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Projection of line to plane

by Theofilius
Tags: line, plane, projection
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Physicsissuef
#37
May26-08, 01:39 AM
P: 908
tiny-tim?
tiny-tim
#38
May26-08, 02:08 AM
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Hint: v = cosθi + sinθj.
Physicsissuef
#39
May26-08, 03:20 AM
P: 908
Isn't mine correct? btw- what is that? Is that vector, where is [tex]\vec{k}[/tex]?
Physicsissuef
#40
May26-08, 02:50 PM
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tiny-Timmie?
tiny-tim
#41
May26-08, 06:41 PM
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Quote Quote by Physicsissuef View Post
Isn't mine correct? btw- what is that? Is that vector, where is [tex]\vec{k}[/tex]?
If you mean
Quote Quote by Physicsissuef View Post
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 learnt about cross-products
Hint: what is [tex]\vec{i}\times(\vec{i}\times\vec{v})[/tex] ?
Physicsissuef
#42
May27-08, 05:09 AM
P: 908
Can we just find AC? What are we now searching for?


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