Projection and Reflection of Vector WRT plane

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SP90
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Homework Statement



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

The Attempt at a Solution



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

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

And for the refection:

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

Are those correct?
 
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SP90 said:

Homework Statement



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

The Attempt at a Solution



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

I think I need [itex]v - (v \cdot n)n = 3i+4j-2k - 7(i-2j+k) = -4i +18j-9k[/itex]
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.
SP90 said:
And for the refection:

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

Are those correct?
Same sign error.
 
Makes sense, thanks for the quick response.

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

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

Thanks for your help
 
Ah, I see.

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

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

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

Is that right now?