Lesson Explainer: Intersection of Planes Mathematics

In this explainer, we will learn how to find the points and lines of intersection between lines and planes in space.

Definition: The General Form of the Equation of a Plane

A plane in 3D space, β„οŠ©, can be described in many different ways. For example, the general equation of a plane is given by π‘Žπ‘₯+𝑏𝑦+𝑐𝑧+𝑑=0.

This plane has a normal vector ⃑𝑛=(π‘Ž,𝑏,𝑐), which defines the plane’s orientation in 3D space. This normal vector is not unique. Any other nonzero scalar multiple of this vector, πœ†βƒ‘π‘›, is also normal to the plane.

The additional constant 𝑑 has no effect on the plane’s orientation but translates the plane 𝑑 units in the direction of the normal vector ⃑𝑛.

For example, for the plane described by the equation π‘₯+2𝑦+3𝑧+10=0, 1π‘₯+2𝑦+3𝑧+10=0, a normal vector ⃑𝑛 to the plane is (1,2,3). Any nonzero scalar multiple of this vector is also a normal vector to the plane, for example, (βˆ’1,βˆ’2,βˆ’3) or (5,10,15).

Definition: Intersection of Planes

Any two planes in β„οŠ© with nonparallel normal vectors will intersect over a straight line.

This line is the set of solutions to the simultaneous equations of the planes: π‘Žπ‘₯+𝑏𝑦+𝑐𝑧+𝑑=0,π‘Žπ‘₯+𝑏𝑦+𝑐𝑧+𝑑=0.

This system of two equations has three unknowns: π‘₯, 𝑦, and 𝑧. Therefore, the system will have either infinitely many solutions or no solutions at all. The former describes the case where the two planes intersect, and the latter describes the case where the two planes are parallel and never intersect.

How To: Finding the General Equation of a Line of Intersection between Two Planes

  1. Eliminate one of the three variables (it does not matter which one, but choose 𝑧 for example) from the two equations, and express one of the two remaining variables explicitly in terms of the other, for example, π‘₯=𝑓(𝑦).
  2. Eliminate the dependent variable, 𝑦, from the original two equations and express the independent variable, π‘₯, in terms of the remaining variable, 𝑧, so π‘₯=𝑔(𝑧).
  3. The general equation of the line of intersection is then given by π‘₯=𝑓(𝑦)=𝑔(𝑧).

Let’s consider an example of finding the line of intersection between two planes:

π‘₯βˆ’4𝑦+3π‘§βˆ’4=0,2π‘₯+2π‘¦βˆ’9𝑧+7=0.(1)(2)

First, we need to eliminate one of the three variables. We can eliminate 𝑧 by multiplying equation (1) by 3 and adding it to equation (2) , which gives 3π‘₯βˆ’12𝑦+9π‘§βˆ’12=0+2π‘₯+2π‘¦βˆ’9𝑧+7=05π‘₯βˆ’10π‘¦βˆ’5=0.

We can rearrange for π‘₯ by adding 10𝑦+5 to both sides and dividing by 5, which gives

π‘₯=2𝑦+1.(3)

Now we need to eliminate the dependent variable, 𝑦, from the original two equations to find an expression for π‘₯ in terms of 𝑧. We can multiply the second equation by 2 and add it to the first, which gives π‘₯βˆ’4𝑦+3π‘§βˆ’4=0+4π‘₯+4π‘¦βˆ’18𝑧+14=05π‘₯βˆ’15𝑧+10=0.

We can rearrange for π‘₯ by adding (15π‘§βˆ’10) to both sides and dividing by 5, which gives π‘₯=3π‘§βˆ’2.

Together with equation (3), we now have two expressions for π‘₯, one in terms of 𝑦 and one in terms of 𝑧: π‘₯=2𝑦+1,π‘₯=3π‘§βˆ’2.

These two equations can be rewritten as one equation with two equalities: π‘₯=2𝑦+1=3π‘§βˆ’2.

This is the general equation of a line in 3D space.

We cannot reduce the system of equations any further than this, or find values for π‘₯, 𝑦, and 𝑧 that uniquely solve the equations, because we have one more unknown than the number of equations. However, we are free to choose any value for one variable, which has corresponding values to the other two variables that solve the equations.

For example, setting π‘₯=1 in the equation above gives 1=2𝑦+1=3π‘§βˆ’2.

From the first part of the equation, we can rearrange to give 𝑦=0, and from the second part, we can rearrange to give 𝑧=1.

Therefore, from setting π‘₯=1, we have 𝑦=0 and 𝑧=1, which gives one point of intersection between the two planes: (1,0,1).

Likewise, we could set π‘₯=2, from which we would obtain 𝑦=12 and 𝑧=43, giving another point of intersection between the two planes ο€Ό2,12,43.

We do not, of course, need to choose π‘₯ for the variable to set as a parameter. We could just as freely choose 𝑦 or 𝑧. For example, choosing 𝑦=1 in the main equation above gives π‘₯=2(1)+1=3π‘§βˆ’2, which in turn can be rearranged to give π‘₯=3 and 𝑧=53, so ο€Ό3,1,53 is another point on the line of intersection between the two planes.

These are just some of the infinitely many solutions to the system of equations that form the line of intersection between the two planes.

The general form is not the only way of describing a line of intersection between two planes. Another way is with a set of parametric equationsβ€”using an external parameter that defines the three variables π‘₯, 𝑦, and 𝑧 separately.

Definition: The Parametric Form of the Equation of a Line in 3D Space

A line in 3D space may be defined by the general set of parametric equations π‘₯=𝑓(𝑑)=π‘₯+π‘Žπ‘‘,𝑦=𝑔(𝑑)=𝑦+𝑏𝑑,𝑧=β„Ž(𝑑)=𝑧+𝑐𝑑, where 𝑑 is a parameter; π‘₯, π‘¦οŠ¦, and π‘§οŠ¦ are the coordinates of a point lying on the line; and π‘Ž, 𝑏, and 𝑐 are the components of the direction vector of the line or parallel to the line.

Since there are infinitely many points on the line, there are infinitely many choices of (π‘₯,𝑦,𝑧) for the parametric equation of the line.

How To: Finding the Parametric Equation of a Line of Intersection between Two Planes

  1. Express one of the three variables in the equations of the two planes as a linear function of a parameter, 𝑑, for example, π‘₯=π‘₯+π‘Žπ‘‘οŠ¦.
  2. Substitute this expression into the original equations of the planes, and solve the system of equations to express the other two variables in terms of the parameter, 𝑑.

Let’s look at an example of constructing a set of parametric equations for a line of intersection given the general equations of two planes.

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 π‘₯+𝑧=3 and 2π‘₯βˆ’π‘¦βˆ’π‘§=βˆ’2.

  1. π‘₯=3+𝑑,𝑦=4+3𝑑,𝑧=βˆ’π‘‘and
  2. π‘₯=3+𝑑,𝑦=8βˆ’3𝑑,𝑧=βˆ’π‘‘and
  3. π‘₯=3+𝑑,𝑦=8+3𝑑,𝑧=βˆ’π‘‘and
  4. π‘₯=1+𝑑,𝑦=3+3𝑑,𝑧=2βˆ’π‘‘and
  5. π‘₯=3+𝑑,𝑦=4βˆ’3𝑑,𝑧=βˆ’π‘‘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 β€œfree 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, π‘₯=π‘₯+π‘Žπ‘‘οŠ¦, and then fix values for π‘₯ and π‘Ž at a stage of the calculation that is convenient, or we can fix the parametrization at the start of our calculation, for example, 𝑧=𝑑, 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 𝑧=βˆ’π‘‘ as it seems likely that we would then land on the correct answer. However, we will set 𝑧=𝑑 and then demonstrate that this will in fact give us an equivalent line of intersection, which we can then β€œtweak” to determine the correct answer from the provided options.

If we substitute our chosen parameter for 𝑧 into the equation for the first plane, we get π‘₯+𝑑=3, which gives π‘₯=3βˆ’π‘‘. If we now substitute π‘₯ and 𝑧 into the equation for our second plane, we get 2(3βˆ’π‘‘)βˆ’π‘¦βˆ’π‘‘=βˆ’2.

If we distribute over the parentheses and simplify, we get 6βˆ’2π‘‘βˆ’π‘‘+2=𝑦, which simplifies to 𝑦=8βˆ’3𝑑. Therefore, the parametric equations for π‘₯, 𝑦, and 𝑧 are π‘₯=3βˆ’π‘‘,𝑦=8βˆ’3𝑑,𝑧=𝑑.and

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 (3,8,0) with direction vector (βˆ’1,βˆ’3,1) 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 3𝑑 and 𝑑 when compared with the direction vector of our line. The remaining options have a direction vector that is a multiple of βˆ’1 of our line and is, therefore, equivalent.

We can then discount option A as it passes through the same π‘₯-coordinate but a different 𝑦-coordinate, which leaves options C and D. Option C is described by the same point, (3,8,0), 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 (3,8,0) is a point on this particular line: substituting the value 𝑑=2 into each of the parametric equations leads us to the point (3,9,0). 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 π‘₯=𝑓(𝑑)=π‘₯+π‘Žπ‘‘,𝑦=𝑔(𝑑)=𝑦+𝑏𝑑,𝑧=β„Ž(𝑑)=𝑧+𝑐𝑑, where 𝑑 is a parameter; π‘₯, π‘¦οŠ¦, and π‘§οŠ¦ are the coordinates of a point lying on the line; and π‘Ž, 𝑏, and 𝑐 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 π‘₯=π‘₯+π‘Žπ‘‘.

Substituting this expression into the equations of the planes gives

π‘₯+π‘Žπ‘‘+𝑧=3,2(π‘₯+π‘Žπ‘‘)βˆ’π‘¦βˆ’π‘§=βˆ’2.(4)(5)

We now have two simultaneous equations for 𝑦 and 𝑧, which can be β€œsolved” to give expressions for 𝑦 and 𝑧 in terms of 𝑑.

From equation (4), we can rearrange to give an expression for 𝑧 in terms of 𝑑: 𝑧=3βˆ’π‘₯βˆ’π‘Žπ‘‘.

And substituting this expression for 𝑧 into equation (5) gives 2(π‘₯+π‘Žπ‘‘)βˆ’π‘¦βˆ’(3βˆ’π‘₯βˆ’π‘Žπ‘‘)=βˆ’2.

Distributing over the parentheses and rearranging for 𝑦 gives an expression for 𝑦 in terms of 𝑑: 𝑦=3(π‘₯+π‘Žπ‘‘)βˆ’1.

We can now choose values for π‘₯ and π‘Ž at our convenience to make the equations as simple as possible.

We cannot choose π‘Ž=0, because the parameter would then be constant and not uniquely define every point on the line, but we can choose any value of π‘₯ we like.

From the list of possible answers, four of them have the parametric equation for π‘₯ as π‘₯=3+𝑑, so let’s try this one. This means that we have π‘₯=3,π‘Ž=1.

Substituting these values of π‘₯ and π‘Ž into the expressions for 𝑦 and 𝑧 gives 𝑦=3(3+𝑑)βˆ’1=8+3𝑑.

And 𝑦=3βˆ’3βˆ’π‘‘=βˆ’π‘‘.

We then have one possible set of parametric equations for π‘₯, 𝑦, and 𝑧: π‘₯=3+𝑑,𝑦=8+3𝑑,𝑧=βˆ’π‘‘,and which matches with answer C.

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

A final way of describing the line of intersection between two planes is with a vector equation.

Definition: The Vector Form of the Equation of a Line in 3D Space

A line in 3D space may be defined in vector form by the general equation βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+𝑑⃑𝑑, where βƒ‘π‘Ÿ=(π‘₯,𝑦,𝑧) is the position vector of a known point on the line, ⃑𝑑 is a nonzero vector parallel to the line, and 𝑑 is a scalar.

How To: Finding the Vector Equation of a Line of Intersection between Two Planes

  1. Find a single point, βƒ‘π‘ŸοŠ¦, that lies in both planes. This can be done by setting the value of one variable, for example, π‘₯=π‘₯, and solving the equations of the two planes to find the corresponding values of the other two variables, 𝑦=π‘¦οŠ¦ and 𝑧=π‘§οŠ¦.
  2. Determine normal vectors to each plane, βƒ‘π‘›οŠ§ and βƒ‘π‘›οŠ¨, by reading off the coefficients from their equations.
  3. Take the cross product of the normal vectors, ⃑𝑑=βƒ‘π‘›Γ—βƒ‘π‘›οŠ§οŠ¨, to give a vector, ⃑𝑑, parallel to the line of intersection between the planes.
  4. The vector equation of the line of intersection is then given by βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+π‘‘βƒ‘π‘‘οŠ¦, where 𝑑 is a scalar.

Let’s look at an example of using the cross product to find the direction vector of the line of intersection between two planes, and then the vector equation of that line.

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 π‘₯+3𝑦+2π‘§βˆ’6=0 and 2π‘₯βˆ’π‘¦+𝑧+2=0.

  1. βƒ‘π‘Ÿ=(0,2,βˆ’12)+𝑑(5,3,βˆ’7)
  2. βƒ‘π‘Ÿ=(0,14,12)+𝑑(2,βˆ’3,2)
  3. βƒ‘π‘Ÿ=(0,2,0)+𝑑(2,βˆ’3,2)
  4. βƒ‘π‘Ÿ=(0,2,0)+𝑑(5,3,βˆ’7)
  5. βƒ‘π‘Ÿ=(0,14,12)+𝑑(5,3,βˆ’7)

Answer

To find the vector equation of the line of intersection between the two planes, we need to find the position vector, βƒ‘π‘ŸοŠ¦, of a point that lies in both planes and then find a nonzero direction vector ⃑𝑑 parallel to the line of intersection. The vector equation of the line is then given by βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+𝑑⃑𝑑, where 𝑑 is a scalar.

Let’s start with finding the position vector, βƒ‘π‘ŸοŠ¦, 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 π‘₯ component of zero, it makes sense to set π‘₯=0.

Let π‘₯=0.

In the equations of the two planes, this gives 3𝑦+2π‘§βˆ’6=0,βˆ’π‘¦+𝑧+2=0.

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 𝑦𝑧-plane, the value of π‘₯ 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 does not lie on the line of intersection.

This is not the case here, so we now have two equations for 𝑦 and 𝑧 that can be solved simultaneously. From the equation of the second plane, 𝑦=𝑧+2.

Substituting this expression for 𝑦 into the equation for the first plane gives 3(𝑧+2)+2π‘§βˆ’6=0.

Distributing over the parentheses and rearranging for 𝑧 gives 𝑧=0.

From the equation above, 𝑦=𝑧+2, so we have 𝑦=2.

So, the position vector of one point on the line of intersection between the planes is βƒ‘π‘Ÿ=(0,2,0).

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 1π‘₯+3𝑦+2π‘§βˆ’6=0,2π‘₯βˆ’1𝑦+1𝑧+2=0.

Therefore, two normal vectors to the planes are ⃑𝑛=(1,3,2) and ⃑𝑛=(2,βˆ’1,1) respectively.

We can now evaluate the cross product βƒ‘π‘›Γ—βƒ‘π‘›οŠ§οŠ¨ by taking the determinant of the matrix: οƒβƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜1322βˆ’11.

Evaluating the determinant, ||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜1322βˆ’11||||=⃑𝑖||32βˆ’11||βˆ’βƒ‘π‘—||1221||+βƒ‘π‘˜||132βˆ’1||=⃑𝑖(3β‹…1βˆ’2β‹…(βˆ’1))βˆ’βƒ‘π‘—(1β‹…1βˆ’2β‹…2)+βƒ‘π‘˜(1β‹…(βˆ’1)βˆ’3β‹…2)=⃑𝑖(3+2)βˆ’βƒ‘π‘—(1βˆ’4)+βƒ‘π‘˜(βˆ’1βˆ’6)=5⃑𝑖+3βƒ‘π‘—βˆ’7βƒ‘π‘˜=(5,3,βˆ’7).

Thus, we have the direction vector for the line of intersection between the two planes: ⃑𝑑=(5,3,βˆ’7).

Hence, the vector equation of the line of intersection between the two planes is given by βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+𝑑⃑𝑑=(0,2,0)+𝑑(5,3,βˆ’7).

This is option D.

Definition: Point of Intersection between a Line and a Plane

A line and a nonparallel plane will intersect at a single point.

This point is the unique solution of the equation of the line and the equation of the plane.

The equation of the plane, π‘Žπ‘₯+𝑏𝑦+𝑐𝑧+𝑑=0, is one equation, and the equation of the line, π‘Žπ‘₯+π‘₯=𝑏𝑦+𝑦=𝑐𝑧+𝑧, can be rewritten as two distinct equations: π‘Žπ‘₯+π‘₯=𝑏𝑦+𝑦,π‘Žπ‘₯+π‘₯=𝑐𝑧+𝑧.

This is a system of three distinct equations for three unknowns and therefore will have either no solutions (if the line and plane are parallel and do not intersect), one unique solution (if the line and plane are not coplanar and intersect), or infinitely many solutions (if the line and plane are coplanar).

As with any system of 𝑛 equations for 𝑛 unknowns, there are multiple methods of solution.

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

Find the point of intersection of the straight line βˆ’3π‘₯=4π‘¦βˆ’2=𝑧+1 and the plane βˆ’3π‘₯+𝑦+𝑧=13.

Answer

The point of intersection (π‘₯,𝑦,𝑧) 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 𝑧: βˆ’3π‘₯=𝑧+1,4π‘¦βˆ’2=𝑧+1.

Rearranging these two equations gives π‘₯ and 𝑦 explicitly in terms of 𝑧: π‘₯=βˆ’13(𝑧+1),𝑦=14(𝑧+3).

Substituting these expressions for π‘₯ and 𝑦 into the equation for the plane gives an equation only in 𝑧, which we can solve for 𝑧: βˆ’3ο€Όβˆ’13(𝑧+1)+14(𝑧+3)+𝑧=13.

Distributing over the parentheses and simplifying gives 𝑧+1+𝑧4+34+𝑧=139𝑧4=454𝑧=5.

Substituting this value for 𝑧 into the equations for π‘₯ and 𝑦, π‘₯=βˆ’13(5+1)=βˆ’2.𝑦=14(5+3)=2.

Therefore, the point of intersection between the line and the plane is (βˆ’2,2,5).

The point of intersection between a line and a plane may also be found given their vector equations.

Definition: The Vector Form of the Equation of a Plane

A plane may be defined by a vector equation of the form βƒ‘π‘›β‹…βƒ‘π‘Ÿ=𝑐, where βƒ‘π‘Ÿ is the position vector of a general point on the plane, ⃑𝑛 is a constant vector that is normal to the plane, and 𝑐 is a constant scalar.

Also recall that the vector equation of a line in β„οŠ is given by βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+𝑑⃑𝑑, where βƒ‘π‘ŸοŠ¦ is the position vector of a point on the line, ⃑𝑑 is any nonzero vector parallel to the line, and 𝑑 is a scalar.

The value of the scalar parameter 𝑑 uniquely defines every point on the line, so the point of intersection between the line and the plane will be given by a unique value of 𝑑. This value of 𝑑 may be found by setting the general position vector βƒ‘π‘Ÿ in the equation of the plane equal to the general position vector βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+π‘‘βƒ‘π‘‘οŠ¦ in the equation of the line, since at the point of intersection (if it exists) the position vectors will be the same.

Therefore, we need to find the value of 𝑑 that solves the equation: βƒ‘π‘›β‹…ο€»βƒ‘π‘Ÿ+𝑑⃑𝑑=𝑐.

Let’s look at an example of using this method to find the point of intersection between a line and a plane in 3D space given their vector equations.

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 βƒ‘π‘Ÿ=(8,2,βˆ’5)+𝑑(βˆ’7,βˆ’9,13) with the plane (9,4,βˆ’5)β‹…βƒ‘π‘Ÿ=βˆ’59.

Answer

If the line and the plane intersect, there must be a unique value of 𝑑 for which the vector βƒ‘π‘Ÿ 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: βƒ‘π‘Ÿ=(8βˆ’7𝑑,2βˆ’9𝑑,βˆ’5+13𝑑).

At the point of intersection, the position vector βƒ‘π‘Ÿ will be the same in both equations, so we can substitute the vector βƒ‘π‘Ÿ from the equation of the line into the equation of the plane. This gives (9,4,βˆ’5)β‹…(8βˆ’7𝑑,2βˆ’9𝑑,βˆ’5+13𝑑)=βˆ’59.

Expanding the scalar product, 9(8βˆ’7𝑑)+4(2βˆ’9𝑑)βˆ’5(βˆ’5+13𝑑)=βˆ’59.

Simplifying and solving for 𝑑, 72βˆ’63𝑑+8βˆ’36𝑑+25βˆ’65𝑑=βˆ’59βˆ’164𝑑=βˆ’164𝑑=1.

This is the value of 𝑑 at the point of intersection between the line and the plane. Substituting this into the equation of the line, βƒ‘π‘Ÿ=(8βˆ’7,2βˆ’9,βˆ’5+13)=(1,βˆ’7,8).

Therefore, the point of intersection between the line and the plane is (1,βˆ’7,8).

For three distinct planes in 3D space, there is a much broader range of possible scenarios.

  1. If all three planes are parallel, there is no intersection between any of them.
  2. If two planes are parallel to each other and a third is not, then this third plane will intersect the other two planes over two separate lines of intersection.
  3. If all three planes are nonparallel to each other, they may intersect at a single point.
  4. Also, if all the planes are non-parallel, they may intersect along a line.
  5. If all three planes are nonparallel, the third plane may also intersect with the other two planes separately, giving three lines of intersection that are parallel to each other.

Let’s look at an example of finding the single point of intersection between three planes in scenario 𝑐 above.

Example 5: Finding the Point of Intersection of Three Planes

Find the point of intersection of the planes βˆ’5π‘₯βˆ’2𝑦+6π‘§βˆ’1=0, βˆ’7π‘₯+8𝑦+π‘§βˆ’6=0, and π‘₯βˆ’3𝑦+3𝑧+11=0.

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 π‘₯, 𝑦, and 𝑧 in terms of a parameter, 𝑑. We can then substitute these expressions for π‘₯, 𝑦, and 𝑧 into the equation for the third plane and solve the resulting equation to give the value of 𝑑. Substituting this value of 𝑑 into the parametric equation for the line will give the π‘₯-, 𝑦-, and 𝑧-coordinates of the point of intersection between all three planes.

Consider the general equations for the first two planes: βˆ’5π‘₯βˆ’2𝑦+6π‘§βˆ’1=0,βˆ’7π‘₯+8𝑦+π‘§βˆ’6=0.

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

Let 𝑧=𝑑.

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

βˆ’5π‘₯βˆ’2𝑦+6π‘‘βˆ’1=0,βˆ’7π‘₯+8𝑦+π‘‘βˆ’6=0.(6)(7)

We now need to eliminate one variable from the equations. Multiplying equation (6) by 4 and adding it to equation (7) gives βˆ’27π‘₯+25π‘‘βˆ’10=0.

Solving for π‘₯, π‘₯=25π‘‘βˆ’1027.

Now, we can substitute this expression for π‘₯ into equation (6) and solve for 𝑦: βˆ’5ο€Ό25π‘‘βˆ’1027οˆβˆ’2𝑦+6π‘‘βˆ’1=0𝑦=βˆ’5+6π‘‘βˆ’12𝑦=37𝑑+2354.

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

The equation of the third plane is given by π‘₯βˆ’3𝑦+3𝑧+11=0.

Substituting in the parametric expressions for π‘₯, 𝑦, and 𝑧, 25π‘‘βˆ’1027βˆ’3ο€Ό37𝑑+2354+3𝑑+11=0.

Now, solving for 𝑑, 50π‘‘βˆ’2054βˆ’ο€Ό111𝑑+6954+162𝑑54+59454=050π‘‘βˆ’20βˆ’(111𝑑+69)+162𝑑+594=0101𝑑+505=0𝑑=βˆ’5.

Substituting this value of 𝑑 into the parametric equations for π‘₯, 𝑦, and 𝑧 gives π‘₯=25π‘‘βˆ’1027=βˆ’125βˆ’1027=βˆ’5,𝑦=37𝑑+2354=βˆ’185+2354=βˆ’3,𝑧=𝑑=βˆ’5.

Therefore, the point of intersection between all three planes is (βˆ’5,βˆ’3,βˆ’5).

Method 2: Cramer’s Rule

We begin by rewriting the system of equations as a matrix equation of the form 𝐴𝑋=𝐡: βˆ’5π‘₯βˆ’2𝑦+6π‘§βˆ’1=0,βˆ’7π‘₯+8𝑦+π‘§βˆ’6=0,π‘₯βˆ’3𝑦+3𝑧+11=0.

Taking the constants βˆ’1, βˆ’6, and 11 to the right-hand side and rewriting the left-hand side as the product of a matrix 𝐴 and the solution matrix, 𝑋=ο€Ώπ‘₯𝑦𝑧, we then have ο€βˆ’5βˆ’26βˆ’7811βˆ’33οŒο€Ώπ‘₯𝑦𝑧=16βˆ’11.

Now, Cramer’s rule tells us that π‘₯=ΔΔ,𝑦=ΔΔ,𝑧=Ξ”Ξ”ο—ο˜ο™ is the unique solution to this system of equations, where Ξ” is the determinant of the matrix of coefficients, 𝐴 and Δ is the determinant of the matrix formed by replacing the column of 𝐴 associated with π‘₯ (the first column) with matrix 𝐡.

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

Since Ξ”, the determinant of the unchanged matrix 𝐴, is common to all three equations, let’s evaluate this first: Ξ”=||||βˆ’5βˆ’26βˆ’7811βˆ’33||||=βˆ’5||81βˆ’33||+2||βˆ’7113||+6||βˆ’781βˆ’3||=βˆ’5(24βˆ’(βˆ’3))+2(βˆ’21βˆ’1)+6(21βˆ’8)=βˆ’135βˆ’44+78=βˆ’101.

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

Now, to find Δ, we find the determinant of the matrix formed by replacing the column of 𝐴 associated with π‘₯ with matrix 𝐡 on the right-hand side: Ξ”=||||1βˆ’26681βˆ’11βˆ’33||||=1||81βˆ’33||+2||61βˆ’113||+6||68βˆ’11βˆ’3||=(8β‹…3βˆ’1β‹…(βˆ’3))+2(6β‹…3βˆ’1β‹…(βˆ’11))+6(6β‹…(βˆ’3)βˆ’8β‹…(βˆ’11))=27+58+420=505.

Substituting this value of Δ into Cramer’s rule, π‘₯=ΔΔ=505βˆ’101=βˆ’5.

We can follow the same procedure for 𝑦 and 𝑧: Ξ”=||||βˆ’516βˆ’7611βˆ’113||||=βˆ’5||61βˆ’113||βˆ’1||βˆ’7113||+6||βˆ’761βˆ’11||=βˆ’5(6β‹…3βˆ’1β‹…(βˆ’11))βˆ’(βˆ’7β‹…3βˆ’1β‹…1)+6(βˆ’7β‹…(βˆ’11)βˆ’6β‹…1)=βˆ’145+22+426=303.

Substituting this value of Ξ”ο˜ into Cramer’s rule, 𝑦=ΔΔ=303βˆ’101=βˆ’3.

And finally, for 𝑧, Ξ”=||||βˆ’5βˆ’21βˆ’7861βˆ’3βˆ’11||||=βˆ’5||86βˆ’3βˆ’11||+2||βˆ’761βˆ’11||+1||βˆ’781βˆ’3||=βˆ’5(8β‹…(βˆ’11)βˆ’6β‹…(βˆ’3))+2(βˆ’7β‹…(βˆ’11)βˆ’6β‹…1)+(βˆ’7β‹…(βˆ’3)βˆ’8β‹…1)=350+142+13=505.

Substituting this value of Δ into Cramer’s rule, 𝑧=ΔΔ=505βˆ’101=βˆ’5.

So, we have π‘₯=βˆ’5, 𝑦=βˆ’3, and 𝑧=βˆ’5. This is the unique solution to the equations of the three planes. Therefore, the point of intersection between the three planes is (βˆ’5,βˆ’3,βˆ’5).

We conclude our discussion of points and lines of intersection between lines and planes in β„οŠ© by noting some key points.

Key Points

  • Two nonparallel planes in β„οŠ© will intersect over a straight line, which is the one-dimensionally parametrized set of solutions to the equations of both planes.
  • The direction vector, ⃑𝑑, of the line of intersection of two planes may be given by the cross product of the normal vectors of the planes, βƒ‘π‘›Γ—βƒ‘π‘›οŠ§οŠ¨.
  • A line and a nonparallel plane in β„οŠ© will intersect at a single point, which is the unique solution to the equation of the line and the equation of the plane.
  • Three nonparallel planes will intersect at a single point if and only if there exists a unique solution to the system of equations of the three planes. When written as a matrix equation, this equates to the determinant of the coefficient matrix being invertible, that is, Ξ”β‰ 0. If the determinant of the coefficient matrix is zero, then the planes do not intersect at a unique point, if at all.

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