14} ) \hat{k} â€“ 6 + 5 \frac{3}{14} = 0 $$

$$ \vec{r}. [ \frac{20}{14} \hat{i} + \frac{23}{14} \hat{j} + \frac{26}{14} \hat{k} ] -\frac{69}{14} = 0 $$ or,

$$ \vec{r} . ( 20 \vec{i} + 23 \vec{j} + 23 \vec{k} ) = 69 $$ is the required equation of the plane in Vector form.

Similarly, even if the equations are given in the Cartesian form, all you need to do is use the given point to replace the values of x, y, and z and find the value of \( \lambda \). Using this value, substitute it back again in the original equation, and you will get the equation of the desired plane.

**Question 2: Why does the intersection of two plans take place in a line?**

**Answer:** Two planes are the same or parallel only if their normal vectors happen to be scalar multiples of each other. However, in case the two planes are not parallel, then their intersection takes place, but rather than intersecting at a single point, the set of points where the intersection takes place results in the formation of a line.

**Question 3: What is meant by the intersection of two planes?**

**Answer:** The intersection of two planes is referred to as a line. This is due to the fact that planes are two-dimensional flat surfaces.

**Question 4: Explain why the intersection of two planes is always a line?**

**Answer:** The intersection of two planes always happens to be a line because if two points lie in a plane, then the entire line that involves those points shall in that plane.

**Question 5: Explain the intersection of two lines?**

**Answer:** The point where the intersection of the lines takes place is known as the point of intersection. If the production of angles is such that they are all right angles, the lines are perpendicular lines.

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