# 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.