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When testing for orthogonal vectors my book always that you have to test vectors individually. Can you test the vectors all at once?

This is supposed to be a vector set. Not sure how to set that up.

$\vec{v_1}=\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}$

$\vec{v_2}=\begin{bmatrix} 2\\ 1\\ -1 \end{bmatrix}$

$\vec{v_3}=\begin{bmatrix} 3\\ -2\\ -1 \end{bmatrix}$

The book shows:

$ \vec{v_1}\cdot\vec{v_2} = 1 \cdot 2 + 2 \cdot 1 + 3 \cdot (-1) = 1$

$ \vec{v_1}\cdot\vec{v_3} = 1 \cdot 3 + 2 \cdot (-2) + 3 \cdot (-1) = -4$

$ \vec{v_2}\cdot\vec{v_3} = 2 \cdot 3 + 1 \cdot (-2) + (-1) \cdot (-1) = 5$

Can you do this?

$ \vec{v_1}\cdot\vec{v_2}\cdot\vec{v_3} = 1 \cdot 2 \cdot 3 + 2 \cdot 1 \cdot (-2) + 3 \cdot (-1) \cdot (-1) = 5$

This shows that its not orthogonal. Are there are special cases where you can not do them all at once?

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2 Answers 2

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No, orthogonality is a pairwise property, so you have to test it individually. Look at $$ \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}. $$ Every pair of these is not orthogonal, but multiplying the elements of all three together gives zero.

But more generally, multiplying three vectors together cannot be done in the way you describe: it's completely meaningless. The dot product, which tests orthogonality, takes two vectors and gives you a scalar. There is such an operation for three vectors (in 3D), called the scalar triple product, but this tells you about linear independence, not orthogonality.

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No your statement does not make since.

The dot product that you are referring to is the standard innner product on $\mathbb{R}^n$. An inner product is a bilinear map $(\cdot,\cdot): V \times V \to \mathbb{K}$ where K is the underlying field of the vector space. Two vectors are defined to be orthogonal if $(v,u)=0$. An inner product is a function of two variables.

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