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# The Cross Product of Vectors - 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.

In Calculus we often use **the cross product of vectors** to find orthogonal vectors and the area of parallelograms in three dimensions. The cross product of vectors is found by identifying the 3x3 determinants, however we substitute one of the rows with symbols that represent unit vectors. Finding 3x3 determinants can be made easier if we understand how to simplify determinants.

One really important application of determinants is that it can be used to define the cross product of vectors so this brings us back to vectors. Let's review one idea though. A vector a 3D vector v whose components are a, b, c can be written in terms of the unit vectors i, j and k where i is the unit vector in the positive x direction, j is the unit vector in the positive y direction and k is the unit vector in the positive z direction so it's in alphabetic order just like xyz, ijk and this is how you would write vector v you'd write it as aah aah a times i plus b times j plus c times k right, the components are still easy to see.

Now let's suppose we have two vectors in this form in this i, j, k form then the cross product is defined v cross w equals the determinant and across the first row, I put the 3 unit vectors i, j and k and then I put across the second row the components of the first vector in the cross product v and across the last row I put the components of the second vector in the cross product w so that's really important the order that you put these two right? The first one in the cross product you put in the second row the second one you put in the third row. Let's try this in an example.

So I have two vectors v equals 4, -5, -3, w equals 2, 2, 1 I first have to put these in i j k form so that's really easy because all you have to do is just fill in the i-5j-3k same for this, 2i plus 2j plus 1k it's just k and then we cross w is the determinant i, j, k so you always put i, j, k across the top and then 4, -5, -3 those are the components of v and then the components of w 2, 2 and 1 and you expand this the way you would a normal determinant so I'll expand across the top row I have i times this minor -5, -3, 2, 1 then I have to subtract j times it's minor 4, -3, 2, 1 then I add k times its minor oops 4, -5, 2, 2 okay and then I just compute these two by two determinants so this one is going to be -5--6 so -5+6 which is 1 that gives me i times 1 minus j this will be 4--6, 4+6 10 so j times 10 plus k I have 8--10 8+10 is 18 so the answer is i-10j+18k and of course you can put that into regular component form if you like, v cross w is 1,-10, 18 so depending on what what form your teacher wants either of these is an acceptable answer.

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###### 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!”

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