## Force Vector Component?

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Boris Lagutin

### Re: Force Vector Component?

Thank you very much, a.s.h!

What I have understood clearly now:

1) if we take a vector coming from the origin to any point on some plane we always get the same number on RHS (doesn't matter how many points on the plane we would take);

2) "d" (number on RHS in the plane equation) shows how far the plane may be from the origin if we consider a vector in 1 as a position vector of the point on the plane.

Are 1 and 2 correct? I hope yes a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Re: Force Vector Component?

Boris Lagutin wrote:Thank you very much, a.s.h!

What I have understood clearly now:

1) if we take a vector coming from the origin to any point on some plane we always get the same number on RHS (doesn't matter how many points on the plane we would take);

2) "d" (number on RHS in the plane equation) shows how far the plane may be from the origin if we consider a vector in 1 as a position vector of the point on the plane.

Are 1 and 2 correct? I hope yes 1) Correct

2) The value of d is a measure of how far the plane is from the origin (if d = 0, the plane passes through the origin, and if d ≠ 0, then the plane does not pass through the origin). However, in general, d is not equal to the distance D of the plane from the origin. I'll derive an expression for D in terms of d in the below posts.
Last edited by a.s.h. on Wed Jul 08, 2015 1:22 pm, edited 3 times in total.
a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Re: Force Vector Component?

For the plane in 3-space Π: ax + by + cz = d (Π is capital Greek letter Pi, commonly used to name a plane), let D be the distance of Π from the origin O (by this, we mean the shortest distance). A normal vector to Π is u = <a,b,c>. Hence a normal unit vector is û = u/|u| = (1/√(a² + b² + c²)) <a,b,c>. Note the line ℓ with direction û passing through the origin has parametric vector equation ℓ: x = λû, λ ∈ ℝ. The point P on Π closest to O is going to be the intersection of ℓ and Π, because the shortest distance of a point to a plane is the perpendicular distance. To find this intersection, we substitute x = λû into the plane equation, which is ux = d, and solve for λ, so we have:

u•(λû) = d

u•(λ(u/|u|)) = d

⇒ (λ/|u|)uu = d

(cont.)
Last edited by a.s.h. on Wed Jul 08, 2015 1:20 pm, edited 1 time in total.
a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Re: Force Vector Component?

(cont.)

⇒ (λ/|u|) |u|² = d (using uu = |u|²)

⇒ λ|u| = d

⇒ λ = d/|u| (rearranging).

So for the point P on the plane closest to O, its position vector is λû, where λ = d/|u| (as we just found).

Hence the required distance D, which is just the length of OP, is

D = |OP|

= |λû|, where λ = d/|u|

= |λ|, since |λû| = |λ||û|=|λ|⋅1 = |λ|. (Note |λ| refers to the absolute value of λ, not a vector length, since λ is a scalar.)

= |d|/|u|, since λ = d/|u| (|d| refers to absolute value of d).

(cont.)
Last edited by a.s.h. on Wed Jul 08, 2015 1:19 pm, edited 2 times in total.
a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Re: Force Vector Component?

Now we have found how d affects the distance from the origin, as we have shown D = |d|/|u|. From this equation, we can see that for a fixed normal vector u, the distance of the plane to the origin is proportional to |d| (the sign of d tells us in which direction the plane is moved away from the origin). Also, D will be equal to |d| if and only if u is a unit normal (length 1).
Boris Lagutin

### Re: Force Vector Component?

Thank you, a.s.h., so much! Your help is priceless. I couldn't grasp the vector and plane conception clearly because I didn't understand that we use position vectors and for all points on a plane "d" (if we insert point's coordinates in a plane equation) is the same (I got confused about vectors and plane equations). Now I can make a next step in multi-variable calculus. Truly I have already done Have a nice day.