# Cross product of vector functions

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

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

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### Jeremy Rowler

Updated on August 01, 2022

• Jeremy Rowler over 1 year

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?

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

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

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

I'm learning this exact same thing in class. Are you in my class??
• Jeremy Rowler about 10 years
I don't know.... U of M?
• Jeremy Rowler about 10 years
But shouldn't it be <40,-40,0>
• Jonathan Aronson about 10 years
no, 5 * 4 = 20.
• Jeremy Rowler about 10 years
I didn't see the way the cos and sin cancel out. But now I see, it makes a lot more sense now