# Finding the dot product.

Finding the dot product of $(-2w) \cdot w$ where $w=(0,-2,-2)$ $$\text{dot product} = \frac{v . u}{\| v \| . \| u \|}$$

So $$-2 (0,-2,-2)=(0,4,4) \\ (0+4+4) = 8 \\ (0,-2,-2)=-4 \\ 8 \times -4=-32$$

$$\| v \|= \sqrt{0^2+(-2)^2+(-2)^2}=2 \sqrt 2 \\ \|-2w \|= \sqrt{0^2+4^2+4^2}= 4 \sqrt 2\\ \frac{-32}{2 \sqrt 2 \times 4 \sqrt 2} = -64$$ Is this correct?

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The dot product of $w = (w_1, w_2, w_3)$ and $-2w = (-2 w_1, -2w_2, -2v_3)$ is $$-2 w_1^2 + -2 w_2^2 + -2 w_3^2 = -2(w_1^2 + w_2^2 + w_3^2).$$ – user2468 May 13 '12 at 16:31

You have several mistakes here:

The formula in your question is for the cosine of the angle between two vectors, not the dot product. The numerator in the right-hand side of the formula is the dot product.

To calculate the dot product of two vectors, you multiply corresponding entries in these vectors and then sum the products. You seem to be summing the components and then multiplying the sums.

So in this case:

\begin{align*} (0, -2, -2) \cdot (0, 4, 4) &= 0\times0 + (-2)\times4 + (-2)\times4 \\ &= -16 \end{align*}

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Ah silly me. Thank you for this. :) – Xabi May 13 '12 at 17:24

What you have written as the dot product isn't the dot product.

The dot product of $v=(v_1,v_2,v_3)$ and $w=(w_1,w_2,w_3)$ is simply $$v\cdot w=v_1w_1+v_2w_2+v_3w_3.$$

In your problem, we have $w=(0,-2,-2)$. Computing $-2w$, we have $$-2w=-2(0,-2,-2)=\bigl(-2\cdot0, -2\cdot(-2), -2\cdot(-2)\bigr)=(0,4,4).$$ Then $$(-2w)\cdot w= (0,4,4)\cdot(0,-2,-2) =0\cdot0+(4)\cdot(-2)+(4)\cdot(-2)=-16.$$

Your expression at the end of your post demands that we find the norm of two vectors. The norm of $w$ is $$\Vert w\Vert=\sqrt{ 0^2+(-2)^2+(-2)^2}=\sqrt {8}=2\sqrt2.$$
The norm of $-2w$ can be calculated in a similar manner; but you could also compute it using the rule $\Vert \alpha v\Vert=|\alpha|\Vert v\Vert$:

$$\Vert -2w\Vert=2\Vert w\Vert=2(2\sqrt2)=4\sqrt2.$$
Finally, if you wish to compute $(-2w)\cdot w\over \Vert w\Vert \Vert 2w\Vert$, just substitute: $${ (-2w)\cdot w\over \Vert w\Vert \Vert 2w\Vert} ={-16\over (2\sqrt2)(4\sqrt2)}=-1.$$

This should come as no suprise, since $(-2w)\cdot w =-2(w\cdot w)=-2\Vert w\Vert^2$ and $\Vert -2w\Vert =2\Vert w\Vert$; we could have arrived at the result with no computations: $${ (-2w)\cdot w\over \Vert w\Vert \Vert 2w\Vert} ={-2\Vert w\Vert^2 \over2\Vert w\Vert\Vert w\Vert}={-2\Vert w\Vert^2 \over2\Vert w\Vert^2}=-1.$$

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