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Cross product of 3 dimensional vectors

$a=<a_1,a_2,a_3>, b=<b_1,b_2,b_3>$ can be defined as

$c=<a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1>$

Length of the vector $c$ can be defined as:

$|c|=\sqrt{\sum_{i=1}^{3} c_{i}^{2}}$

or equivalently:

$|c|=|a| |b| sin (\phi)$, where $\phi$ is the angle between $a$ and $b$.

Now when I compute cross product in vector form: $a\times b = a \times (b1+b2) = a \times b1 + a \times b2$ holds ($b1, b2 \in \mathbb{R}^{3}$, $b=b1+b2$), but when doing this with formula for scalar: $|c|=|a| |b| sin (\phi)$, it linearity does not hold.

Example:

$a=<1, 2, 3>$

$b=<-3, 3, 1>$

$b1=<-2, 1, 0>$

$b2=<-1, 2, 1>$

I calculate value of sinus based on dot product: $sin \phi = \sqrt{(1-cos^{2}\phi)} = \sqrt{(1 - \frac {<a, b>^{2}}{|a|^{2}|b|^{2}})}$

$|a \times (b1 + b2)| = |a \times b| = |a| |b| sin(a, b) = 3.74 * 4.36 * 0.93 = 15.2$

$|a \times b1 + a \times b2)| = ~ 15.3$

What have I misunderstood? What is the problem? What can cause it?

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  • $\begingroup$ First of all, edit your question to be consistent when writing about b, b1, b2, and d. Then, report the values of \phi between a and b, a and d, a and (b+d) to check if your problem is not there. $\endgroup$ – toliveira Mar 8 '17 at 17:53
  • $\begingroup$ Show us how you arrived at those last two values. $\endgroup$ – amd Mar 8 '17 at 17:55
  • $\begingroup$ 15.2 and 15.3 are basically the same number. Are you sure you're not just getting round-off error? $\endgroup$ – Dylan Mar 8 '17 at 20:16
  • $\begingroup$ @Dylan , I am sure, because when I put other values for coordinates for b, b1, b2 difference is bigger. $\endgroup$ – Nathan Mar 8 '17 at 20:19
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Somewhere in your calculations you are making a formula error. I don't know where, since you haven't shown your calculations. But the formulas you've shown here are correct. However, the values are not. Both values are $\sqrt{230}\approx 15.16575$.

To see this, note that $$a \times b\phantom{2} = (-7, -10, \phantom{-}9)\\a \times b1 = (-3, \phantom{1}-6, \phantom{-}5)\\ a\times b2 = (-4, \phantom{1}-4, \phantom{-}4)$$ So obviously $a \times b = (a\times b1) + (a\times b2)$. Now $$|a \times b| = \sqrt{(-7)^2 + (-10)^2 + 9^2}= \sqrt{230}$$ This alone tells where the problem is. You have mis-calculated either $a\times b1, a\times b2$, their sum, or $|a\times b|$, since $\sqrt{230} \approx 15.16575 < 15.3$.

But for completeness, note that $$|a|^2 = 14, |b|^2 = 19, \langle a, b\rangle^2 = 6^2 = 36$$ So $$\cos^2\phi = \frac{36}{14\cdot 19}\\\sin^2\phi = 1 - \frac{36}{14\cdot 19} = \frac{230}{14\cdot 19}$$ Therefore $$|a\times b|^2 = |a|^2|b|^2\sin^2\phi = 14\cdot 19\cdot\frac{230}{14\cdot 19} = 230$$ giving $|a\times b| = \sqrt{230}$, just as before.

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  • $\begingroup$ thanks for the detailed answer, what I am trying to check and understand is: $|a \times b|^{2} = |a|^{2} |b|^{2} sin^{2} \phi = 230$ $|a \times b_{1}|^{2} = |a|^{2} |b_{1}|^{2} sin^{2} (angle(a, b_{1})) = 70$ $|a \times b_{2}|^{2} = |a|^{2} |b_{2}|^{2} sin^{2} (angle(a, b_{2})) = 48$ And why here it does not work like: $230 = |a \times b| = |a \times b_{1}| + |a \times b_{2}| = 118$ ? $\endgroup$ – Nathan Mar 9 '17 at 8:02
  • $\begingroup$ What is the reason that linearity doesn't hold for lengths? $\endgroup$ – Nathan Mar 9 '17 at 8:08
  • $\begingroup$ You surely have seen this many times. in general $|x + y| \ne |x| + |y|$. Algebraicly, this follows from the fact that $\sqrt{r + s} \ne \sqrt{r} + \sqrt{s}$. Geometrically, $x+y$ is the third side of a triangle formed by $x$ and $y$ translated by $x$. Having one side of a triangle equal to the sum of the other two sides only occurs when the other two sides point in opposite directions, so we don't have a true triangle at all. The best you can say is the triangle inequality: $|x + y| \le |x| + |y|$. $\endgroup$ – Paul Sinclair Mar 9 '17 at 17:43

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