# Area of a triangle (cross product lesson)

1. Jul 16, 2014

### Lebombo

1. The problem statement, all variables and given/known data

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.

2. Jul 16, 2014

### 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)

3. Jul 16, 2014

### SteamKing

Staff Emeritus

4. Sep 12, 2015

### Lebombo

Thank you! Appreciate you guys