Equation of Line Projection on Plane - Homework Help

In summary, the conversation was about finding the equation of the projection of a given line onto a given plane. The participants discussed various methods and equations, and finally the person solving the task shared their solution, but it did not match the answer given in the textbook.
  • #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)
 
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  • #37
tiny-tim?
 
  • #38
:smile: Hint: v = cosθi + sinθj. :smile:
 
  • #39
Isn't mine correct? btw- what is that? Is that vector, where is [tex]\vec{k}[/tex]?
 
  • #40
tiny-Timmie?
 
  • #41
Physicsissuef said:
Isn't mine correct? btw- what is that? Is that vector, where is [tex]\vec{k}[/tex]?

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

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

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