Answer:
Step-by-step explanation:
From the given information:
We can see that:
[tex]x + 2y + 3z = 0 --- (1) \\ \\ 2x + 4y + 6z = 0 --- (2)[/tex]
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;
[tex]y = [1 \ 1 \ 1]^T[/tex]
Here, the normal vector [tex]n = [1\ 2\ 3]^T[/tex] is related to the plane x + 2y + 3z = 0
Suppose θ is the angle between the plane W and the point [tex]y = [1 \ 1 \ 1]^T[/tex], then the distance is can be expressed as:
[tex]||y|cos \theta| = \dfrac{n*y}{|n|}[/tex]
[tex]||y|cos \theta| = \dfrac{[1 \ 2\ 3 ]^T [1 \ 1 \ 1] ^T}{\sqrt{1^2+2^2+3^2}}[/tex]
[tex]||y|cos \theta| = \dfrac{[1+ 2+ 3 ]}{\sqrt{1+4+9}}[/tex]
[tex]||y|cos \theta| = \dfrac{6}{\sqrt{14}}[/tex]
[tex]||y|cos \theta| = 3\sqrt{\dfrac{2}{7}}[/tex]
