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Two numbers $3$ and $4$ their multiplication is each one from the first number is repeated a number of times as the second number i.e. $3$ times $4$ is $(1+1+1)$ times four meaning $1+1+1+1 + 1+1+1+1 + 1+1+1+1$ for example we can represent $3$ by any countable thing like three bags each of them has four apples then the total is $12$ .. Excuse me for this I know it's obvious but I want an explanation like this one I've given

Assuming we have two one dimensional vectors $ 3\hat{i}$ and $3 \hat{j}$ then Could you give me explain it to me in the previous way I explained multiplication of scalars ..

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  • $\begingroup$ "The cross product a × b is defined as a vector c that is perpendicular to both a and b, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.". What about this definition is not clear to you? $\endgroup$ – CuriousOne Jun 26 '15 at 19:00
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    $\begingroup$ At some point in physics you have to give up on getting an "explain it like I'm five" explanation. $\endgroup$ – David Hammen Jun 26 '15 at 19:02
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    $\begingroup$ Then you have what the definition says you have: a vector that identifies the third dimension and that is proportional to the area spanned by the two others. If you were building a house and the two vectors are the directions of the walls pointing away from a corner, then the cross product would point in the direction in which you should build the walls towards the ceiling and it would tell you the floor area. :-) $\endgroup$ – CuriousOne Jun 26 '15 at 19:24
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    $\begingroup$ Can you perform the equivalent of finger counting in this case? Absolutely. What everybody here seems to struggle with is why you would want to. Now, there is a very famous book that does something similar to finger counting in the case of tensor algebra: "Gravitation" by Misner, Thorne, Wheeler has some extremely nice graphics of this kind, but it would probably not be very useful for you at this moment. :-) $\endgroup$ – CuriousOne Jun 26 '15 at 19:39
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    $\begingroup$ Let me dissuade you from thinking multiplication is repeated addition. There are a number of mathematics educators who strongly disagree with the notion that multiplication is "just" repeated addition. How do you add 1 to itself zero times? -3 times? 22/7 times? pi times? What if neither item is a positive integer? $\endgroup$ – David Hammen Jun 26 '15 at 20:19
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Let's turn this around. If you wanted to know (for general vectors) what vector is perpendicular to both of them, you would have to solve two equations with three unknowns. You know that the dot product of two vectors is zero if they are perpendicular, so you would write

$$\vec{x}\cdot \vec{a} = 0\\ \vec{x}\cdot \vec{b} = 0$$

But if you are working in three dimensions, that leaves you with two equations and three unknowns - so you can't solve it.

We could add a third constraint. For example, we could decide that the length of the third vector has to be equal to the area of the parallelogram described by the first two vectors. We know the area is $A=|\vec{a}|\cdot|\vec{b}|\cdot \sin\theta$. That gives me the length of vector $\vec{x}$ and my third equation.

Now I have three equations and three unknowns. I could write them out in full- the first two are easy:

$$x_1 a_1 + x_2 a_2 + x_3 a_3 = 0\\ x_1 b_1 + x_2 b_2 + x_3 b_3 = 0$$

The third one is harder. Following http://heaveninthebackyard.blogspot.com/2011/12/derivation-of-cross-product-formula.html, if you set the length of vector $\vec{x}$ equal to 1, you could solve the equations but they are quite messy:

$$\vec{x} = ±\frac{1}{Z}(a_3b_2 - a_2 b_3, a_1b_3 - a_3 b_1, a_2 b_1 - a_1 b_2)$$

Then some more manipulation tells you that if you set the length equal to the magnitude of the area instead, the expression simplifies to the one we know and love.

Feel free to read through the lines and lines of derivation at the link above... it is long because it assumes very little advanced algebra - but that seems to be what you are asking for.

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    $\begingroup$ In German they call that the pedestrian way... and some like to crawl a little longer before they walk. :-) $\endgroup$ – CuriousOne Jun 26 '15 at 23:36
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Find me a vector that's mutually orthogonal to two other vectors

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