1. ## reflection of curve

Hi
I have the following question:
there is a plane in R^3 given by:
2x+z+1 = 0

Let \psi:R^3 -> R^3 be the reflection along the plane, and let C be a curve in R^3 given by:
x = t
y = t^2
z = t^3

Find the equation of \psi(C)

I know the relection equation and I have got down to the following:

(t, t^2, t^3) - (2/5)(2t + t^2 + 1)(2, 0, 1)

How do I simplify this down to the x, x and z components?

Thanks

2. Originally Posted by srose34
Hi
I have the following question:
there is a plane in R^3 given by:
2x+z+1 = 0

Let \psi:R^3 -> R^3 be the reflection along the plane, and let C be a curve in R^3 given by:
x = t
y = t^2
z = t^3

Find the equation of \psi(C)

I know the relection equation and I have got down to the following:

(t, t^2, t^3) - (2/5)(2t + t^2 + 1)(2, 0, 1)

How do I simplify this down to the x, x and z components?

Thanks
1. I didn't check if your calculations are OK.

2. I changed the notation so you can see what is a scalar factor and what is a vector.

$\displaystyle \langle t, t^2, t^3 \rangle - (2/5)(2t + t^2 + 1)\langle2, 0, 1\rangle$

3. You'll get:

$\displaystyle \begin{array}{l}x = t - \frac25 (2t+t^2+1) \cdot 2 \\ y = t^2 - \frac25 (2t+t^2+1) \cdot 0 \\ z = t^3 - \frac25 (2t+t^2+1) \cdot 1\end{array}$

4. Simplify the RHSs.

3. ok, thanks that helps.