Lesson Video: The Perpendicular Distance between Points and Planes Mathematics

In this video, 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.

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Video Transcript

In this video, 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.

In order to find the shortest distance between a point and a plane, we first need to determine exactly what is meant by the shortest distance between these two geometric objects. Let’s begin by considering the plane with general equation π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero, together with a point with coordinates π‘₯ sub one, 𝑦 sub one, 𝑧 sub one.

To find the shortest distance between these two objects, let’s firstly consider the distance between point 𝑃 and a point 𝑅 that lies on the plane. We can show that this is not the shortest distance between point 𝑃 and the plane by constructing the following right triangle. We choose a point 𝑄 on the plane such that the line segment 𝑃𝑄 is perpendicular to the plane. We can then see that the line segment 𝑃𝑅 is the hypotenuse of the right triangle, which means that it is longer than the other sides. In particular, it means that the length of 𝑃𝑄 is shorter than 𝑃𝑅. And as we can construct this triangle for any point 𝑅 that lies on the plane, then the line segment 𝑃𝑄 is the shortest distance between the point 𝑃 and the plane.

In this video, we will just quote the formula that can be used to calculate this distance which we will call 𝐷. The shortest, or perpendicular, distance from some point 𝑃 with coordinates π‘₯ sub one, 𝑦 sub one, 𝑧 sub one and a plane with equation π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero is equal to the absolute value of π‘Žπ‘₯ sub one plus 𝑏𝑦 sub one plus 𝑐𝑧 sub one plus 𝑑 all divided by the square root of π‘Ž squared plus 𝑏 squared plus 𝑐 squared.

Whilst it is outside of the scope of this lesson, this formula can be derived using our knowledge of right angle trigonometry together with the scalar, or dot, product of two vectors, in this case, the vectors 𝑃𝑄 and 𝑃𝑅. Let’s now consider an example where we can use this formula to calculate the distance between a point and a plane.

Find the distance between the point negative five, negative eight, negative six and the plane negative two π‘₯ plus 𝑦 plus two 𝑧 equals seven.

In this question, we are asked to find the distance between a point and a plane. We recall that the distance between a point and a plane means the perpendicular distance since this is the shortest distance between the two objects. There is a formula that helps us to do this. The perpendicular distance 𝐷 between a point π‘₯ sub one, 𝑦 sub one, 𝑧 sub one and the plane π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero is given by 𝐷 is equal to the absolute value of π‘Žπ‘₯ sub one plus 𝑏𝑦 sub one plus 𝑐𝑧 sub one plus 𝑑 all divided by the square root of π‘Ž squared plus 𝑏 squared plus 𝑐 squared.

We are given a point with coordinates negative five, negative eight, negative six. This means that π‘₯ sub one is equal to negative five, 𝑦 sub one equals negative eight, and 𝑧 sub one is equal to negative six. We note that the equation of the plane is given in a slightly different format to what is required. By subtracting seven from both sides of the equation, we have negative two π‘₯ plus 𝑦 plus two 𝑧 minus seven equals zero. Since π‘Ž, 𝑏, and 𝑐 are the coefficients of π‘₯, 𝑦, and 𝑧, respectively, we have π‘Ž is equal to negative two, 𝑏 is equal to one, and 𝑐 is equal to two. 𝑑 is equal to the constant term negative seven.

We can now substitute these values into the formula. The numerator simplifies to the absolute value of 10 plus negative eight plus negative 12 plus negative seven. And the denominator is the square root of four plus one plus four. And this, in turn, is equal to the absolute value of negative 17 over root nine. The absolute value of a number is its distance from zero, so the absolute value of negative 17 is 17. And since the square root of nine is three, we have 𝐷 is equal to 17 over three. And we can therefore conclude that the distance between the point negative five, negative eight, negative six and the plane negative two π‘₯ plus 𝑦 plus two 𝑧 equals seven is 17 over three length units.

In this question, we calculated the distance between a point and a plane. We will now consider how we can adapt our formula to find the distance between a straight line and a plane. If the equation of our plane is given in vector form as opposed to general form, we can still use the same formula to find the shortest distance between a point and a plane. We can use the same process to find the shortest distance between a straight line and a plane, recalling that if a line and a plane are not parallel and not coincident, then they intersect, which means that the distance between them would be zero.

Then, if they are parallel and distinct, we can show that the shortest distance between them is the perpendicular distance between any point on the line and the plane. In the diagram shown, we choose an arbitrary point 𝑃 with coordinate π‘₯ sub one, 𝑦 sub one, 𝑧 sub one that lies on the line, together with an arbitrary point 𝑅 that lies on the plane. Once again, we see that the line segment 𝑃𝑅 is the hypotenuse of a right triangle, which means that the length of the line segment 𝑃𝑄 is the shortest distance to the plane.

This can be summarized as follows. The shortest distance 𝐷 between a parallel line and a plane where π‘₯ sub one, 𝑦 sub one, 𝑧 sub one is any point on the line and the plane has the equation the scalar product of vector 𝐫 and the vector 𝐚, 𝐛, 𝐜 is equal to negative 𝑑 is given by capital 𝐷 is equal to the absolute value of π‘Žπ‘₯ sub one plus 𝑏𝑦 sub one plus 𝑐𝑧 sub one plus 𝑑 all divided by the square root of π‘Ž squared plus 𝑏 squared plus 𝑐 squared. We will now look at an example where we need to calculate this distance.

Find the perpendicular distance between the line 𝐫 which is equal to one, two, four plus 𝑑 multiplied by negative two, one, four and the plane which is equal to the scalar product of 𝐫 and two, zero, one equals one.

This question involves finding the perpendicular, or shortest, distance between a line and a plane. Both the line and plane are currently given in vector form. When considering a line and a plane, there are two possibilities. Firstly, the line is parallel to the plane. Or secondly, it intersects the plane. If the line does intersect the plane, then the shortest distance between them is equal to zero. This means that the first question we need to ask is, do the line and plane intersect or are they parallel? Let’s begin by considering the equation of the line. This can be rewritten as 𝐫 is equal to one minus two 𝑑, two plus 𝑑, four plus four 𝑑.

We can now substitute this vector into the equation of the plane. We have the dot, or scalar, product of one minus two 𝑑, two plus 𝑑, four plus four 𝑑 and two, zero, one is equal to one. Finding the dot product gives us the following equation, and this simplifies to two minus four 𝑑 plus four plus four 𝑑 equals one. On the left-hand side, the four 𝑑’s cancel, and we are left with six equals one. This is incorrect and is therefore not true for any value of 𝑑. And we can therefore conclude that the line and plane do not intersect and must therefore be parallel.

The shortest distance between the line and the plane can therefore be found by taking any point 𝑃 that lies on the line and finding the perpendicular distance to the plane. In order to find the position vector of any point that lies on the line, we can substitute any value of 𝑑 into our equation. For example, when 𝑑 is equal to zero, 𝐫 is equal to one, two, four. This means that the point with coordinates one, two, four lies on the line. And we will let this be point 𝑃.

We will now recall the formula that enables us to calculate the perpendicular, or shortest, distance from a point to a plane. When the equation of the plane is written in vector form, as in this case, then the perpendicular distance capital 𝐷 is equal to the absolute value of π‘Žπ‘₯ sub one plus 𝑏𝑦 sub one plus 𝑐𝑧 sub one plus 𝑑 all divided by the square root of π‘Ž squared plus 𝑏 squared plus 𝑐 squared. The values of π‘₯ sub one, 𝑦 sub one, and 𝑧 sub one are one, two, and four, respectively. From the vector equation of the plane, we have π‘Ž is equal to two, 𝑏 is equal to zero, and 𝑐 is equal to one. Since negative 𝑑 is equal to one, 𝑑 is equal to negative one.

Substituting in our values, we have the distance 𝐷 is equal to the absolute value of two multiplied by one plus zero multiplied by two plus one multiplied by four plus negative one all divided by the square root of two squared plus zero squared plus one squared. This simplifies to the absolute value of five divided by the square root of five, which in turn is equal to five over root five. We can then rationalize the denominator by multiplying the numerator and denominator by root five. This is equal to five root five over five, which simplifies to root five. The perpendicular distance between the given line and plane is root five length units.

Our final example in this video will involve using the formula to find the distance between two parallel planes. Let’s firstly consider how this can be done. We will begin by taking an arbitrary point 𝑃 on one of the planes. We can then calculate the perpendicular distance between this point and the other plane as before. In the example that follows, the equations of the planes will be given in general form. However, it is important to note that we can use the same formula when the equations are given in vector form.

Find the distance between the two planes negative π‘₯ minus two 𝑦 minus two 𝑧 is equal to negative two and negative two π‘₯ minus four 𝑦 minus four 𝑧 is equal to three.

In this question, we are asked to find the distance between two planes. This means that we need to find the perpendicular, or shortest, distance between the two planes. When dealing with two planes, if they are not parallel, they will intersect and the distance between them will therefore be equal to zero. This means that the first question we need to ask is, are the planes parallel? One way to do this is to consider the normal vectors of the two planes. These are equal to the coefficients of π‘₯, 𝑦, and 𝑧 when the equation of the plane is written in general form. The first plane has normal vector negative one, negative two, negative two. And the second plane has normal vector negative two, negative four, negative four. As these two vectors are scalar multiples of one another, we can conclude that the two planes are parallel.

Let’s now recall how we can find the distance between two parallel planes. If we pick a point 𝑃 on one of the planes, we can calculate the distance 𝐷 between the two planes by calculating the distance between point 𝑃 and the other plane. This satisfies the formula 𝐷 is equal to the absolute value of π‘Žπ‘₯ sub one plus 𝑏𝑦 sub one plus 𝑐𝑧 sub one plus 𝑑 all divided by the square root of π‘Ž squared plus 𝑏 squared plus 𝑐 squared, where point 𝑃 has coordinates π‘₯ sub one, 𝑦 sub one, 𝑧 sub one and the plane has equation π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero.

We begin by finding the coordinates of any point that lies on the first plane. One way of doing this is to choose the point where π‘₯ equals zero and 𝑦 equals zero. Substituting these values into the equation of the first plane, we have negative zero minus two multiplied by zero minus two 𝑧 is equal to negative two. This simplifies to negative two 𝑧 is equal to negative two. And dividing through by negative two, we have 𝑧 is equal to one. This means that the coordinates of one point that lies on the first plane are zero, zero, one. And we now have values of π‘₯ sub one, 𝑦 sub one, and 𝑧 sub one that we can substitute into our formula.

By considering the equation of the second plane, we see that π‘Ž is equal to negative two and both 𝑏 and 𝑐 are equal to negative four. These are the coefficients of π‘₯, 𝑦, and 𝑧, respectively. Noting that in order to find 𝐷, we need the equation of the plane to be equal to zero, we see that 𝐷 is equal to negative three. Substituting our values into the formula, we have the distance capital 𝐷 is equal to the absolute value of negative two multiplied by zero plus negative four multiplied by zero plus negative four multiplied by one plus negative three all divided by the square root of negative two squared plus negative four squared plus negative four squared. This is equal to the absolute value of negative seven over the square root of 36, which in turn is equal to seven over six.

We can therefore conclude that the distance between the two given planes is seven over six length units.

We will now summarize the key points from this video. The distance 𝐷 between the point π‘₯ sub one, 𝑦 sub one, 𝑧 sub one and the plane π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero is given by capital 𝐷 is equal to the absolute value of π‘Žπ‘₯ sub one plus 𝑏𝑦 sub one plus 𝑐𝑧 sub one plus 𝑑 all divided by the square root of π‘Ž squared plus 𝑏 squared plus 𝑐 squared. We can use the same formula when the equation of the plane is given in vector form such that the dot, or scalar, product of vector 𝐫 and vector 𝐚, 𝐛, 𝐜 is equal to negative 𝑑, where 𝐚, 𝐛, 𝐜 is a vector normal to the plane.

We also saw that the distance between a line parallel to a plane and that plane is equal to the distance between any point on the line and the plane. In a similar way, the distance between two parallel planes is equal to the distance between any point on either plane and the other plane. It is important to note that when talking about this distance 𝐷, we mean the shortest, or perpendicular, distance between the two objects.

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