find the distance from the point to the plane. (0,-1,0), 2 x+y+2 z=4

Name: Problem 43, Chapter 12: Thomas Calculus
Uploaded: 2020-02-01T21:49:13-08:00
Duration: 6 min 6 s
Channel: Robert Fisher
Description: find the distance from the point to the plane. (0,-1,0), 2 x+y+2 z=4

Find the distance from the point to the plane. ...

Key Concepts

Distance from a Point to a Plane

This concept refers to the shortest distance between a given point and a plane in three-dimensional space. It is typically determined by projecting the point onto the plane along a line perpendicular to it, using a formula that involves the coefficients of the plane’s equation.

Plane Equation

A plane in 3D space can be represented by a linear equation in the form ax + by + cz = d, where (a, b, c) constitute the normal vector to the plane. This equation provides a way to describe a flat, infinitely extending surface in space.

Normal Vector

The normal vector is a vector that is perpendicular to a plane and is derived from the coefficients of the plane's equation. It is essential for calculating distances and angles, as it defines the direction perpendicular to the plane.

Distance Formula

In analytic geometry, the distance from a point to a plane is calculated using the formula |ax0 + by0 + cz0 - d| divided by the magnitude of the normal vector. This formula ensures that the computed distance is the shortest path from the point to the plane.

Transcript

00:01 We're looking to find the distance between the point 0 negative 1, 0, and the plane 2x plus y plus 2z equals 4.

00:11 So we have a plane and we have some point at 0 negative 1 ,0, and we are looking for this distance.

00:23 The distance to the plane is perpendicular to the plane, orthogonal to the plane, which means this distance.

00:32 Is on the normal.

00:36 So although we are only looking for this part or something longer, but something along this normal.

00:46 If i knew another point on this plane, say this point here, p1, i could find a vector going up here and this vector here would be the normal vector, same as this vector.

01:11 And if i project, you know, find what the projection of this vector here is on the normal, that would be this height for d.

01:23 So what we know is if i have some vector along here, we'll call it vector u.

01:34 And i have some other vector here we call v.

01:39 If they're separated by angle theta, this distance here, which is the projection.

01:44 On vector u of vector v would be v times the cosine of the angle and between them.

01:58 The dot product u dot v is magnitude of u, magnitude of v, times the cosine of theta.

02:12 You notice this part and this part are shared, and so the projection on vector u of v is going to be the dot product of u and v divided by this vector u that we're projecting on...

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