Finding the projection of a line onto a plane

[tiny-tim]

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!

 

[Physicsissuef]

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?

 

[tiny-tim]

Hi Physicsissuef!

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?

 

[Physicsissuef]

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.

 

[tiny-tim]

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.

 

[Physicsissuef]

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)

 

[Physicsissuef]

tiny-tim?

 

[tiny-tim]

Hint: v = cosθi + sinθj. ​

 

[Physicsissuef]

Isn't mine correct? btw- what is that? Is that vector, where is $$\vec{k}$$?

 

[Physicsissuef]

tiny-Timmie?

 

[tiny-tim]

Isn't mine correct? btw- what is that? Is that vector, where is $$\vec{k}$$?

If you mean …

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 $$\vec{i}\times(\vec{i}\times\vec{v})$$ ?

 

[Physicsissuef]

Can we just find AC? What are we now searching for?