Projection and Reflection of Vector WRT plane

[SP90]

Homework Statement

Given a plane \Pi with normal n=i-2j+k and a vector v=3i+4j-2k calculate the projection of v onto \Pi and the reflection of v with respect to \Pi.

The Attempt at a Solution

I need to check that I'm doing this is right.

I think I need v - (v \cdot n)n = 3i+4j-2k - 7(i-2j+k) = -4i +18j-9k

And for the refection:

v - 2(v \cdot n)n = 3i+4j-2k - 14(i-2j+k) = -11i +32j-16k

Are those correct?

 

[vela]

Homework Statement

Given a plane \Pi with normal n=i-2j+k and a vector v=3i+4j-2k calculate the projection of v onto \Pi and the reflection of v with respect to \Pi.

The Attempt at a Solution

I need to check that I'm doing this is right.

I think I need v - (v \cdot n)n = 3i+4j-2k - 7(i-2j+k) = -4i +18j-9k

You made a sign error.

The projection of v onto the plane should be perpendicular to the plane, right? So what should your answer dotted with $\vec{n}$ equal? That's how you can check your answer.

And for the refection:

v - 2(v \cdot n)n = 3i+4j-2k - 14(i-2j+k) = -11i +32j-16k

Are those correct?

Same sign error.

 

[SP90]

Makes sense, thanks for the quick response.

I get 10i-10k+5k which, when dotted with n obviously gives 0.

Correcting the sign error for the second yields 17i-24j+12k.

Thanks for your help

 

[vela]

That's not correct either. You need to normalize the normal vector before you use it in your formulas.

 

[SP90]

Ah, I see.

So instead I use \hat{n}=\frac{1}{\sqrt{6}}(1i-2j+1k)

and that gives Pv=\frac{1}{6}(25i+10j-5k).

And Tv=\frac{1}{6}(32i-4j+2k).

Is that right now?

 

[vela]

Yes, those are correct.