Calculus - The Cross Product

This video shows how to use the cross product to determine if two vectors are parallel, find perpendicular vectors, areas of parallelograms, and volume of a parallelepiped. We determine if three vectors are coplanar and work an application problem finding torque.

[umarbashir]

Sir, how to solve determinents of trigonometric variables

Det(sin40°. Cos60°)

These type of questions we apply the basic technique but enable to solve it lastly

[thomasolson]

I think I found something interesting about the altitudes of cos and sin with these vectors.
a:=sqrt(3)/3 <1, 1, 1>
b:=sqrt(3)/3 <1, -1, -1>
I should point out that these are unit vectors and are involved in the cube.
a.b/(|a|*|b|)=side bc =cos(θ)=-1/3 (remove neg sign for triangle solver)
|a×b|/(|a|*|b|)=side ab = sin(θ)=2*sqrt(2)*(1/3)
Anyway, I put those into a triangle solver and noticed the altitudes are the same
alt side bc = 1/3
alt side ab = 2sqrt(2)/3
alt side ac = bc*ab
I realized a few days ago that this behavior is what I see when I take a R^3 vector and make it into an R^2 vector by sqrt(y^2+z^2).
cos(θ)=x/sqrt(x^2+y^2+z^2)

So, <-1/3, 2*sqrt(2)*(1/3)> must be some kind of vector in R^3 <x, y, z>. This particular ones is easy to find. Solve for x.

x=sqrt(2)/2
I got lucky here because I used a vector from the cube. y and z were equal, so this worked out.
c:=<-1/3, 2/3, 2/3>
I have another vector for ya.
d:=(x, y, z) 🢡<(x^2-y^2-z^2)/(x^2+y^2+z^2), (2*x*y)/(x^2+y^2+z^2), (2*x*z)/(x^2+y^2+z^2)>
d(1, 1, 1)=<-1/3, 2/3, 2/3>
This d function is algebra for the sphere. You can find it in the Mandlebulb as part of a resultant vector. The algebra for the circle has meaning in angle multiplication/addition. So d is like some kind of double angle but in 3D where no such thing exists. d never returns to a zero vector like an R^2 polytope would (if done recursively). I can't tell you if d(1, 1, 1) is a vector between a and b, but I think it might be.