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So, we are currently studying algebraic vectors in my math class, and I noticed an interesting property I couldn't explain that my teacher didn't want to explain since it has to do with material outside the corriculum, and didn't want to confuse students.

What I found was that the area of a triangle ABC define by the vectors AB and AC is equal to a half of the magnitude of the cross product of AB and AC (0.5 * |AC| * |AB| * sin(θ) with θ being the angle between AB and AC).

Is this coincidental or is there some relation between the area of the triangle and the cross product?

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That is not a coincidence. The magnitude of the cross product of two vectors equals the area of the parallelogram spanned by these two vectors. And since a triangle is half of a parallelogram, your relation follows.

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    $\begingroup$ Oh, I didn't know that! Thanks! Any easy to explain reason as to why the magnitude of the cross product is as such? $\endgroup$ Commented Dec 16, 2016 at 19:23
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    $\begingroup$ By definition, I guess? By following the definition from Wikipedia: "The cross product $a \times b$ 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. The cross product is defined by the formula $a \times b =\|a\| \cdot \|b\| \sin (\theta) n$ $\endgroup$
    – Pawel
    Commented Dec 16, 2016 at 21:12
  • $\begingroup$ Oh, I only knew the absin(theta) definition and my physics teacher gave me a definition I didn't understand using matrices. Didn't realize the cross product can be defined this way. Thanks! $\endgroup$ Commented Dec 17, 2016 at 16:49
  • $\begingroup$ Area of the parallelogram is calculated by $A=a \cdot h$, where $h$ is the height of the parallelogram and $a$ is the base. But if you draw it on a piece of paper, you will notice that from trigonometry, $h=b \cdot \sin(\theta)$. Thus, the area is $A=ab \sin(\theta).$ $\endgroup$
    – Pawel
    Commented Dec 17, 2016 at 16:58

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