Cross product of vector functions

Question

I was trying to make sense of a problem when I stumbled upon this on yahoo answers. I was just wondering if it was correct. If it is, can you please maybe explain why?

${\bf r}'(t) = \langle -5 \cos t, -5 \cos t, -4 \sin t \rangle$

${\bf r}''(t) = \langle 5 \sin t, 5 \sin t, -4 \cos t \rangle$.

${\bf r}'(t) \times {\bf r}''(t) = \langle 20, -20, 0 \rangle$.

Answer 1

$${\bf r}'(t) \times {\bf r}''(t) = \langle -5 \cos t, -5 \cos t, -4 \sin t \rangle \times \langle 5 \sin t, 5 \sin t, -4 \cos t \rangle = \begin{vmatrix} {\bf i} & {\bf j} & {\bf k} \\ -5 \cos t & -5 \cos t & -4 \sin t \\ 5 \sin t & 5 \sin t & -4 \cos t \end{vmatrix}$$ $$= \begin{vmatrix} -5 \cos t & -4 \sin t \\ 5 \sin t & -4 \cos t \end{vmatrix} {\bf i}+ \begin{vmatrix} -5 \cos t & -4 \sin t \\ 5 \sin t & -4 \cos t \end{vmatrix} {\bf j}+ \begin{vmatrix} -5 \cos t & -5 \cos t \\ 5 \sin t & 5 \sin t \end{vmatrix} {\bf k} = \langle (-5\cos t)(-4\cos t)-(5\sin t)(-4\sin t), -\left( (-5\cos t)(-4 \cos t)-(5\sin t)(-4\sin t) \right), (-5\cos t)(5\sin t)-(5\sin t)(-5\cos t) \rangle$$ $$=\langle 20\cos^2 t +20\sin^2 t, -20\cos^2 t-20\sin^2 t, -25\sin t\cos t+25\sin t \cos t\rangle = \langle 20, -20, 0 \rangle$$

Using $20\cos^2 t +20\sin^2 t = 20(\cos^2 t+\sin^2 t) = 20(1) = 20$ etc.

Answer 2

This follows from the pythagorean identity.

Answer 3

Yeah, it is right. In general:

(AxB)_x = a_yb_z- a_zb_y
(AxB)_y = a_zb_x- a_xb_z
(AxB)_z = a_xb_y- a_yb_x

In this particular case:

(r'(t)xr''(t))_x=(-5cost)(-4cost)-(5sint)(-4cost)=20(sin²t+cos²t)=20

(r'(t)xr''(t))_y=(-4sint)(-5sint)-(-5cost)(-4cost)=-20(sin²t+cos²t)=-20

(r'(t)xr''(t))_z=(-5cost)(5sint)-(-5cost)(5sint)=0

Hope this helps :)