Cross product equation with sine, I don't understand the unit vector "n" [closed]

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$\vec{a} \times \vec{b}=\mid \vec{a} \mid \mid \vec{b} \mid \sin(\theta)\, \vec{n}$

I know this is probably an incredibly simple question, and I know that "n" is the unit vector at right angles to both equations, and that a unit vector has a magnitude of 1, its just for the purposes of this equation, I just don't understand how to find it for two given vectors.

I'm used to calculating the cross product by multiplying the numbers e.g. $i-j+k$, and I just found out that I need to learn this equation. This post must seem really rambly and unrefined, but I just can't find a conclusive answer, and I'm really desperate for some clarification.

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3 Answers

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If you have two curves $\vec x(t),\vec y(t):\mathbb{R}\to\mathbb{R}^3$, the unit normal vector to them at some point $t$ is $$\hat n=\widehat{\vec x\times\vec y}=\frac{\vec x\times\vec y}{\lVert\vec x\times\vec y\rVert}=\frac{\vec x\times\vec y}{\lVert\vec x\rVert\cdot\lVert\vec y\rVert\cdot\sin\theta}=\frac{{\hat x}\times{\hat y}}{\sin\theta},$$ where $\theta$ is the angle of incidence between $\vec x$ & $\vec y$ and the "hat" (^) represents the unit version of any vector: $$\vec x=(x_2,x_2,x_3)\implies\hat x=\frac{\vec x}{\lVert\vec x\rVert}=\frac{(x_1,x_2,x_3)}{\sqrt{x_1^2+x_2^2+x_3^2}}.$$ So in your case (with your clarification),

$$\hat n=\frac{\vec a\times\vec b}{\lVert \vec a\times\vec b \rVert}.$$

First find the cross product $\vec a\times\vec b$ in the numerator; then take its norm, to get the denominator. Then divide by that norm (a scalar multiplication) to get the unit version of $\vec a\times\vec b$. You can then calculate the angle $\theta$, if you need it, from $$\vec a\times\vec b=\left(\lVert\vec a\rVert\cdot\lVert\vec b\rVert\cdot\sin\theta\right)\hat n$$ by taking the inverse sine of $$\sin\theta=\frac{\lVert\vec a\times\vec b\rVert}{\lVert\vec a\rVert\cdot\lVert\vec b\rVert}.$$

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If you have the coordinates of two vectors and all you need to do is find the coordinates of their cross product, it would be silly to use the "$\sin\theta$" equation to find the answer. It is much simpler to add products of the coordinates in the way you already know.

But knowing that $$ \mathbf a \times \mathbf b = \lVert \mathbf a \rVert \lVert \mathbf b \rVert (\sin\theta) \mathbf n, $$ where $\mathbf n$ is a unit vector perpendicular to both $\mathbf a$ and $\mathbf b$, you can find the angle $\theta$ between two vectors (given by their coordinates) by computing the cross-product without using the "$\sin\theta$" formula, then plugging that result into the equation above, taking the magnitude of both sides (which means $\mathbf n$ drops out, since $\lVert\mathbf n\rVert = 1$), and solving for $\theta$. (Of course, to find $\theta$ uniquely you must have some other information, since there are generally two values of $\theta$ between $0$ and $\pi$ that both yield the same value of $\sin\theta$.)

On the other hand, if you happen to be given the magnitudes of two vectors and the angle between them, you can immediately apply the equation above without having to figure out how to convert the magnitudes and angle into suitable coordinates of the two vectors.

So the equation does not always have to be used the same way. It is simply one of several useful facts that can help you find out something about two vectors when you know some other things about them.

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You know how to find the cross product, but you don't know how to normalize the vector?

If $\mathbf v = x\mathbf i+y\mathbf j+z\mathbf k$ then $||\mathbf v|| = (x^2+y^2+z^2)^{1/2}$, and you are looking for $\dfrac {\mathbf v}{||\mathbf v||}.$

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