It was my understanding that for two vectors to be orthogonal then their scalar product must be zero.

I know that the scalar product of two vectors $\vec u$ and $\vec v$ is given by $$\vec{u}\cdot\vec{v} = |\vec{u}||\vec{v}|\cos(\theta)$$

After years of using this formula I thought I had a firm handle on the concept of orthogonality in the context of vectors.

Then recently I found two examples of pairs of vectors that I thought were not orthogonal due to a non-zero scalar product, but I was told otherwise.

The first pair of vectors is $\vec u=\begin{bmatrix}3 \\-2 \\1 \end{bmatrix}$ and $\vec v=\begin{bmatrix}1 \\4 \\2 \end{bmatrix}$ $$\vec u \cdot \vec v =\begin{bmatrix}3 \\-2 \\1 \end{bmatrix}\cdot\begin{bmatrix}1 \\4 \\2 \end{bmatrix}=3\times 1-2\times 4 +1\times 2=3-8+2=-3\ne 0$$

The second pair the vectors are $\vec p = \dfrac{1}{\sqrt{13}}\begin{bmatrix}-3 \\-2 \end{bmatrix}$ and $\vec q = \dfrac{1}{\sqrt{13}}\begin{bmatrix}3 \\-2 \end{bmatrix}$

The scalar product is $$\vec p \cdot \vec q=\begin{bmatrix}-3 \\-2 \end{bmatrix}\cdot\begin{bmatrix}3 \\-2 \end{bmatrix}=-9 + 4=-5\ne 0$$

If I'm not missing something very simple here then has there been some kind of mistake and both the sets of vectors I mentioned are not orthogonal as I thought?

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    $\begingroup$ The question on the slide was woefully mistaken. Both blue and orange are correct. $p$ and $q$ are certainly not orthogonal either. $\endgroup$ – Cameron Williams Oct 19 '16 at 20:54
  • $\begingroup$ @Cameron Thanks for the confirmation, I had to check that's all. I know that this is a bad question. $\endgroup$ – BLAZE Oct 19 '16 at 20:56

No (unless one of the vectors is the 0 vector). Generally the dot product of two non zero vectors is zero iff the two vectors are perpendiculars.

One way to think of why it is that way - is to think of the dot product between two vectors $\vec a$ and $\vec b $ as the multiplication of the size of $\vec a$ and the size of the projection of $\vec b$ of on $\vec a$. Now if the two vectors are not perpendiculars, then the size of the projection of $\vec b$ on a (lets just call it B) is a non zero scalar (can be negative). and as the size of $\vec a$ (call it A) is positive then the multiplication of two non zero numbers A*B cannot be zero.

On the other hand if the vectors are prependiculars, then the projection of $\vec b$ on $\vec a$ is just a single dot, with size $B=0$. and there for $A*B=A*0=0$

  • $\begingroup$ Many thanks for your answer and confirmation. $\endgroup$ – BLAZE Oct 19 '16 at 21:23

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