# Projection of line to plane

## Homework Statement

Hello! Find the equation of the projection of the line $$\frac{x}{4}=\frac{y-4}{3}=\frac{z+1}{-2}$$ of the plane x-y+3z+8=0.

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

## 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: $$\frac{x}{1}=\frac{y-4}{-1}=\frac{z+1}{3}$$

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

So we just need to find the point...

That point is $$M(\frac{1}{11},4-\frac{1}{11},-1+3\frac{1}{11})$$

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

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tiny-tim
Homework Helper
Hello Theofilius! 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? I don't know, but recently I figured out that $$\vec{a} \circ \vec{n} \neq 0$$. So, my way of solving the task is wrong...

tiny-tim
Homework Helper
So, my way of solving the task is wrong...
(Well, we both knew that! )

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? Like http://img220.imageshack.us/img220/8030/72569812kh7.jpg" [Broken] P is parallel to L

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tiny-tim
Homework Helper
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? tiny-tim
Homework Helper
Hi Theofilius! 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? I understand now. So ABC is triangle. I don't know for what relationship you mean? Probably all lines are intersect each other... tiny-tim
Homework Helper
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. What should I do now? Should I find the intersection point of the plane and the line?

HallsofIvy
Homework Helper

But I don't know those vectors...

What are the coordinates of AB BC and AC?

tiny-tim
Homework Helper
Take two fixed vectors, AB and BC.

Make any linear combination of those vectors …

what do you get? 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:

$$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$$

Does this make any sense?

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tiny-tim
Homework Helper
Hi Theofilius! Yes, that will certainly work, once you've found those two points of intersection.

Have a go! Unfortunately I don't get same solution as the text book results... I get for
$$\vec{a}(\frac{57}{11}, \frac{20}{11} , \frac{17}{11})$$
and they get (7,4,-1)

the vector a is the vector parallel to the line that we are finding...

tiny-tim
Homework Helper
Theofilius, how can we help by telling you where you've gone wrong, if you don't show us any working? 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:

$$\frac{x}{1}=\frac{y-4}{-1}=\frac{z+1}{3}$$

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

$$\frac{x-4}{1}=\frac{y-7}{-1}=\frac{z+3}{3}$$

The interesection point with the line L and the plane:

$$(Aa_1+Ba_2+Ca_3)t+Ax_1+By_1+Cz_1+D=0$$

$$at+b=0$$

$$t=\frac{-b}{a}$$

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

$$x=t$$

$$y=4-t$$

$$z=-1+3t$$

$$a=1*1+(-1)(-1)+3*3=1+1+9=11$$

$$b=0*1+4*(-1)+(-1)*3+8=-4-3+8=1$$

So $$t=\frac{-1}{11}$$

Now we are substituting for t

$$x=\frac{-1}{11}$$

$$y=4-\frac{-1}{11}=\frac{45}{11}$$

$$z=-1+3*\frac{-1}{11}=\frac{-14}{11}$$

The first point which is lying on the plane is:

$$M_1(\frac{-1}{11},\frac{45}{11},\frac{-14}{11})$$

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

$$x=4+t$$

$$y=7-t$$

$$z=-3+3t$$

a=11

b=4-7-9=-12

$$t=\frac{12}{11}$$

Now we are substituting for t:

$$x=4+\frac{12}{11}=\frac{56}{11}$$

$$y=7-\frac{12}{11}=\frac{65}{11}$$

$$z=-3-\frac{12}{11}=\frac{-45}{11}$$

The second point which is lying on the plane is:

$$M_2(\frac{56}{11},\frac{65}{11},\frac{-45}{11})$$

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

$$\frac{x+\frac{1}{11}}{\frac{56}{11}+\frac{1}{11}}=\frac{y-\frac{45}{11}}{\frac{65}{11}-\frac{45}{11}}=\frac{z+\frac{14}{11}}{\frac{-45}{11}+\frac{14}{11}}$$

$$\frac{x+\frac{1}{11}}{\frac{57}{11}}=\frac{y-\frac{45}{11}}{\frac{20}{11}}=\frac{z+\frac{14}{11}}{\frac{-31}{11}}$$

How should I check if their answer is correct?

Did I solve correctly?

tiny-tim
Homework Helper
$$z=-3+3t$$

$$t=\frac{12}{11}$$

Now we are substituting for t:

$$z=-3-\frac{12}{11}=\frac{-45}{11}$$
Hi!

I haven't checked the rest yet, but shouldn't that be $$z=-3-\frac{36}{11}$$ ?

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. tiny-tim
Homework Helper
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. Hi Theofilius! Well, I get (7,4,-1), the same as the book, and I get it in about 4 lines.
Hi Theofilius! 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? 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. Take two fixed vectors, AB and BC.

Make any linear combination of those vectors …

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

there should be an easy way of finding the third line, shouldn't there? 