Lesson Explainer: The Perpendicular Distance between Points and Planes Mathematics

In this explainer, we will learn how to calculate the perpendicular distance between a plane and a point, between a plane and a straight line parallel to it, and between two parallel planes using a formula.

To find the shortest distance between a point and a line, we first need to determine exactly what is meant by the shortest distance between these two geometric objects. To do this, we first note that if a point 𝑃(𝑥,𝑦,𝑧) lies on the plane 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0, then the distance between these objects will be zero. So, we will assume our point does not lie on the plane.

To find the shortest distance between these objects, let us first consider the distance between 𝑃 and a point 𝑅 on our plane.

We can show that this is not the shortest distance between 𝑃 and the plane by constructing the following right triangle.

We choose point 𝑄 on our plane so that the line segment 𝑃𝑄 is perpendicular to the plane. We can then see that 𝑃𝑅 is the hypotenuse of a right triangle, which means it must be longer than the other sides. In particular, this means that the length of 𝑃𝑄 is shorter than 𝑃𝑅. We can construct this triangle for any point 𝑅 on our plane, so the 𝑃𝑄 must be the shortest distance between the point 𝑃 and the plane.

We call this the perpendicular distance between the point and the plane because 𝑃𝑄 is perpendicular to the plane. We could find this distance by finding the coordinates of 𝑄; however, there is an easier method.

To calculate this distance, we will start by setting 𝑅𝑃𝑄=𝜃 and ||𝑃𝑄||=𝐷. We will also introduce the vectors 𝑃𝑅 and 𝑃𝑄.

We see in our diagram that 𝐷 is the length of the side adjacent to angle 𝜃 in a right triangle; this tells us that coslengthofadjacentsidelengthofhypotenuse𝜃==𝐷𝑃𝑅.

In particular, we get the equation 𝑃𝑅𝜃=𝐷.cos

We can construct another equation involving the expression 𝑃𝑅𝜃cos by recalling the following property about vectors.

Definition: The Dot Product of Two Vectors

If 𝜃 is the measure of the angle between two vectors 𝑢 and 𝑣, then 𝑢𝑣=𝑢𝑣𝜃.cos

Applying this property to the vectors 𝑃𝑄 and 𝑃𝑅, we get 𝑃𝑄𝑃𝑅=𝑃𝑄𝑃𝑅𝜃.cos

We can rearrange this equation to get 𝑃𝑄𝑃𝑅𝑃𝑄=𝑃𝑅𝜃=𝐷.cos

However, we cannot evaluate this expression directly because we do not know the coordinates of 𝑄. We can get around this by recalling that 𝑃𝑄 is perpendicular to the plane and we can find another vector perpendicular to the plane.

Recall that the vector (𝑎,𝑏,𝑐) is perpendicular to the plane 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0. Therefore, since both (𝑎,𝑏,𝑐) and 𝑃𝑄 are perpendicular to the plane, we must have (𝑎,𝑏,𝑐)𝑃𝑄.

Remember, we say two vectors are parallel if they are a non zero scalar multiples of each other; we will call this scalar 𝑘: 𝑃𝑄=𝑘(𝑎,𝑏,𝑐).

We can use this to find the perpendicular distance 𝐷. First, we can substitute this expression into our equation for the dot product and then simplify: 𝐷=𝑃𝑄𝑃𝑅𝑃𝑄=𝑘(𝑎,𝑏,𝑐)𝑃𝑅𝑘(𝑎,𝑏,𝑐)=𝑘(𝑎,𝑏,𝑐)𝑃𝑅|𝑘|(𝑎,𝑏,𝑐).

Next, we need to be careful to simplify 𝑘|𝑘|, since we do not know if 𝑘 is negative or positive. Instead, we know that 𝐷 is a length and must therefore be positive. This means we can take the absolute value of both sides of this equation: 𝐷=𝑘(𝑎,𝑏,𝑐)𝑃𝑅|𝑘|(𝑎,𝑏,𝑐),|𝐷|=|||𝑘(𝑎,𝑏,𝑐)𝑃𝑅|𝑘|(𝑎,𝑏,𝑐)|||.

We can then simplify this equation by using the properties of the absolute value: 𝐷=||𝑘(𝑎,𝑏,𝑐)𝑃𝑅|||𝑘|(𝑎,𝑏,𝑐)=|𝑘|||(𝑎,𝑏,𝑐)𝑃𝑅|||𝑘|(𝑎,𝑏,𝑐)=||(𝑎,𝑏,𝑐)𝑃𝑅||(𝑎,𝑏,𝑐).

We could leave our expression for 𝐷 in this form; however, we can simplify this further. Recall that 𝑅 is any point on our plane; let us say 𝑅=(𝑥,𝑦,𝑧).

We can then find the components of 𝑃𝑅: 𝑃𝑅=(𝑥,𝑦,𝑧)(𝑥,𝑦,𝑧)=(𝑥𝑥,𝑦𝑦,𝑧𝑧).

Then, we can substitute this expression into our equation for 𝐷 and evaluate the dot product: 𝐷=||(𝑎,𝑏,𝑐)𝑃𝑅||(𝑎,𝑏,𝑐)=|(𝑎,𝑏,𝑐)(𝑥𝑥,𝑦𝑦,𝑧𝑧)|(𝑎,𝑏,𝑐)=|𝑎(𝑥𝑥)+𝑏(𝑦𝑦)+𝑐(𝑧𝑧)|(𝑎,𝑏,𝑐)=|𝑎𝑥+𝑏𝑦+𝑐𝑧(𝑎𝑥+𝑏𝑦+𝑐𝑧)|(𝑎,𝑏,𝑐).

Finally, we will use the fact that 𝑅 lies on the plane 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0; this means 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0.

We can rearrange this to see 𝑎𝑥+𝑏𝑦+𝑐𝑧=𝑑.

Substituting this into our equation for 𝐷 and simplifying, we get 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧(𝑎𝑥+𝑏𝑦+𝑐𝑧)||(𝑎,𝑏,𝑐)=|𝑑(𝑎𝑥+𝑏𝑦+𝑐𝑧)|(𝑎,𝑏,𝑐)=|(𝑑+𝑎𝑥+𝑏𝑦+𝑐𝑧)|(𝑎,𝑏,𝑐)=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|(𝑎,𝑏,𝑐)=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐.

We can summarize this result as follows.

Definition: Distance between a Point and a Plane

The shortest distance (or perpendicular distance), 𝐷, between the point (𝑥,𝑦,𝑧) and the plane 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0 is given by 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐.

Let us see an example of how we can use this formula to find the perpendicular distance between a point and a plane given in the general form.

Example 1: Finding the Distance between a Point and a Plane

Find the distance between the point (5,8,6) and the plane 2𝑥+𝑦+2𝑧=7.

Answer

We want to find the distance between a point and a plane. To do this, we first need to recall that the distance between a point and a plane means the perpendicular distance, since this is the shortest distance between these two objects.

To find the perpendicular distance, we need to recall the following formula.

The perpendicular distance, 𝐷, between the point (𝑥,𝑦,𝑧) and the plane 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0 is given by 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐.

We have (𝑥,𝑦,𝑧)=(5,8,6) and we need to rewrite our equation for the plane 2𝑥+𝑦+2𝑧=72𝑥+𝑦+2𝑧7=0.

So, 𝑎=2, 𝑏=1, 𝑐=2, and 𝑑=7.

Substituting these values into our formula, we get 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐=|(2)(5)+(1)(8)+(2)(6)7|(2)+1+2=|17|9=173.lengthunits

We can add length units to this value since it represents a length.

Hence, we were able to show that the distance between the point (5,8,6) and the plane 2𝑥+𝑦+2𝑧=7 is 173lengthunits.

In our next example, we will see how we can apply this formula to find the distance between a point and a plane given in vector form.

Example 2: Finding the Distance between a Point and a Plane

Find the distance between the point (2,1,3) and the plane 𝑟(2,2,1)=3.

Answer

We want to find the distance between a point and a plane. To do this, we need to recall that the distance, 𝐷, between the point (𝑥,𝑦,𝑧) and the plane 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0 is given by 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐.

We cannot apply this formula directly because our plane is given in vector form. Therefore, to apply our formula we want to convert the plane into the general form for the equation of our plane.

To do this, we substitute 𝑟=(𝑥,𝑦,𝑧) into the vector equation of our plane: 𝑟(2,2,1)=3(𝑥,𝑦,𝑧)(2,2,1)=32𝑥+2𝑦+𝑧=3.

Then, we subtract 3 from both sides of the equation: 2𝑥+2𝑦+𝑧3=0.

Now that we have the equation of our plane in the general form, we can apply our formula for the distance. We have (𝑥,𝑦,𝑧)=(2,1,3), 𝑎=2, 𝑏=2, 𝑐=1, and 𝑑=3; substituting these values in, we get 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐=|(2)(2)+(2)(1)+(1)(3)3|(2)+2+1=|6|9=2;lengthunits we can add length units to this value because we know it represents a length.

Hence, we were able to show that the distance between the point (2,1,3) and the plane 𝑟(2,2,1)=3 is 2lengthunits.

In the previous example, we found the distance between a point and a plane given in the vector form by finding the equation of our line in the general form. We can use this process to find a formula for the distance between a point and a plane in the vector form.

We can always rewrite the plane with equation 𝑟(𝑎,𝑏,𝑐)=𝑑 into the form 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0. Applying our formula for the perpendicular distance gives us the following result.

Theorem: Distance between a Point and a Plane in Vector Form

The shortest distance (or perpendicular distance), 𝐷, between the point (𝑥,𝑦,𝑧) and the plane 𝑟(𝑎,𝑏,𝑐)=𝑑 is given by 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐.

We could have used this result to directly evaluate the distance given to us in the second example.

We can use this same process to determine the distance between a line and a plane. First, if the line and plane are not parallel or not distinct, then they intersect, so the distance between them is 0. Second, if they are parallel and distinct then, we can show that the shortest distance between them is the perpendicular distance between any point on the line and the plane. Consider the distance between an arbitrary point 𝑃 on a line and another arbitrary point 𝑅 on the plane parallel to the line.

We can show that 𝑃𝑅 is the hypotenuse of a right triangle, so this distance is always larger than the perpendicular distance between the point 𝑃 and the plane. Finally, since the line and plane are parallel, the distance between them is constant, so we can choose any point 𝑃 on our line and the distance will be the same, which means we can use the formula for the distance between a point and a plane.

Theorem: Distance between a Line and a Plane in Vector Form

The shortest distance (or perpendicular distance), 𝐷, between a parallel line and a plane, where (𝑥,𝑦,𝑧) is any point on the line and the plane has the equation 𝑟(𝑎,𝑏,𝑐)=𝑑, is given by 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐.

In our next two examples, we will see how we can apply this process to find the distance between a line parallel to a plane and said plane.

Example 3: Finding the Distance between a Line and a Plane

Find the perpendicular distance between the line 𝑟=(1,2,4)+𝑡(2,1,4) and the plane 𝑟(2,0,1)=1.

Answer

The question asks us to find the perpendicular distance between a line and a plane. We need to determine if they intersect; to do this, we first rewrite the line as 𝑟=(12𝑡,2+𝑡,4+4𝑡).

Then, we substitute this into the equation of the plane: (12𝑡,2+𝑡,4+4𝑡)(2,0,1)=12(12𝑡)+0(2+𝑡)+1(4+4𝑡)=124𝑡+4+4𝑡=16=1.

This equation will not be true for any value of 𝑡, so the line and plane do not intersect. Hence, they are parallel.

Alternatively, we can show that the line and plane are parallel by showing that the normal vector to the plane and the direction vector of the line are perpendicular; we can do this by computing their dot product: (2,1,4)(2,0,1)=(2×2)+(1×0)+(4×1)=4+4=0.

Hence, the line is perpendicular to the normal vector of the plane, and so the line and plane are parallel.

We recall that the distance between a line and a plane is given by the distance between any point on the line and the plane. We know that the point (1,2,4) lies on the line, since this is the position vector when 𝑡=0, and that the perpendicular distance, 𝐷, between the point (𝑥,𝑦,𝑧) and the plane 𝑟(𝑎,𝑏,𝑐)=𝑑 is given by 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐.

We substitute 𝑎=2, 𝑏=0, 𝑐=1, 𝑑=1, 𝑥=1, 𝑦=2, and 𝑧=4 into the formula to get 𝐷=|(2)(1)+(0)(2)+(1)(4)1|(2)+(0)+(1)=|5|5=5.

Hence, the distance between the line and plane, to one decimal place, is 5 length units.

Example 4: Finding the Distance between a Line and a Plane

Find the distance between the line 𝑥12=𝑦24=𝑧+52 and the plane 3𝑥2𝑦+𝑧=2. Give your answer to one decimal place.

Answer

The question asks us to find the perpendicular distance between a line and a plane. We need to determine whether they intersect. We first check if the line and plane are parallel. For the line and plane to be parallel, the direction vector of the line must be perpendicular to the normal vector of the plane. We can check this by computing their dot product. The direction vector of the line is (2,4,2) and the normal vector of the plane is (3,2,1), giving us (2,4,2)(3,2,1)=(2×3)+(4×(2))+(2×1)=68+2=0.

Since this is equal to zero, the line is perpendicular to the normal vector of the plane, which means they are parallel.

We recall that the distance between a parallel line and a plane is the same as the distance between any point on the line and the plane. Setting each part of the Cartesian equation of the line equal to zero and solving tells us that (1,2,5) lies on the line. We also know that the distance, 𝐷, between the point (𝑥,𝑦,𝑧) and the plane 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0 is given by 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐.

Substituting 𝑥=1, 𝑦=2, 𝑧=5, 𝑎=3, 𝑏=2, 𝑐=1, and 𝑑=2 into this formula gives us 𝐷=|3(1)+(2)(2)+(1)(5)+2|3+(2)+1=|4|141.1.lengthunits

Hence, the distance between the line and plane to one decimal place is 1.1 length units.

We can also use the formulae to determine the distance between two parallel planes. To do this, we could try finding the distance between an arbitrary point on each plane, which we will call 𝑃 and 𝑅.

However, if we compare this to the perpendicular distance, we can see that 𝑃𝑅 is the hypotenuse of a right triangle, which means it is longer than the perpendicular distance. This is true for any two points we choose. In other words, the shortest distance between two parallel planes is the perpendicular distance. In fact, since parallel planes stay the same distance apart, we can choose any point 𝑃 to be our starting point. Thus, we can use our formulae for the distance between a point and a plane to determine the distance between two parallel planes.

In our final example, we will see how to apply this process to find the distance between two parallel planes.

Example 5: Finding the Distance between Two Planes

Find the distance between the two planes 𝑥2𝑦2𝑧=2 and 2𝑥4𝑦4𝑧=3.

Answer

We want to find the distance between two planes. To do this, we will start by checking whether the two planes are parallel, since we can then apply the formula for the perpendicular distance.

We recall that two planes are parallel if the normal vectors to each plane are parallel. The normal vector to each plane is given by the coefficients, so the normal vectors of the two planes are (1,2,2) and (2,4,4), which are scalar multiples of each other. Hence, the planes are parallel.

We want to find a point on one of our planes; to do this, we can substitute 𝑥=0 and 𝑦=0 into the equation of our first plane: 02(0)2𝑧=22𝑧=2𝑧=1.

This means that the point (0,0,1) lies on the first plane. To find the distance between the two planes, we will find the distance between the point (0,0,1) and the plane 2𝑥4𝑦4𝑧=3.

We recall that the perpendicular distance, 𝐷, between the point (𝑥,𝑦,𝑧) and the plane 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0 is given by 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐.

To apply this, we need to rewrite the equation of our plane by subtracting 3 from both sides of the equation: 2𝑥4𝑦4𝑧3=0.

This gives us 𝑎=2, 𝑏=4, 𝑐=4, and 𝑑=3. Substituting these values and our point into our formula gives us 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐=|(2)(0)+(4)(0)+(4)(1)+(3)|(2)+(4)+(4)=|7|36=76.lengthunits

We can add length units to this value because we know it represents a length.

Hence, we were able to show that the distance between two planes 𝑥2𝑦2𝑧=2 and 2𝑥4𝑦4𝑧=3 is 76lengthunits.

Let us finish by recapping some of the important points of this explainer.

Key Points

  • The distance, 𝐷, between the point (𝑥,𝑦,𝑧) and the plane 𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0 is given by 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐.
  • The distance, 𝐷, between the point (𝑥,𝑦,𝑧) and the plane 𝑟(𝑎,𝑏,𝑐)=𝑑 is given by 𝐷=|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|𝑎+𝑏+𝑐.
  • The distance between a line parallel to a plane and said plane is equal to the distance between any point on the line and the plane.
  • The distance between two parallel planes is equal to the distance between any point on either plane and the other plane.
  • The perpendicular distance between a point and a plane is the shortest distance between these two objects.

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