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# Defining the angle between vectors

Introducing the idea of an angle between two vectors. Created by Sal Khan.

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• I don't see how this proof is valid in dimensional spaces other than R2. He defined the angles using a sketch of a triangle in 2D, and then used the law of cosines which wasn't proved to be valid (or meaningful) outside of 2D and 3D space. Isn't there a whole lot of hand-waiving in this proof? •   The definition of an angle between vectors is the angle between two sides of a triangle in 2D with lengths ||a||,||b||,||a-b||. Such a triangle is 2D by construction even for v∈ℝⁿ because ||v|| is a scalar.
• What is the difference between Orthogonal and Perpendicular ? He said that Zero Vector is Orthogonal to everything but not Perpendicular ? I don't get it. o.O • Two vectors are perpendicular if they are not the zero vector AND their dot product is zero. They are only orthogonal if one or both of them are the zero vector and their dot product is zero. The definition of perpendicular relies on the angle between the vectors being 90 degrees, and with the zero vector, there's no intuitive way of thinking about the angle. The orthogonal case deals with the zero vector, and it is orthogonal to every vector because the zero vector dotted with anything is zero.
• If we wanted to solve for the angle between two known vectors, would it not be easier to rearrange the equation Sal explains at ? Instead you could have it as θ= arccos((a・b) /(||a||*||b||) • "Find two unit vectors that make an angle of 60 degrees with u = (3; 4)."

I've been trying everything for this question, but I've yet to come across an answer. Would anybody be able to explain how to approach this problem? I know that ||u|| = 5 and that another vector v, ||v|| = 1 , but I can't seem to accomplish anything more than this. Please help!!! • Here's an approach which doesn't require us to know anything besides the angle formula and the length formula for vectors:

You want to find vectors v' = [x, y] (where v' means the transpose of v) such that:
u . v = |u||v|cos(60)
or, in other words,
u . v = 5*|v|*1/2
We also want the vector v to have length one, i.e. |v|=1.

In total, we get two constraints on v:
(i) 3x + 4y = 5/2 (from your u, the dot-product formula, and the above equation)
(ii) x^2 + y^2 = 1 (since the length (the square root of this) should be 1)
Right. Now we have to solve this for x and y.

Equation (i) gives us that
x = 5/6 - 4y/3
Plugging this into Equation (ii) gives us
(5/6 - 4y/3)^2 + y^2 = 1,
or (by multiplying out and simplifying a bit):
25y^2/9 - 20y/9 + 25/36 -1 = 0
We're looking for the roots of this second order polynomial. They are
y = 1/10 (4 - 3sqrt(3)) and y = 1/10 (4+3sqrt(3))
where sqrt(3) means the square root of 3.

Alright, now we know all the possible values for y. What are the corresponding x-values? Use Equation (ii) to get
x = 5/6 - 4/3 (1/10 (4 - 3sqrt(3))) and x = 5/6 - 4/3 (1/10 (4+3sqrt(3)),
which simplifies to
x = 1/10(3 + 4sqrt(3)) and x = 1/10(3-4sqrt(3)).

Our two vectors are:
v = [x,y]
with these values for x and y.

Tell me if anything was unclear. :)
• At Sal tells us that the zero vector is orthogonal to all other vectors.
If the zero vector is orthogonal to all other vectors, doesn't that arise from R0 being orthogonal to R1 in the same way that in constructing the second dimension of R2 one sets the x any y axes orthogonal to each other? • By definition, orthogonal is the name given to the relationship between two vectors described when their dot product is 0. The dot product of the 0 vector with any other vector is 0, so by defintion the 0 vector is orthogonal to every other vector. In other words it just has the right properties for that definition (which someone made up).. there is no more to it than that.
• At why sal changed the direction of vector a-b to b-a ? • I don't understand how |a| can be less than or equal to |b| + |a-b|.
I get how it has to be less than, but not how it can be equal to, and still be a triangle. It's completely flat, and has no height whatsoever, so why is that a valid triangle? • What is the difference between 'Orthogonal' and 'Orthonormal' ?
(1 vote) • Does the word "orthogonal" mean "dot product = 0"? Does the word "perpendicular" have any usage in R^n, n>3? • You are exactly correct, orthogonal means the dot product is zero. Orthogonal is the generalization of the notion of perpendicular, which is usually used only in two dimensions.

In the comments Matthew Daly points out the technical distinction: "The distinction between the two is that (by definition) the zero vector is orthogonal to all vectors but perpendicular to none of them."

In higher dimensions there is more than line that is orthogonal to another so you have to be careful about that too.

I have heard physics books refer to a line perpendicular to a plane in three dimensions but once again, orthogonal is the more mathematical word. 