Linear Algebra 46, Distance from the Origin to a Plane

Linear Algebra 46, Distance from the Origin to a Plane

[aharontam]

I love your videos.
Just want to say that the proof of distance is quite easy to show:
1. First we can claim that the distance of any point outside the hyperplane (Y1, Y2, ... Yn) to the plane is perpendicular line from the that point to the plane which means it has the direction of  the normal vector.
2. If this vector crosses the hyperplane at point (X1, X2, , , , Xn), we can express by the formula: X=(X1, X2, , , , Xn)-(Y1, Y2, , , , Yn). And if this outside point is the origin then X is simply X=(X1, X2, , , , Xn)-0; which is a vector from origin to the point of contact on the hyperplane.
3. The length of this vector is X*X=L^2
4. Since X is a point on the plane: it satisfies: X*N=b, N being the Normal vector
5. squaring both sides yields: X*X*N*N = b^2
6. L^2*N*N=b^2 ==> L^2*(a1^2+a2^2+....) = b^2
7. L=|b|/sqrt(a1^2+a2^2+....)

[guibarta]

thank you for the videos..woud you ever upload videos about universal algebra(closure systems and direct and subdirect decomposition of algebras)?