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Let say I have 2 vectors $(1, 0, 0)$ and $(0, 2, 0)$, and I want to find a third vector that is orthogonal to both of them. I can do a cross product and get $(0, 0, 2)$. However, I know there are infinite vector in the following form $(0, 0, x)$ where $x$ $\in$ R that are orthogonal to the other two. My question is what is the difference between the orthogonal vector results from cross product and any other orthogonal vector? Why the cross product gives only 1 specific orthogonal vector? What is the significance of this vector? Thank you!

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  • $\begingroup$ Don't understand "I can do a cross product and get (0, 0, 2)". It will be helpful if you can explain more. $\endgroup$
    – user295959
    Commented Jan 1, 2016 at 6:06
  • $\begingroup$ What he means is that the cross product (outer product) of (1,0,0) and (0,2,0) is orthogonal to both of them. $\endgroup$ Commented Jan 1, 2016 at 9:06

3 Answers 3

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The vector you get by performing the cross product is the unique vector orthogonal to both of your original vectors that

  1. has a length equal to the magnitude of the area of the parallelogram (actually rectangle in this case) with sides $(1,0,0)$ and $(0,2,0)$ and
  2. forms a right-handed set with $(1,0,0)$ and $(0,2,0)$

If you don't care about either of those two properties, then you could just choose any vector of the form $(0,0,c)$. But sometimes those properties are useful.

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If $\vec{a}$ and $\vec{b}$ are two vectors then the vector $\vec{a} \times \vec{b}$ is another vector perpendicular, or normal, the plane formed by $\vec{a}$ and $\vec{b}$. As you rightly suspected, there are an infinite number of vectors normal to the plane formed by $\vec{a}$ and $\vec{b}$. They are all of the form $\alpha(\vec{a} \times \vec{b})$, where $\alpha$ is a scalar. Since all of these "point" to the same direction, it is customary to choose a vector that has unit magnitude. It is given by \begin{equation} \vec{c} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}, \end{equation} where $|\cdot|$ denotes the magnitude of a vector.

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  • $\begingroup$ It's certainly not "customary" do this unitization in any scenario where computational performance is important. The division and the square root operation are often unnecessary, and they will take much longer than the cross product computation. $\endgroup$
    – bubba
    Commented Jan 1, 2016 at 10:18
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The vector product is a common operation in modern vector analysis.

  1. It gives you a vector with length $$\lVert a \times b \rVert = \lVert a \rVert \, \lVert b \rVert \sin \angle(a, b)$$
  2. Then $c = a \times b$ has the properties $0 = c \cdot a = c \cdot b$, using the scalar product, so it is orthogonal to both $a$ and $b$. However this and the above is true as well for $-c$.
  3. So the last property is that $a, b, c$ are arranged like in the right-hand-rule.

right hand rule (Source)

All three properities can be derived from the connection to the vector product with the determinant: $$ u \cdot (a \times b) = u_i \epsilon_{ijk} a_j b_k = \epsilon_{ijk} u_i a_j b_k = \text{det}(u, a, b) $$

The significance of the vector product comes from its use in physics, especially mechanics and electro dynamics.

E.g. angular momentum $L = r \times p$ or Lorentz force $F_L = q v \times B$.

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