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# Calculating dot and cross products with unit vector notation

Calculating the dot and cross products when vectors are presented in their x, y, and z (or i, j, and k) components. Created by Sal Khan.

## Want to join the conversation?

• Sal said at in the video that he is going to do a proof can some one tell me where is it?
• He was trying to prove that the dot product is easier and more compatible than the cross product
• I don't understand how he determines the sign (plus or minus) of the components of the resulting vector of the cross product. Could someone explain please?? Would really appreciate it!
• At Sal points out that the signs are + - +.
No matter what are the values of the vectors involved, this pattern stays the same because of how matrices are defined.

This means that the i component is +[value of the first matrix's determinant],

the value of the j component is -[value of the second matrix's determinant]

and, finally, the value of the k component is +[value of the third matrix's determinant]

You can learn more heading to the linear algebra section or checking out the 3Blue1Brown linear algebra course.
• In the value -35i in the cross product should have been -45i. (-24-21=-45).
• he says it's 'plus minus plus', is it always this way or is it because of the values in this problem?
• No it is always in that way.the first row will be + - +
second will be - + -
third will be + - +
• I have a question. For the cross product, after you find out which direction it points to using the right hand rule, how do we actually write this down on paper? Using the symbol Sal showed us? Or replace n with the positive or negative unit vector?
• Well a cross product would give you two possible vectors, each pointing in the opposite direction of the other, and each orthogonal to the two vectors you crossed. If the vector your calculated, ie. <x, y, z> is going in the correct direction based on the right hand rule, you can leave it positive. If you need it's opposite, multiply it by a negative scalar, and your answer will be -<x, y, z> which is also <-x, -y, -z>.
• Why is the j value taken as negative
(1 vote)
• Great question. The reason is that the cross product uses the "right-hand rule", and the unit vectors i, j, and k, have a "right-handed" orientation. What this means is that i X j = k (this is why k is taken as positive).

Now, draw (or imagine) the three coordinate axes in space, so that it's true that i X j = k (that means that when you point your fingers along the x-axis (i) and then curl them 90 degrees to the y-axis (j), then your thumb must point along the z-axis (k) ). We want to check the other two cross products of unit vectors. You will see from these axes, that j X k = i, which is why i is taken as positive.
Here is an image: https://services.math.duke.edu/education/ccp/materials/mvcalc/vectors/hand1a.gif

But you can also see that i X k = -j (that is, when you point your fingers along the x-axis (i) and then curl them 90 degrees to the z-axis (k), then your thumb will point along the NEGATIVE y-axis (-j) ). This is why j is taken as negative.

Ultimately, it is related to the fact that the "cyclic permutations" of 123 are 231 and 312. (meaning, if I write 123 again and again, 123123123, then the only "three in a row" patterns are 123, 231, and 312). We can translate these, using x~i~1, y~j~2, z~k~3:
123: i X j = k
231: j X k = i
312: k X i = j
But the three OTHER permutations of 1, 2, and 3 are 321, 213, 132, which are the reverse of the above, and that confirms what we should already know -- that reversing the order of a cross product gives us the OPPOSITE result:
213: j X i = -k
321: k X j = -i
132: i X k = -j
You should be able to confirm all six of these using the right-hand rule on your x-y-z axes.
• Did Sal get to making the proof video for the cross product with unit vector notation? I would very much like to see that proof.
• There is no prof yet on khan academy
but sal is thinking about making one.
(1 vote)
• what if the sum is 0? does it mean it is perpendicular?
(1 vote)
• if the sum is 0 it means that the vectors are equal in magnitude but opposite in direction .