Finding the Shortest Distance from a Point to a Line in 3-Dimensional Space

[ibysaiyan]

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+\lambda B

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 :

(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 \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! :)

 

[ibysaiyan]

Can anyone give a feedback over my working . Thanks.

 

[ibysaiyan]

Anyone ? =]

 

[HallsofIvy]

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+\lambda B

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 \lambda B (-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 \lambda= 0 r= 5i+ 4j+ 2k which is A. When \lambda= 1, r= 5i+ 4j+ 2k- 3i- 5j- 4k= 2i- j- 2k which is C.<br /> <br /> [/quote]Part e )<br /> <br /> Now I think there are two ways of finding out the shortest distance between a given point and a line.<br /> The question&#039;s : <br /> <br /> My working:<br /> Let H be the point which&#039;s perpendicular to the line and is at a shorter distance.[/quote]<br /> I have no idea what you could mean by saying that a point is perpendicular to a line. I <b>think</b> you mean a point on the line such that the line HB is perpendicular to the line.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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)<br /> 2i−j−2k<br /> <br /> Therefore, Vector BH = H + (-B) = ( 1-3\lambda , -5\lambda +9 ,-4\lambda -1) </div> </div> </blockquote> I&#039;m confused. B was 2i- j- 2k. Where did i+ ij- k come from?<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> The condition which arises is that BH is perpendicular to direction vector of the equation i.e : -3i -5j-4k<br /> <br /> In other words dot product = 0 ,<br /> which gives me the value of lamba = 22/25<br /> <br /> In the end the distance of the length OH which i have found out is :about 2.7 to s.f)<br /> <br /> Is my working right ? also I wanted to ask that.. on my summary booklet, there&#039;s a much easier way to find this length : The formula is :<br /> d = | (p-a) x B^ | (equation 1)<br /> <br /> Ironically, I know how to derive the above equation (1)<br /> BUT i don&#039;t understand it :(<br /> <br /> <br /> <br /> Any help , feedback will be appreciated! :) </div> </div> </blockquote> How could you possibly derive an equation if you don&#039;t understand it?<br /> <b>I</b> don&#039;t understand it because I have no idea what &quot;p&quot; is. Is &quot;a&quot; the same as &quot;A&quot; before? What in the world is that &quot;^&quot;? But precisely because I don&#039;t understand what those mean, I <br /> couldn&#039;t possibly derive it! <br /> <br /> What I would do is start by calculating the equation of the <b>plane</b> 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.

 

[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.