WEBVTT
00:00:00.680 --> 00:00:16.120
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.
00:00:16.600 --> 00:00:25.600
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.
00:00:26.520 --> 00:00:38.480
Letβs begin by considering the plane with general equation ππ₯ plus ππ¦ plus ππ§ plus π equals zero, together with a point with coordinates π₯ sub one, π¦ sub one, π§ sub one.
00:00:39.200 --> 00:00:48.520
To find the shortest distance between these two objects, letβs firstly consider the distance between point π and a point π
that lies on the plane.
00:00:49.040 --> 00:00:57.200
We can show that this is not the shortest distance between point π and the plane by constructing the following right triangle.
00:00:58.040 --> 00:01:05.000
We choose a point π on the plane such that the line segment ππ is perpendicular to the plane.
00:01:05.600 --> 00:01:13.320
We can then see that the line segment ππ
is the hypotenuse of the right triangle, which means that it is longer than the other sides.
00:01:13.880 --> 00:01:17.520
In particular, it means that the length of ππ is shorter than ππ
.
00:01:18.000 --> 00:01:30.480
And as we can construct this triangle for any point π
that lies on the plane, then the line segment ππ is the shortest distance between the point π and the plane.
00:01:31.160 --> 00:01:38.440
In this video, we will just quote the formula that can be used to calculate this distance which we will call π·.
00:01:39.160 --> 00:02:06.600
The shortest, or perpendicular, distance from some point π with coordinates π₯ sub one, π¦ sub one, π§ sub one and a plane with equation ππ₯ plus ππ¦ plus ππ§ plus π equals zero is equal to the absolute value of ππ₯ sub one plus ππ¦ sub one plus ππ§ sub one plus π all divided by the square root of π squared plus π squared plus π squared.
00:02:07.160 --> 00:02:25.160
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 ππ and ππ
.
00:02:25.960 --> 00:02:33.680
Letβs now consider an example where we can use this formula to calculate the distance between a point and a plane.
00:02:34.360 --> 00:02:44.680
Find the distance between the point negative five, negative eight, negative six and the plane negative two π₯ plus π¦ plus two π§ equals seven.
00:02:45.320 --> 00:02:49.760
In this question, we are asked to find the distance between a point and a plane.
00:02:50.160 --> 00:03:02.000
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.
00:03:02.960 --> 00:03:05.480
There is a formula that helps us to do this.
00:03:06.120 --> 00:03:31.800
The perpendicular distance π· between a point π₯ sub one, π¦ sub one, π§ sub one and the plane ππ₯ plus ππ¦ plus ππ§ plus π equals zero is given by π· is equal to the absolute value of ππ₯ sub one plus ππ¦ sub one plus ππ§ sub one plus π all divided by the square root of π squared plus π squared plus π squared.
00:03:32.280 --> 00:03:36.960
We are given a point with coordinates negative five, negative eight, negative six.
00:03:37.360 --> 00:03:44.080
This means that π₯ sub one is equal to negative five, π¦ sub one equals negative eight, and π§ sub one is equal to negative six.
00:03:44.680 --> 00:03:50.000
We note that the equation of the plane is given in a slightly different format to what is required.
00:03:50.480 --> 00:04:00.280
By subtracting seven from both sides of the equation, we have negative two π₯ plus π¦ plus two π§ minus seven equals zero.
00:04:00.880 --> 00:04:11.600
Since π, π, and π are the coefficients of π₯, π¦, and π§, respectively, we have π is equal to negative two, π is equal to one, and π is equal to two.
00:04:11.880 --> 00:04:14.560
π is equal to the constant term negative seven.
00:04:15.320 --> 00:04:17.840
We can now substitute these values into the formula.
00:04:18.280 --> 00:04:25.560
The numerator simplifies to the absolute value of 10 plus negative eight plus negative 12 plus negative seven.
00:04:26.200 --> 00:04:30.880
And the denominator is the square root of four plus one plus four.
00:04:31.640 --> 00:04:36.080
And this, in turn, is equal to the absolute value of negative 17 over root nine.
00:04:36.800 --> 00:04:45.800
The absolute value of a number is its distance from zero, so the absolute value of negative 17 is 17.
00:04:46.360 --> 00:04:52.800
And since the square root of nine is three, we have π· is equal to 17 over three.
00:04:53.520 --> 00:05:08.720
And we can therefore conclude that the distance between the point negative five, negative eight, negative six and the plane negative two π₯ plus π¦ plus two π§ equals seven is 17 over three length units.
00:05:09.600 --> 00:05:13.280
In this question, we calculated the distance between a point and a plane.
00:05:13.720 --> 00:05:20.440
We will now consider how we can adapt our formula to find the distance between a straight line and a plane.
00:05:20.960 --> 00:05:33.600
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.
00:05:33.960 --> 00:05:52.120
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.
00:05:52.840 --> 00:06:05.040
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.
00:06:05.680 --> 00:06:18.840
In the diagram shown, we choose an arbitrary point π with coordinate π₯ sub one, π¦ sub one, π§ sub one that lies on the line, together with an arbitrary point π
that lies on the plane.
00:06:19.440 --> 00:06:31.440
Once again, we see that the line segment ππ
is the hypotenuse of a right triangle, which means that the length of the line segment ππ is the shortest distance to the plane.
00:06:32.280 --> 00:06:34.320
This can be summarized as follows.
00:06:34.920 --> 00:07:06.880
The shortest distance π· between a parallel line and a plane where π₯ sub one, π¦ sub one, π§ sub one is any point on the line and the plane has the equation the scalar product of vector π« and the vector π, π, π is equal to negative π is given by capital π· is equal to the absolute value of ππ₯ sub one plus ππ¦ sub one plus ππ§ sub one plus π all divided by the square root of π squared plus π squared plus π squared.
00:07:07.640 --> 00:07:11.000
We will now look at an example where we need to calculate this distance.
00:07:11.760 --> 00:07:29.680
Find the perpendicular distance between the line π« which is equal to one, two, four plus π‘ multiplied by negative two, one, four and the plane which is equal to the scalar product of π« and two, zero, one equals one.
00:07:30.920 --> 00:07:38.040
This question involves finding the perpendicular, or shortest, distance between a line and a plane.
00:07:38.600 --> 00:07:43.200
Both the line and plane are currently given in vector form.
00:07:43.720 --> 00:07:48.120
When considering a line and a plane, there are two possibilities.
00:07:48.880 --> 00:07:52.640
Firstly, the line is parallel to the plane.
00:07:53.160 --> 00:07:56.520
Or secondly, it intersects the plane.
00:07:57.200 --> 00:08:02.600
If the line does intersect the plane, then the shortest distance between them is equal to zero.
00:08:03.480 --> 00:08:11.200
This means that the first question we need to ask is, do the line and plane intersect or are they parallel?
00:08:12.080 --> 00:08:14.600
Letβs begin by considering the equation of the line.
00:08:15.160 --> 00:08:23.480
This can be rewritten as π« is equal to one minus two π‘, two plus π‘, four plus four π‘.
00:08:24.200 --> 00:08:27.680
We can now substitute this vector into the equation of the plane.
00:08:28.120 --> 00:08:37.760
We have the dot, or scalar, product of one minus two π‘, two plus π‘, four plus four π‘ and two, zero, one is equal to one.
00:08:38.480 --> 00:08:50.480
Finding the dot product gives us the following equation, and this simplifies to two minus four π‘ plus four plus four π‘ equals one.
00:08:51.360 --> 00:08:57.640
On the left-hand side, the four π‘βs cancel, and we are left with six equals one.
00:08:58.160 --> 00:09:02.920
This is incorrect and is therefore not true for any value of π‘.
00:09:03.200 --> 00:09:11.520
And we can therefore conclude that the line and plane do not intersect and must therefore be parallel.
00:09:12.520 --> 00:09:23.200
The shortest distance between the line and the plane can therefore be found by taking any point π that lies on the line and finding the perpendicular distance to the plane.
00:09:23.840 --> 00:09:30.720
In order to find the position vector of any point that lies on the line, we can substitute any value of π‘ into our equation.
00:09:31.320 --> 00:09:37.000
For example, when π‘ is equal to zero, π« is equal to one, two, four.
00:09:37.600 --> 00:09:43.720
This means that the point with coordinates one, two, four lies on the line.
00:09:44.400 --> 00:09:47.320
And we will let this be point π.
00:09:48.160 --> 00:09:54.960
We will now recall the formula that enables us to calculate the perpendicular, or shortest, distance from a point to a plane.
00:09:55.600 --> 00:10:13.040
When the equation of the plane is written in vector form, as in this case, then the perpendicular distance capital π· is equal to the absolute value of ππ₯ sub one plus ππ¦ sub one plus ππ§ sub one plus π all divided by the square root of π squared plus π squared plus π squared.
00:10:13.440 --> 00:10:18.560
The values of π₯ sub one, π¦ sub one, and π§ sub one are one, two, and four, respectively.
00:10:19.560 --> 00:10:26.600
From the vector equation of the plane, we have π is equal to two, π is equal to zero, and π is equal to one.
00:10:27.280 --> 00:10:31.280
Since negative π is equal to one, π is equal to negative one.
00:10:32.120 --> 00:10:49.880
Substituting in our values, we have the distance π· 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.
00:10:50.400 --> 00:10:58.160
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.
00:10:58.680 --> 00:11:05.120
We can then rationalize the denominator by multiplying the numerator and denominator by root five.
00:11:05.360 --> 00:11:11.320
This is equal to five root five over five, which simplifies to root five.
00:11:11.920 --> 00:11:17.880
The perpendicular distance between the given line and plane is root five length units.
00:11:19.080 --> 00:11:26.640
Our final example in this video will involve using the formula to find the distance between two parallel planes.
00:11:27.640 --> 00:11:29.960
Letβs firstly consider how this can be done.
00:11:30.680 --> 00:11:34.480
We will begin by taking an arbitrary point π on one of the planes.
00:11:35.400 --> 00:11:41.680
We can then calculate the perpendicular distance between this point and the other plane as before.
00:11:42.640 --> 00:11:47.720
In the example that follows, the equations of the planes will be given in general form.
00:11:48.280 --> 00:11:55.960
However, it is important to note that we can use the same formula when the equations are given in vector form.
00:11:56.760 --> 00:12:09.440
Find the distance between the two planes negative π₯ minus two π¦ minus two π§ is equal to negative two and negative two π₯ minus four π¦ minus four π§ is equal to three.
00:12:10.200 --> 00:12:13.920
In this question, we are asked to find the distance between two planes.
00:12:14.560 --> 00:12:19.520
This means that we need to find the perpendicular, or shortest, distance between the two planes.
00:12:20.560 --> 00:12:28.320
When dealing with two planes, if they are not parallel, they will intersect and the distance between them will therefore be equal to zero.
00:12:29.240 --> 00:12:34.640
This means that the first question we need to ask is, are the planes parallel?
00:12:35.360 --> 00:12:40.320
One way to do this is to consider the normal vectors of the two planes.
00:12:41.160 --> 00:12:48.800
These are equal to the coefficients of π₯, π¦, and π§ when the equation of the plane is written in general form.
00:12:49.680 --> 00:12:54.560
The first plane has normal vector negative one, negative two, negative two.
00:12:55.400 --> 00:13:00.640
And the second plane has normal vector negative two, negative four, negative four.
00:13:01.280 --> 00:13:10.680
As these two vectors are scalar multiples of one another, we can conclude that the two planes are parallel.
00:13:11.600 --> 00:13:16.960
Letβs now recall how we can find the distance between two parallel planes.
00:13:17.880 --> 00:13:29.000
If we pick a point π on one of the planes, we can calculate the distance π· between the two planes by calculating the distance between point π and the other plane.
00:13:29.720 --> 00:13:54.520
This satisfies the formula π· is equal to the absolute value of ππ₯ sub one plus ππ¦ sub one plus ππ§ sub one plus π all divided by the square root of π squared plus π squared plus π squared, where point π has coordinates π₯ sub one, π¦ sub one, π§ sub one and the plane has equation ππ₯ plus ππ¦ plus ππ§ plus π equals zero.
00:13:55.480 --> 00:13:58.480
We begin by finding the coordinates of any point that lies on the first plane.
00:13:59.000 --> 00:14:04.960
One way of doing this is to choose the point where π₯ equals zero and π¦ equals zero.
00:14:05.880 --> 00:14:16.280
Substituting these values into the equation of the first plane, we have negative zero minus two multiplied by zero minus two π§ is equal to negative two.
00:14:16.920 --> 00:14:20.560
This simplifies to negative two π§ is equal to negative two.
00:14:21.320 --> 00:14:25.920
And dividing through by negative two, we have π§ is equal to one.
00:14:26.720 --> 00:14:32.960
This means that the coordinates of one point that lies on the first plane are zero, zero, one.
00:14:33.640 --> 00:14:39.000
And we now have values of π₯ sub one, π¦ sub one, and π§ sub one that we can substitute into our formula.
00:14:39.600 --> 00:14:47.320
By considering the equation of the second plane, we see that π is equal to negative two and both π and π are equal to negative four.
00:14:47.800 --> 00:14:51.680
These are the coefficients of π₯, π¦, and π§, respectively.
00:14:52.480 --> 00:15:01.560
Noting that in order to find π·, we need the equation of the plane to be equal to zero, we see that π· is equal to negative three.
00:15:02.000 --> 00:15:23.640
Substituting our values into the formula, we have the distance capital π· 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.
00:15:24.120 --> 00:15:31.880
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.
00:15:32.480 --> 00:15:39.640
We can therefore conclude that the distance between the two given planes is seven over six length units.
00:15:40.560 --> 00:15:44.480
We will now summarize the key points from this video.
00:15:45.040 --> 00:16:07.520
The distance π· between the point π₯ sub one, π¦ sub one, π§ sub one and the plane ππ₯ plus ππ¦ plus ππ§ plus π equals zero is given by capital π· is equal to the absolute value of ππ₯ sub one plus ππ¦ sub one plus ππ§ sub one plus π all divided by the square root of π squared plus π squared plus π squared.
00:16:08.080 --> 00:16:24.000
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 π« and vector π, π, π is equal to negative π, where π, π, π is a vector normal to the plane.
00:16:24.880 --> 00:16:34.400
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.
00:16:34.920 --> 00:16:45.040
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.
00:16:45.600 --> 00:16:55.360
It is important to note that when talking about this distance π·, we mean the shortest, or perpendicular, distance between the two objects.