Find the distance from (4,-2,6) to each of the following. (a) The x y-plane (b) The y z-plane (c

Find the distance from (4,-2,6) to each of the following....

Key Concepts

Cartesian Coordinate System

The Cartesian coordinate system in three dimensions defines points using three coordinates (x, y, and z), and includes important geometric objects such as planes (xy-plane, xz-plane, and yz-plane) and axes (x-axis, y-axis, and z-axis). Understanding this system is crucial as it allows us to describe locations and compute distances in space by considering the individual coordinate values.

Distance from a Point to a Plane

In three-dimensional geometry, the distance from a point to a plane is measured along the perpendicular (normal) direction to that plane. For coordinate planes like the xy-plane, this distance is simply the absolute value of the coordinate not present in the plane (in this case, the z-coordinate). This concept similarly applies to the other coordinate planes by taking the appropriate absolute coordinate value.

Distance from a Point to an Axis

The distance from a point to a coordinate axis is determined by projecting the point onto the plane perpendicular to that axis, and then calculating the Euclidean distance in that plane. This involves ignoring the coordinate that runs along the axis and using the remaining two coordinates to compute the distance using the Pythagorean theorem.

Pythagorean Theorem in 3D Geometry

The Pythagorean theorem is essential in determining distances in two dimensions, and by extension in three-dimensional geometry when working with projections. When calculating the distance from a point to an axis, we apply this theorem to the two coordinates in the plane perpendicular to the axis, ensuring that we measure the shortest distance (the perpendicular distance) from the point to the axis.

Transcript

00:02 Let's look at the point 4 -2 -6 and find its distances to each of the coordinate planes and each of the coordinate axes.

00:10 We're going to start with the x, y, plane.

00:19 Now, i'm going to do this the hard way the first time, just so we can see an easier way to do this and why that would work.

00:26 The distance between a point and a plane is the distance between the point and a point on the plane.

00:37 Along a perpendicular line from the point to the plane.

00:42 Well, that point on the plane has a name, and it's the projection.

00:47 And the coordinates of a projection of a point to the x, y plane, it's going to be the point with the same x and y coordinates, but with a z coordinate of zero.

01:02 So to find the distance a point is from the x, y, plane, we're going to find the distance it is from the point to its projection.

01:11 And so we're going to use the distance formula for this.

01:16 And then when we're done, we're going to see a much easier way that this could have been solved and why that way would work.

01:24 So let's subtract x coordinates 4 minus 4 squared, plus subtracting y coordinates negative 2 minus negative 2 squared, plus subtracting z coordinates 6 minus 0 squared.

01:40 Well, this gives me the square root of 0 squared.

01:43 Plus 0 squared plus 6 squared, which is the square root of 36, which is 6.

01:53 And it's not an accident or coincidence that this just happens to be the z coordinate.

02:02 If we remember how we find points to begin with, the z coordinate says how far a point is above or below the x, y plane.

02:14 The projection is going to have the same x and y coordinates and those are always going to subtract to zero and so the z coordinate is always going to tell us how far point away is from the y z plane from the xy plane excuse me when i want to look at how far point is from the y z plane again the projection is going to be the point that has an x coordinate of zero the y and z coordinates are going to be the same.

02:51 And so the only thing that's going to contribute to the distance, this point is from the y, z plane, is the x coordinate four.

02:58 Now, we're going to have to make sure that we remember to take the absolute value of that coordinate.

03:06 Excuse me, that will matter with the next problem, because lengths are always a positive quantity, non -negative quantity, excuse me.

03:15 So the projection of this point along the y, z plane, is going to be four, which is what the x -cordinates tells us how far our point is in front or in back of the y -z plane.

03:30 If i then look at the x -z plane, the x -z plane, i said z but managed to write y.

03:43 But look at the x -z plane, the y coordinate is going to tell us that.

03:49 This tells us that our point is two units to the left of the x -z -pain.

03:57 Plane.

03:58 So numerically, we're going to solve this by taking the absolute value of negative 2, and that gives us 2...

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