##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- FREE study tips and eBooks on various topics

# Area With the Cross Product - Concept

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Finding the area of a parallelogram in two dimensions involves the area determinant of a 2x2 matrix, but if we're given a parallelogram in three dimensions we can use the **cross product area**. The cross product area is a technique often used in vector calculus. The cross product is found using methods of 3x3 determinants, and these methods are necessary for finding the cross product area.

I want to show you how the cross product can actually give you the area of a parallelogram. First, let me show you a property of the cross product. For any two 3D vectors u and v, the magnitude of their cross product equals the product of their magnitude times the sine of theta and theta of course is the angle between vectors u and v. Now, this result probably looks very familiar to you it's very similar to the result about the dot product. The dot product u dot v equals the magnitude of u dot times the magnitude of v cosine theta so it's very similar let's remember that the dot product gives you a scalar value right this is a scalar and this does look very similar to this but we actually have to take the magnitude of u cross v in order to get this result so u the cross product goes the sine the dot product go goes with cosine.

Alright, now let's imagine a situation we have two vectors if we put the two vectors end to end, they will always determine a parallelogram in space or in a plane, and this parallelogram is going to have an area. Now let's say that this is theta the angle between the two vectors what would the area be? Well the area will be the base times the altitude so the area of our parallelogram, that's my symbol for parallelogram, equals b times h base times height and the base in this case is the magnitude of vector v. What's the height? Let's draw it in, let's say that this is our height okay this is a right angle we can use right right angle right triangle trigonometry to figure out the h is this length times the sine of theta. This length is the magnitude of u and there we have it, this is our height so magnitude of u sine theta this is exactly the same as magnitude of u magnitude of v sine theta so that's the area of a parallelogram it's the same thing as this, that's an area of a parallelogram since these two things are equal it means that this also gives you the area of the parallelogram and this is usually easier to calculate so the important lesson here u and v will always form a parallelogram and the magnitude of u cross v is the area of that parallelogram.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

##### Concept (1)

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete