The Dot Product of Vectors - Problem 3


Let’s do another exercise with the dot product. Let v equal 3,6 w equal 2,-5. Compute v.v. So let’s see what happens when you dot a vector with itself.

Remember we multiply components. So it would be 3 times 3, or 9 plus 6 times 6 or 36 and that’s 45. Let’s compare that to the magnitude of v². Remember the magnitude of v is the square root of 3² plus 6² that’s 9 plus 36. And that had to be squared. It’s again the square root of 45² and that’s 45. So we actually got the same answer.

Let’s see if that happens again. W is 2,-5, let’s calculate w.w. We get 2 times 2, plus -5 times -5 and that’s going to be, I'll move this down, 4 plus 25, 29. So how does that compare with the magnitude of w²?

The magnitude of w is the square root of 2², 4, plus -5², 25. And I have to square that. Again it's root 29² and that’s 29. It looks like these two will give me the same result, that’s sort of interesting. That there’s a relationship between the dot product, dotting a vector with itself and the vector's magnitude. Let’s prove this result.

If u equals x,y prove that the magnitude of u² is u.u. So we start with the magnitude of u². Just start with one side and work out until you get the other side. U is x,y so let me write this as replacing. The magnitude of x,y is the square root of (x² plus y²) squared. And you know that the square root squared, is just going to give me x² plus y². Now let me observe that that’s x times x plus y times y.

And that means that this is exactly the dot product of the vector x,y with the vector x,y, which is precisely u.u. And that’s our result. So it’s always true, the magnitude of a vector squared is equals that vector dotted with itself.

vector components component form dot product scalar product commutative