# Projection of line to plane

• Theofilius

## 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 itself 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 textbook results I get tottaly different solution: $$\frac{x+9}{7}=\frac{y+1}{4}=\frac{z}{-1}$$

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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...

Theofilius said:
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:
tiny-tim said:
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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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?

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...

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?

But I don't know those vectors...

What are the coordinates of AB BC and AC?

HallsofIvy said:

Take two fixed vectors, AB and BC.

Make any linear combination of those vectors …

what do you get?

Can we just don't complicate? Let's 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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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 textbook 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...

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?

Theofilius said:
$$z=-3+3t$$

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

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 textbook results. I think only one solution is possible to this task. So either mine, or their is correct.

Theofilius said:
Even, if it is like that, my final answer will not be even close to my textbook 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.
tiny-tim said:
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?
tiny-tim said:
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.
HallsofIvy said:
tiny-tim said:

Take two fixed vectors, AB and BC.

Make any linear combination of those vectors …

what do you get?
Theofilius said:
Can we just don't complicate?

Your method was complicated, and it gave the wrong result.

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?

Why my way of solving is not correct?

Theofilius said:
Why my way of solving is not correct?

Because you made a mistake somewhere, and it needs fixing.

Theofilius, the fact that you made a mistake should remind you:

i) the more complicated something is, the more likely a mistake is

ii) you are relying too much on formulas, instead of just looking at the problem as a whole and using ordinary common-sense.

Now, do it the way HallsofIvy and I keep advising you to!

I think logically, that's how I found the solution of this task. :Smile: I just implement formulas. Ok, I will give a try with your method.

AB have coordinates (4,3-2) and BC(-1,1,3). So I need to find CA?

How will I find it?

tiny-tim ?

Comon Tim, help man.

I think the Theofilus method is correct, just can't figure the error.

Physicsissuef said:
I think the Theofilus method is correct, just can't figure the error.

Hi Physicsissuef!

I'm sure the method is correct … but it's horribly long, so it's not surprising there's a mistake in it, and I've no intention of ploughing through it to find it.

Don't copy that method … nothing should be that complicated … always go for something simpler, if possible!

I actually don't understand too much your method. I am sure it is translation problem. Is AB the parallel vector to the line? Is BC the normal vector of the plane?

Hi Physicsissuef!
Physicsissuef said:
I actually don't understand too much your method. I am sure it is translation problem. Is AB the parallel vector to the line? Is BC the normal vector of the plane?

Yes, and CA is the projection.

So what is the angle between BC and CA? And how can CA be expressed as a combination of AB and BC?

The angle between BC and CA is 90 degrees. So
$$-b_1+b_2+3b_3=0$$

We need 2 conditions more. I don't know what do you mean by combination?

I think that we can make linear combination of vectors only if they are parallel.

Physicsissuef said:
The angle between BC and CA is 90 degrees. So
$$-b_1+b_2+3b_3=0$$

We need 2 conditions more. I don't know what do you mean by combination?

I think that we can make linear combination of vectors only if they are parallel.

Combination just means ordinary vector addition.