Projection of line to plane

by Theofilius
Tags: line, plane, projection
 P: 910 tiny-tim?
 Sci Advisor HW Helper Thanks P: 26,157 Hint: v = cosθi + sinθj.
 P: 910 Isn't mine correct? btw- what is that? Is that vector, where is $$\vec{k}$$?
 P: 910 tiny-Timmie?
HW Helper
Thanks
P: 26,157
 Quote by Physicsissuef Isn't mine correct? btw- what is that? Is that vector, where is $$\vec{k}$$?
If you mean …
 Quote by Physicsissuef 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 learnt about cross-products …
Hint: what is $$\vec{i}\times(\vec{i}\times\vec{v})$$ ?
 P: 910 Can we just find AC? What are we now searching for?

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