Equation of Line Projection on Plane - Homework Help

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SUMMARY

The discussion centers on finding the equation of the projection of the line defined by \(\frac{x}{4}=\frac{y-4}{3}=\frac{z+1}{-2}\) onto the plane described by \(x-y+3z+8=0\). Participants analyze the geometric relationships between the line, the normal to the plane, and the projection, ultimately determining that the projection must be a linear combination of the vectors associated with the line and the normal. The correct projection equation is derived as \(\frac{x+9}{7}=\frac{y+1}{4}=\frac{z}{-1}\), which differs from the initial calculations presented by the user seeking help.

PREREQUISITES
  • Understanding of vector mathematics and projections
  • Familiarity with parametric equations of lines
  • Knowledge of planes in three-dimensional space
  • Ability to perform linear combinations of vectors
NEXT STEPS
  • Study vector projections and their applications in geometry
  • Learn about parametric equations and how to derive them for lines and planes
  • Explore the concept of linear combinations in vector spaces
  • Investigate the geometric interpretation of normal vectors and their significance in projections
USEFUL FOR

Students studying geometry, particularly in three dimensions, as well as educators and tutors assisting with vector mathematics and projection problems.

  • #31
Physicsissuef said:
I think the Theofilus method is correct, just can't figure the error.

Hi Physicsissuef! :smile:

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. :rolleyes:

Don't copy that method … nothing should be that complicated … always go for something simpler, if possible! :smile:
 
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  • #32
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?
 
  • #33
Hi Physicsissuef! :smile:
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? :smile:
 
  • #34
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.
 
  • #35
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.
 
  • #36
I understand.

AB+BC=AC
(4,3,-2)+(1,-1,3)=AC
(5,2,1)=AC

Now we can make the equation of the line.

But we haven't got the same vector as the result. In the posts above, the parallel vector is (7,4,-1)
 
Last edited:
  • #37
tiny-tim?
 
  • #38
:smile: Hint: v = cosθi + sinθj. :smile:
 
  • #39
Isn't mine correct? btw- what is that? Is that vector, where is \vec{k}?
 
  • #40
tiny-Timmie?
 
  • #41
Physicsissuef said:
Isn't mine correct? btw- what is that? Is that vector, where is \vec{k}?

If you mean …
Physicsissuef said:
AB+BC=AC
(4,3,-2)+(1,-1,3)=AC
(5,2,1)=AC
… then of course it's correct … but what good is that? … as you say …
But we haven't got the same vector as the result. In the posts above, the parallel vector is (7,4,-1)

If \vec{v}\,=\,\cos\theta\vec{i}+\sin\theta\vec{j} then \vec{j}\,=\,\left(\frac{\vec{v}\,-\,\cos\theta\vec{i}}{\sin\theta}\right)

Alternatively, if you've learned about cross-products …
Hint: what is \vec{i}\times(\vec{i}\times\vec{v}) ? :smile:
 
  • #42
Can we just find AC? What are we now searching for?
 

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