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Projection of line to plane

  • Thread starter Theofilius
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Homework Statement



Hello! :smile:

Find the equation of the projection of the line [tex]\frac{x}{4}=\frac{y-4}{3}=\frac{z+1}{-2}[/tex] of the plane x-y+3z+8=0.

So the line projects its self on the plane...

Homework Equations





The Attempt at a Solution



First I find equation of line which passes throught the line and it has vector "a" parallel to the normal vector of the plane...

The equation of that line is: [tex]\frac{x}{1}=\frac{y-4}{-1}=\frac{z+1}{3}[/tex]

The equation of the line we need to find is probably: [tex]\frac{x-x_1}{4}=\frac{y-y_1}{3}=\frac{z-z_1}{-2}[/tex]

So we just need to find the point...

That point is [tex]M(\frac{1}{11},4-\frac{1}{11},-1+3\frac{1}{11})[/tex]

But in my text book results I get tottaly different solution: [tex]\frac{x+9}{7}=\frac{y+1}{4}=\frac{z}{-1}[/tex]

I don't know what is the problem. Please help.
 
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Answers and Replies

  • #2
tiny-tim
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Hello Theofilius! :smile:

Call the line L, the normal to the plane N, and the projection P.

Hint: what mathematical relationship is there between the three lines L N and P? :smile:
 
  • #3
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I don't know, but recently I figured out that [tex]\vec{a} \circ \vec{n} \neq 0[/tex]. So, my way of solving the task is wrong...
 
  • #4
tiny-tim
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So, my way of solving the task is wrong...
(Well, we both knew that! :rolleyes: )

Do you know how to draw a projection?

If so, draw the diagram, and then:
what mathematical relationship is there between the three lines L N and P? :smile:
 
  • #5
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Like http://img220.imageshack.us/img220/8030/72569812kh7.jpg" [Broken] P is parallel to L
 
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  • #6
tiny-tim
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hmm … useful tip … you're looking for triangles and suchlike, so try to get everything to join up.

Diagrams are supposed to help you … to stop you having to "leave things to the imagination".

So draw it again, for yourself, with the line actually meeting the plane.

And mark in any right-angles!

Then what do you notice about L N and P? :smile:
 
  • #8
tiny-tim
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Hi Theofilius! :smile:

Let's just describe it:

Let the line L meet the plane at the point A.

Let a normal to the plane meet the line L at B and the plane at C.

Then ABC is a right-angled triangle, isn't it?

L is AB, N is BC, P is CA.

So what's the relationship? :smile:
 
  • #9
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I understand now. So ABC is triangle. I don't know for what relationship you mean? Probably all lines are intersect each other... :smile:
 
  • #10
tiny-tim
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ABC is a triangle, so all three lines are in the same plane.

So the vector of the projection, CA, must be a linear combination of AB and BC, both of which are known. :smile:
 
  • #11
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What should I do now? Should I find the intersection point of the plane and the line?
 
  • #12
HallsofIvy
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You are not talking about just lines- you are talking about vectors. Vectors add and subtract.
 
  • #13
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But I don't know those vectors...
 
  • #14
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What are the coordinates of AB BC and AC?
 
  • #15
tiny-tim
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Theofilius, follow HallsofIvy's hint:
You are not talking about just lines- you are talking about vectors. Vectors add and subtract.
Take two fixed vectors, AB and BC.

Make any linear combination of those vectors …

what do you get? :smile:
 
  • #16
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Can we just don't complicate? Lets make like this. Look at http://i29.tinypic.com/5bxr44.jpg" I will choose points from the line, and make 2 new normal lines out from the points. After that I will find the two points which are intersecting with the plane... Then I will use this formula:

[tex]\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}[/tex]

Does this make any sense?
 
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  • #17
tiny-tim
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Hi Theofilius! :smile:

Yes, that will certainly work, once you've found those two points of intersection.

Have a go! :smile:
 
  • #18
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Unfortunately I don't get same solution as the text book results... I get for
[tex]\vec{a}(\frac{57}{11}, \frac{20}{11} , \frac{17}{11})[/tex]
and they get (7,4,-1)

the vector a is the vector parallel to the line that we are finding...
 
  • #19
tiny-tim
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Theofilius, how can we help by telling you where you've gone wrong, if you don't show us any working? :confused:
 
  • #20
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The first point on the line L:

A(0,4,-1) ;

the second point on the line L:

B(4,7,-3)

The line that is passing throught A(0,4,-1) and is normal to the plane:

[tex]\frac{x}{1}=\frac{y-4}{-1}=\frac{z+1}{3}[/tex]

The line that is passing throught B(4,7,-3) and is normal to the plane:

[tex]\frac{x-4}{1}=\frac{y-7}{-1}=\frac{z+3}{3}[/tex]

The interesection point with the line L and the plane:

[tex](Aa_1+Ba_2+Ca_3)t+Ax_1+By_1+Cz_1+D=0[/tex]

[tex]at+b=0[/tex]

[tex]t=\frac{-b}{a}[/tex]

The parametric equation of the line which is passing through A:

[tex]x=t[/tex]

[tex]y=4-t[/tex]

[tex]z=-1+3t[/tex]

[tex]a=1*1+(-1)(-1)+3*3=1+1+9=11[/tex]

[tex]b=0*1+4*(-1)+(-1)*3+8=-4-3+8=1[/tex]

So [tex]t=\frac{-1}{11}[/tex]

Now we are substituting for t

[tex]x=\frac{-1}{11}[/tex]

[tex]y=4-\frac{-1}{11}=\frac{45}{11}[/tex]

[tex]z=-1+3*\frac{-1}{11}=\frac{-14}{11}[/tex]

The first point which is lying on the plane is:

[tex]M_1(\frac{-1}{11},\frac{45}{11},\frac{-14}{11})[/tex]


The parametric equation of the line which is passing throught B:

[tex]x=4+t[/tex]

[tex]y=7-t[/tex]

[tex]z=-3+3t[/tex]

a=11

b=4-7-9=-12

[tex]t=\frac{12}{11}[/tex]

Now we are substituting for t:

[tex]x=4+\frac{12}{11}=\frac{56}{11}[/tex]

[tex]y=7-\frac{12}{11}=\frac{65}{11}[/tex]

[tex]z=-3-\frac{12}{11}=\frac{-45}{11}[/tex]

The second point which is lying on the plane is:

[tex]M_2(\frac{56}{11},\frac{65}{11},\frac{-45}{11})[/tex]

Now the equation of the projection line that we are looking for will be:

[tex]\frac{x+\frac{1}{11}}{\frac{56}{11}+\frac{1}{11}}=\frac{y-\frac{45}{11}}{\frac{65}{11}-\frac{45}{11}}=\frac{z+\frac{14}{11}}{\frac{-45}{11}+\frac{14}{11}}[/tex]

[tex]\frac{x+\frac{1}{11}}{\frac{57}{11}}=\frac{y-\frac{45}{11}}{\frac{20}{11}}=\frac{z+\frac{14}{11}}{\frac{-31}{11}}[/tex]

How should I check if their answer is correct?
 
  • #21
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Did I solve correctly?
 
  • #22
tiny-tim
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[tex]z=-3+3t[/tex]

[tex]t=\frac{12}{11}[/tex]

Now we are substituting for t:

[tex]z=-3-\frac{12}{11}=\frac{-45}{11}[/tex]
Hi!

I haven't checked the rest yet, but shouldn't that be [tex]z=-3-\frac{36}{11}[/tex] ?
 
  • #23
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Even, if it is like that, my final answer will not be even close to my text book results. I think only one solution is possible to this task. So either mine, or their is correct. :smile:
 
  • #24
tiny-tim
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Even, if it is like that, my final answer will not be even close to my text book results. I think only one solution is possible to this task. So either mine, or their is correct. :smile:
Hi Theofilius! :smile:

Well, I get (7,4,-1), the same as the book, and I get it in about 4 lines.
Hi Theofilius! :smile:

Let's just describe it:

Let the line L meet the plane at the point A.

Let a normal to the plane meet the line L at B and the plane at C.

Then ABC is a right-angled triangle, isn't it?

L is AB, N is BC, P is CA.

So what's the relationship? :smile:
ABC is a triangle, so all three lines are in the same plane.

So the vector of the projection, CA, must be a linear combination of AB and BC, both of which are known. :smile:
You are not talking about just lines- you are talking about vectors. Vectors add and subtract.
Theofilius, follow HallsofIvy's hint:


Take two fixed vectors, AB and BC.

Make any linear combination of those vectors …

what do you get? :smile:
Can we just don't complicate?
Your method was complicated, and it gave the wrong result.

This time, follow the advice you've already been given …

you have three vectors in the same plane, and you know two of them and all the angles …

there should be an easy way of finding the third line, shouldn't there? :smile:
 
  • #25
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Why my way of solving is not correct?
 

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