3-dimensional vector problem

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Homework Statement


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.


Homework Equations


r=A+[itex]\lambda B[/itex]


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 [itex]\lambda B[/itex] (-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 :
(e) Find the nearest distance from the
point B to this line.

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 [itex]\lambda [/itex] , -5[itex]\lambda [/itex] +4, -4[itex]\lambda [/itex] +2k)

Therefore, Vector BH = H + (-B) = ( 1-3[itex]\lambda [/itex] , -5[itex]\lambda [/itex] +9 ,-4[itex]\lambda [/itex] -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:

Answers and Replies

  • #2
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Can anyone give a feedback over my working . Thanks.
 
  • #3
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Anyone ? =]
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement


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.


Homework Equations


r=A+[itex]\lambda B[/itex]


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

and any of the given points, this should give me the equation of the line. Which's : r = 5i+4j+2k [itex]\lambda B[/itex] (-3i -5j-4k)
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 [itex]\lambda= 0[/itex] [itex]r= 5i+ 4j+ 2k[/itex] which is A. When [itex]\lambda= 1[/itex], [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.

Since H is on the line , then using the equation of the line I get OH = ( 5-3 [itex]\lambda [/itex] , -5[itex]\lambda [/itex] +4, -4[itex]\lambda [/itex] +2k)
2i−j−2k

Therefore, Vector BH = H + (-B) = ( 1-3[itex]\lambda [/itex] , -5[itex]\lambda [/itex] +9 ,-4[itex]\lambda [/itex] -1)
I'm confused. B was 2i- j- 2k. Where did i+ ij- k come from?

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! :)
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
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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.
 

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