# 3-dimensional vector problem

1. Oct 20, 2011

### ibysaiyan

1. The problem statement, all variables and given/known data
The set of questions is:
The points A, B, C have position vectors
a = 5i+4j+2k
b = 4i−5j+3k
c = 2i−j−2k

(d) Find the vector equation for the line
passing through A and C
(e) Find the nearest distance from the
point B to this line.

2. Relevant equations
r=A+$\lambda B$

3. The attempt at a solution

My working is as following:
First I need to find out vector AC [direction vector] which is : -3i -5j -4k, so when I substitute in -3i -5j -4k into vector B and any of the given points, this should give me the equation of the line. Which's : r = 5i+4j+2k $\lambda B$ (-3i -5j-4k)

Part e )

Now I think there are two ways of finding out the shortest distance between a given point and a line.
The question's :
My working:
Let H be the point which's perpendicular to the line and is at a shorter distance.
Since H is on the line , then using the equation of the line I get OH = ( 5-3 $\lambda$ , -5$\lambda$ +4, -4$\lambda$ +2k)

Therefore, Vector BH = H + (-B) = ( 1-3$\lambda$ , -5$\lambda$ +9 ,-4$\lambda$ -1)
The condition which arises is that BH is perpendicular to direction vector of the equation i.e : -3i -5j-4k

In other words dot product = 0 ,
which gives me the value of lamba = 22/25

In the end the distance of the length OH which i have found out is :about 2.7 to s.f)

Is my working right ? also I wanted to ask that.. on my summary booklet, there's a much easier way to find this length : The formula is :
d = | (p-a) x B^ | (equation 1)

Ironically, I know how to derive the above equation (1)
BUT i don't understand it :(

Any help , feedback will be appreciated! :)

Last edited: Oct 20, 2011
2. Oct 21, 2011

### ibysaiyan

Can anyone give a feedback over my working . Thanks.

3. Oct 22, 2011

### ibysaiyan

Anyone ? =]

4. Oct 22, 2011

### HallsofIvy

Staff Emeritus
What? You were asked to find the line that includes A and C. That has nothing to do with point B!

That "B" should not be in there. Other wise, yes, that is correct. Notice that you could check by seeing that A and C are both on that line: when $\lambda= 0$ $r= 5i+ 4j+ 2k$ which is A. When $\lambda= 1$, [itex]r= 5i+ 4j+ 2k- 3i- 5j- 4k= 2i- j- 2k which is C.

[/quote]Part e )

Now I think there are two ways of finding out the shortest distance between a given point and a line.
The question's :

My working:
Let H be the point which's perpendicular to the line and is at a shorter distance.[/quote]
I have no idea what you could mean by saying that a point is perpendicular to a line. I think you mean a point on the line such that the line HB is perpendicular to the line.

I'm confused. B was 2i- j- 2k. Where did i+ ij- k come from?

How could you possibly derive an equation if you don't understand it?
I don't understand it because I have no idea what "p" is. Is "a" the same as "A" before? What in the world is that "^"? But precisely because I don't understand what those mean, I
couldn't possibly derive it!

What I would do is start by calculating the equation of the plane perpendicular to the line AC containing the point B. Here, you already have the direction vector, -3i- 5j- 4k, and point B is (4, -5, 3) so that plane is -3(x- 4)- 5(y+ 5)- 4(z- 3)= 0 or 3x+ 5y+ 4z= -1. Find the point where line AC intersects the plane and then find the distance from that point to B.

5. Oct 27, 2011

### ibysaiyan

Again thanks for your help hallsofIvy. I plan on going through vector problems over the weekend. I want to make an aplogy over my late reply, workload has kept me busy BUT I am loving it, this is physics <3 Beautiful from every aspect.