D is the midpoint of the edge AB of the triangle ABC, then the vector CD equals

As shown, D is the midpoint on edge AB of triangle ABC, then vector CD equals A.-BC+1/2BAB.-BC-1/2BAC.BC-1/2BAD.BC+1/2BAD As shown, D is the midpoint on edge AB of triangle ABC, then vector CD equals A.-BC+1/2BAB.-BC-1/2BAC.BC-1/2BAD.BC+1/2BA

How to find the normal vector of plane.

Plane normal vector. X+3y+2z+1=0 Why (1.3.2) is the normal vector of this plane.

In fact, a plane has countless normal vectors, these normal vectors are parallel.
Any plane: ax+by+cz+d=0, take a set of numbers x0, y0, z0 to satisfy the equation, then:
Ax0+by0+cz0+d=0, the two formulas are subtracted: a (x-x0)+b (y-y0)+c (z-z0)=0, this is the point normal equation of the plane
A plane representing a passing point (x0, y0, z0) with n=(a, b, c) as the normal. Ax+by+cz+d=0 is the general equation of the plane
Remember: the coefficients of x, y and z in the equation are a normal vector of the plane
Your equation is like this, so a normal vector of the plane: n=(1,3,2), but not unique
Like 3n =(3,9,6) is also.

When the vector a, b vectors a, b |a+b|=|a-b| What is the positional relation of vectors a, b |a+b|=|a-b| When the vector a, b form the position relation |a+b|=|a-b|, find the graph

D is the midpoint of the edge AB of the triangle ABC, then the vector CD equals