Lesson Explainer: Intersection of Planes Mathematics

Example 1: Finding the Parametric Equation of the Line of Intersection of Two Planes

Find the parametric equations of the line of intersection between the two planes and .

Answer

One approach to solving this question is to choose a parametric equation to represent one of our variables. We can do this as we do, in fact, have a “free variable.” In terms of how we go about choosing the parametric equation for the variable, we can do this in a couple of different ways. We can either choose a general parametrization, for example, , and then fix values for and at a stage of the calculation that is convenient, or we can fix the parametrization at the start of our calculation, for example, , and adjust our answer at the end as required. We will demonstrate both methods here.

Method 1: Directly Fixing a Parametrization

If we reference the options presented in the question, it might seem sensible to set as it seems likely that we would then land on the correct answer. However, we will set and then demonstrate that this will in fact give us an equivalent line of intersection, which we can then “tweak” to determine the correct answer from the provided options.

If we substitute our chosen parameter for into the equation for the first plane, we get which gives . If we now substitute and into the equation for our second plane, we get

If we distribute over the parentheses and simplify, we get which simplifies to . Therefore, the parametric equations for , , and are

As we can see from the question, this is not actually one of our options, but it must be equivalent to one of the options. We have described a line that passes through the point with direction vector and need to identify which of the options is equivalent to this form of the line. To do this, we can first compare the direction vectors of each of the lines and then identify which of the points described also lies on our line.

In this particular case, this is not too difficult to do. We can quickly discount options B and E due to the inconsistent signs of and when compared with the direction vector of our line. The remaining options have a direction vector that is a multiple of of our line and is, therefore, equivalent.

We can then discount option A as it passes through the same -coordinate but a different -coordinate, which leaves options C and D. Option C is described by the same point, , so it must be a solution.

Finally, we need to check whether option D is also a solution. We can do this by checking whether is a point on this particular line: substituting the value into each of the parametric equations leads us to the point . Therefore, this is not a valid equation for the line of intersection.

Therefore, the answer is option C.

Method 2: Using a General Parametrization

Recall that the general form for the set of parametric equations for a line in 3D space is given by where is a parameter; , , and are the coordinates of a point lying on the line; and , , and are the components of the direction vector of the line or parallel to the line.

To find the set of parametric equations for the line of intersection, we set an expression for one variable in terms of the parameter, substitute this expression into the equations of the planes, and then rearrange the resulting equations to find expressions for the other two variables in terms of the parameter.

Let

Substituting this expression into the equations of the planes gives

We now have two simultaneous equations for and , which can be “solved” to give expressions for and in terms of .

From equation (4), we can rearrange to give an expression for in terms of :

And substituting this expression for into equation (5) gives

Distributing over the parentheses and rearranging for gives an expression for in terms of :

We can now choose values for and at our convenience to make the equations as simple as possible.

We cannot choose , because the parameter would then be constant and not uniquely define every point on the line, but we can choose any value of we like.

From the list of possible answers, four of them have the parametric equation for as , so let’s try this one. This means that we have

Substituting these values of and into the expressions for and gives

And

We then have one possible set of parametric equations for , , and : which matches with answer C.

This confirms the answer that we found in method one, option C.

Example 2: Finding the Vector Equation of the Line of Intersection of Two Planes

Find the vector equation of the line of intersection between the two planes and .

Answer

To find the vector equation of the line of intersection between the two planes, we need to find the position vector, , of a point that lies in both planes and then find a nonzero direction vector parallel to the line of intersection. The vector equation of the line is then given by where is a scalar.

Let’s start with finding the position vector, , of a point that lies in both planes. We begin by choosing one variable as a parameter and setting it to a value of our choice.

Since all of the possible answers given have a constant vector with an component of zero, it makes sense to set .

Let .

In the equations of the two planes, this gives

If we do not have given possible answers, it is possible that our choice of value for a variable will be invalid. For instance, if the line of intersection lies parallel to the -plane, the value of will be constant along the line and probably not equal to the value chosen. If this is the case, however, it will be obvious on replacing the value we have chosen in the equations of the two planes, since there will be no solutions for a point in both planes with a value that lie on the line of intersection.

This is not the case here, so we now have two equations for and that can be solved simultaneously. From the equation of the second plane,

Substituting this expression for into the equation for the first plane gives

Distributing over the parentheses and rearranging for gives

From the equation above, , so we have

So, the position vector of one point on the line of intersection between the planes is .

We now need to find a direction vector parallel to the line of intersection between the two planes. We can do this by taking the cross product (or cross product) of the normal vectors of each plane.

We can find normal vectors to the two planes simply by reading off the coefficients of the variables in their equations

Therefore, two normal vectors to the planes are and respectively.

We can now evaluate the cross product by taking the determinant of the matrix:

Evaluating the determinant,

Thus, we have the direction vector for the line of intersection between the two planes:

Hence, the vector equation of the line of intersection between the two planes is given by

This is option D.

Example 3: Finding the Intersection of a Line and a Plane given Their General Equations

Find the point of intersection of the straight line and the plane .

Answer

The point of intersection between a line and a plane will be given by the unique solution to the system of equations of the straight line and the plane. There are multiple methods of solution. For this example, we will solve the equations algebraically.

We begin by rewriting the equation of the line as two distinct equations, both involving :

Rearranging these two equations gives and explicitly in terms of :

Substituting these expressions for and into the equation for the plane gives an equation only in , which we can solve for :

Distributing over the parentheses and simplifying gives

Substituting this value for into the equations for and ,

Therefore, the point of intersection between the line and the plane is .

Example 4: Finding the Coordinates of the Intersection Point of a Straight Line and a Plane

Find the coordinates of the point of intersection of the straight line with the plane .

Answer

If the line and the plane intersect, there must be a unique value of for which the vector is equal in both the equation of the line and the plane.

We begin by rewriting the vector equation of the line in terms of one vector:

At the point of intersection, the position vector will be the same in both equations, so we can substitute the vector from the equation of the line into the equation of the plane. This gives

Expanding the scalar product,

Simplifying and solving for ,

This is the value of at the point of intersection between the line and the plane. Substituting this into the equation of the line,

Therefore, the point of intersection between the line and the plane is .

Example 5: Finding the Point of Intersection of Three Planes

Find the point of intersection of the planes , , and .

Answer

In this example, it is given that there is a single point of intersection between the three planes. Since a point of intersection satisfies the equations of all three planes, there is a unique solution to the system of three equations.

Like any system of linear equations, there are multiple methods of solution.

Method 1: Geometric Approach

One method to find the point of intersection between the three planes is to first find the line of intersection between the first two planes and then find the point of intersection between this line and the third plane.

We can do this by finding the parametric equation for the line of intersection between the first two planes, expressing , , and in terms of a parameter, . We can then substitute these expressions for , , and into the equation for the third plane and solve the resulting equation to give the value of . Substituting this value of into the parametric equation for the line will give the -, -, and -coordinates of the point of intersection between all three planes.

Consider the general equations for the first two planes:

We can find the parametric equation for the line of intersection between these two planes by setting one variable equal to parameter and then solving the resulting equations to give expressions for the other two variables in terms of .

Let .

Substituting this expression for into the equations of the two planes gives

We now need to eliminate one variable from the equations. Multiplying equation (6) by 4 and adding it to equation (7) gives

Solving for ,

Now, we can substitute this expression for into equation (6) and solve for :

So, we now have the set of , , and values that lie on the line of intersection between the first two planes expressed in terms of parameter . If we now substitute these expressions for , , and into the equation of the third plane, we can solve for , giving the value of at the point of intersection between all three planes.

The equation of the third plane is given by

Substituting in the parametric expressions for , , and ,

Now, solving for ,

Substituting this value of into the parametric equations for , , and gives

Therefore, the point of intersection between all three planes is .

Method 2: Cramer’s Rule

We begin by rewriting the system of equations as a matrix equation of the form :

Taking the constants , , and 11 to the right-hand side and rewriting the left-hand side as the product of a matrix and the solution matrix, we then have

Now, Cramer’s rule tells us that is the unique solution to this system of equations, where is the determinant of the matrix of coefficients, and is the determinant of the matrix formed by replacing the column of associated with (the first column) with matrix .

It is worth noting here that the three planes will intersect at a single point if and only if the determinant of the matrix, , is nonzero. This is equivalent to the existence of a unique solution to the system of equations.

Since , the determinant of the unchanged matrix , is common to all three equations, let’s evaluate this first:

Although this was given in the question, we have now confirmed that the three planes must intersect at a single point, since the determinant is nonzero.

Now, to find , we find the determinant of the matrix formed by replacing the column of associated with with matrix on the right-hand side:

Substituting this value of into Cramer’s rule,

We can follow the same procedure for and :

Substituting this value of into Cramer’s rule,

And finally, for ,

Substituting this value of into Cramer’s rule,

So, we have , , and . This is the unique solution to the equations of the three planes. Therefore, the point of intersection between the three planes is .