Find the closest point in the plane spanned by (1,2,-1)^T,(0,-1,3)^T to the point (1,1,1)^T. Wha

Find the closest point in the plane spanned by...

Key Concepts

Span of Vectors

In linear algebra, a span refers to the set of all possible linear combinations of a given set of vectors. When you are given two vectors and asked to find a plane, you are really considering all vectors that can be formed by scaling and adding those vectors, which creates a subspace of the entire vector space.

Orthogonal Projection

The orthogonal projection is the process of 'dropping' a perpendicular from a point to a subspace (such as a plane) in order to find the point in the subspace that is closest to the original point. This concept is critical in many areas including solving least squares problems and minimizing distances in Euclidean spaces.

Normal Vector

A normal vector is a vector that is perpendicular to a plane or subspace. In the context of finding the distance from a point to a plane, the normal vector is used to determine the direction along which the distance is measured and to compute the orthogonal projection by removing the component of the point that is orthogonal to the plane.

Distance in Euclidean Space

Distance in Euclidean space is typically measured using the norm (or magnitude) of the difference between two vectors. When finding the distance from a point to a plane, the distance is given by the norm of the vector connecting the point to its orthogonal projection onto the plane, which represents the shortest path between the point and the plane.

Transcript

00:01 Okay, so trying to get closest to the point one -one -one with giving this equation x plus 2y plus 3z equals 13 so if we want our f -of -x y equation x y -z equation it's gonna look like x minus 1 squared plus y minus 1 squared plus z minus 1 squared given that our points are 1 -11 1 to find a distance to that, it would be x minus the point squared, y minus the point squared, and z minus the point squared.

00:47 So that's why it's going to be minus one and not like a plus one.

00:52 And then our g of x, y, z is going to be equal to x plus 2y plus 3 z minus 13 is equal to 0.

01:05 Now, del of f of x y z is going to be equal to lambda times, del o g x y z so we're gonna have two times x minus one and i hat plus two times y minus one and j hat plus two times z minus one the k hat it's going to be equal to lambda i hat plus two lambda j hat plus 3 lambda k hat so that means that we're going to have i'm going to simplify a little bit so 2x minus 2 is going to be equal to lambda 2 y minus 2 is going to be equal to 2 lambda and then 2 z minus 2 is going to be equal to 3 lambda all right and then i'm going to solve each one of these for the variable itself, x, y, z.

02:32 So for x, i'm going to get x is equal to lambda plus 2 over 2.

02:39 Y is going to be 2 lambda plus 2 over 2.

02:47 We could simplify that if we want.

02:49 Y could also be equal to lambda plus 1.

02:53 And then z is going to be equal to 3 lambda plus 2 over 2.

03:09 Okay, so now that we have all these, we can plug it back into our g of x, y, z equation.

03:16 So we're going to have x, lambda, plus 2 over 2, plus y, plus 2, so there's going to be 2 times land up plus 2 over 2, sorry, 2 land up plus 2 over 2, plus 3 z, 3 times 3 lambda plus 2 over 2 over 2.

03:51 Is going to be equal to 13.

03:57 So these are going to cancel.

04:01 But what i'm going to do is i'm going to make all these denominators common.

04:07 So i'm going to have actually i'm not going to cancel these out...

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