Question

Let W be the solution set to the homogeneous system x + 2y + 3z = 0 2x + 4y + 6z = 0 Then W is a subspace of R3. Compute The Distance Between Y =[1 1 1] And W.

Answer 1

Explanation:

From the given information:

We can see that:

`x + 2y + 3z = 0 --- (1) \\ \\ 2x + 4y + 6z = 0 --- (2)`

From equation (1), if we multiply it by 2, we will get what we have in equation (2).

It implies that,

x + 2y + 3z = 0 ⇔ 2x + 4y + 6z = 0

And, W satisfies the equation x + 2y + 3z = 0

i.e.

W = {(x,y,z) ∈ R³║x+2y+3z = 0}

Now, to determine the distance through the plane W and point is;

`y = [1 \ 1 \ 1]^T`

Here, the normal vector
`n = [1\ 2\ 3]^T` is related to the plane x + 2y + 3z = 0

Suppose θ is the angle between the plane W and the point
`y = [1 \ 1 \ 1]^T`, then the distance is can be expressed as:

`||y|cos \theta| = (n*y)/(|n|)`

`||y|cos \theta| = ([1 \ 2\ 3 ]^T [1 \ 1 \ 1] ^T)/(√(1^2+2^2+3^2))`

`||y|cos \theta| = ([1+ 2+ 3 ])/(√(1+4+9))`

`||y|cos \theta| = (6)/(√(14))`

`||y|cos \theta| = 3\sqrt{(2)/(7)}`