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.

Step-by-step explanation:

From the given information:

We can see that:

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;

Here, the normal vector is related to the plane x + 2y + 3z = 0

Suppose θ is the angle between the plane W and the point , then the distance is can be expressed as:

asdsaddsa

step-by-step explanation:

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm