Area of a triangle (cross product lesson)

[Lebombo]

Homework Statement

Youtube: Lec 2 | MIT 18.02 Multivariable Calculus, Fall 2007 (Video time frame: between 11:00 minutes and 12:30 minutes)

Find the area of a triangle.

Area = $\frac{1}{2}(base)(height)$ = $\frac{1}{2}|a||b|sinθ$

The lecturer says to first find cosine of the angle using dot product. Next, solve for sine using [sin$^{2}θ + cos^{2}θ$ = 1]. And then substitute into area formula.

However, he doesn't actually work this out, so I don't know what formula its supposed to produce or how to actually arrive there. This is a section on cross products so this might be a preface to the cross product formula and I'm interest to know where he was actually going with this. Any info would be very appreciated. Thanks.

 

[Nathanael]

The magnitude of the cross product ($\frac{1}{2}|a||b|sinθ$) is equal to the area of the parrallelagram made by the two vectors.

This means the area of a triangle is equal to half of the magnitude of the cross product of two of the sides (with the third side being the diagonal of the parrallelgram which connects the ends of the first two sides)

 

[SteamKing]

The magnitude of the cross product ($\frac{1}{2}|a||b|sinθ$) is equal to the area of the parrallelagram made by the two vectors.

The magnitude of the cross product of vectors A and B is equal to the area of the parallelogram made by the two vectors. Half of this magnitude is the area of the triangle formed by the two vectors.

http://en.wikipedia.org/wiki/Cross_product

Check the 'Properties' section in the article above for an illustration.

 

[Lebombo]

Thank you! Appreciate you guys