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

## Force Vector Component?

**Moderator:** msmod

### Re: Force Vector Component?

1) CorrectBoris 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

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.

### Re: Force Vector Component?

For the plane in 3-space Π:

⇒

⇒ (λ/|

(cont.)

*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**= λ**û**, 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**= λ**û**into the plane equation, which is**u**•**x**=*d*, and solve for λ, so we have:**u**•(λ**û**) =*d*⇒

**u**•(λ(**u**/|**u**|)) =*d*⇒ (λ/|

**u**|)**u**•**u**=*d*(cont.)

Last edited by a.s.h. on Wed Jul 08, 2015 1:20 pm, edited 1 time in total.

### Re: Force Vector Component?

(cont.)

⇒ (λ/|

⇒ λ|

⇒ λ =

So for the point

Hence the required distance

= |λ

= |λ|, since |λ

= |

(cont.)

⇒ (λ/|

**u**|) |**u**|² =*d*(using**u**•**u**= |**u**|²)⇒ λ|

**u**| =*d*⇒ λ =

*d*/|**u**| (rearranging).So for the point

*P*on the plane closest to*O*, its position vector is λ**û**, where λ =*d*/|**u**| (as we just found).Hence the required distance

*D*, which is just the length of**OP**, is*D*= |**OP**|= |λ

**û**|, where λ =*d*/|**u**|= |λ|, since |λ

**û**| = |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.

### 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.

Have a nice day.