Is the plane normal vector parallel to the same plane

Is the plane normal vector parallel to the same plane

Yes, or coincidence

How to multiply a space vector? How to calculate a module of a space vector and how to multiply a module

Space vectors are expressed by coordinates. Vector multiplication is the product of abscissa of two vectors plus the product of ordinate plus the product of z-axis coordinates. For example, if AB vector coordinates are (A1, B1, C1) and CD vector coordinates are (A2, B2, C2), then vector AB multiplied by vector CD equals to A1A2 + b1b2 + C1C2. The module of vector is the abscissa under the root sign, For example, the module of the vector AB whose coordinate axis is (a, B, c) AB is A2 + B2 + C2 under the root sign. The module has no direction but only size. The multiplication by touching is equivalent to the multiplication of primary school numbers, and the direct multiplication is OK

Why does vector a multiply vector B directly by them instead of by modulus?
The example is to find the angle between vector a = 2E1 + E2 and vector b = 2e2-3e1
Analytic solution to the vector E1 point multiplication E2 = 1 / 2
The next step is to multiply the vector a by the vector b = (2E1 = E2) (2e2-3e1). I just didn't understand here. Isn't it the cos angle of the modular multiplication B of the modular multiplication B of the vector a by the X vector b = a? How can we directly multiply the vectors a and B here? This is different from the definition=

a·b=(2e1+e2)·(2e2-3e1) =-6|e1|^2+2|e2|^2-4e1·e2
This is the distributive law of the scalar product of vectors
Of course, we can also use the formula: a · B = | 2E1 + E2 | * | 2e2-3e1 | * cos
But the calculation is too complicated and flexible

Multiplication of vector modules
Given the vector PA = (- 6-x, - y), the vector Pb = (6-x, - y), help me to simplify | PA | times | Pb | = 36

|PA|=√[(x+6)^2+y^2] |PB|=√[(x-6)^2+y^2]
|PA|*|PB|=√[(x^2+y^2+36)^2-144x^2]=36
x^4+y^4-72x^2+72y^2+2x^2y^2=0