Vectors Need Help

1. Feb 27, 2015

amy098yay

1. The problem statement, all variables and given/known data
using the 3 vectors (1, 1, 0) ,(1, 0, −1) and (0, 1, 1) I had tried to show that https://ucdsb.elearningontario.ca/content/enforced/4850117-BL_1415Sem2__MAT_MCV4UU-948314_1_ELO/MCV4UPU01/MCV4UPU01A06/images/vec-a.gif?_&d2lSessionVal=Y3hirJUTSYjH76OEZwqHIBATE&ou=4850117×(https://ucdsb.elearningontario.ca/content/enforced/4850117-BL_1415Sem2__MAT_MCV4UU-948314_1_ELO/MCV4UPU01/MCV4UPU01A06/images/vec-b.gif?_&d2lSessionVal=Y3hirJUTSYjH76OEZwqHIBATE&ou=4850117 × https://ucdsb.elearningontario.ca/content/enforced/4850117-BL_1415Sem2__MAT_MCV4UU-948314_1_ELO/MCV4UPU01/MCV4UPU01A06/images/vec-c.gif?_&d2lSessionVal=Y3hirJUTSYjH76OEZwqHIBATE&ou=4850117) is not equal to (https://ucdsb.elearningontario.ca/content/enforced/4850117-BL_1415Sem2__MAT_MCV4UU-948314_1_ELO/MCV4UPU01/MCV4UPU01A06/images/vec-a.gif?_&d2lSessionVal=Y3hirJUTSYjH76OEZwqHIBATE&ou=4850117 × https://ucdsb.elearningontario.ca/content/enforced/4850117-BL_1415Sem2__MAT_MCV4UU-948314_1_ELO/MCV4UPU01/MCV4UPU01A06/images/vec-b.gif?_&d2lSessionVal=Y3hirJUTSYjH76OEZwqHIBATE&ou=4850117) × https://ucdsb.elearningontario.ca/content/enforced/4850117-BL_1415Sem2__MAT_MCV4UU-948314_1_ELO/MCV4UPU01/MCV4UPU01A06/images/vec-c.gif?_&d2lSessionVal=Y3hirJUTSYjH76OEZwqHIBATE&ou=4850117.
My attempt at the solution is in this pdf file, where did I go wrong and how can I fix it?

3. The attempt at a solution

Attached Files:

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2. Feb 28, 2015

Simon Bridge

If I read you correctly, you got $\vec a \times (\vec b \times \vec c) = (0,1,0)$ and $(\vec a \times \vec b) \times \vec c = (-2,2,-2)$, therefore $\vec a \times (\vec b \times \vec c) \neq (\vec a \times \vec b) \times \vec c$ ... which is what you wanted to demonstrate.

3. Feb 28, 2015

Staff: Mentor

Your values for b x c = <1, -1, 1> and a x b = <-1, 1, -1> look good, but you're doing something wrong when you calculate a x (b x c) and (a x b) x c. For the first triple product you should be evaluating this pseudodeterminant:
$$\begin{vmatrix} i & j & k \\ 1 & 1 & 0 \\ 1 & -1 & 1\end{vmatrix}$$
Here the 2nd row has the coordinates of a and the 3rd row has the coordinates of b x c
For the second triple product you need to evaluate this one:
$$\begin{vmatrix} i & j & k \\ -1 & 1 & -1 \\ 0 & 1 & -1\end{vmatrix}$$
Here the 2nd row has the coordinates of a x b and the 3rd row has the coordinates of c.