Normal vector of 2- 1 plane, an elective course of mathematics in senior two.

Solution: Vector AB = (1, -4,1)-(-2,0,3) = (3,4,2);

Vector AC = (0,3,6)-(-2,0,3) = (2,3,3).

Vector AB∈α, vector AC∈α,

Let n=(x, y, z) be the normal vector α, ∫ vector n⊥α, ∴ vector n⊥ vector AB, and vector n⊥ vector AC of the plane.

That is, vector n. vector AB=0, vector n. vector AC=0.

That is (x, y, z). (3,-4,-2)=0.

3x-4y-2z=0。

(x，y，z)。 (2,3,3)=0,

2x+3y+3z=0。

If z=3, then 3x-4y-6 = 0 (1) and 2x+3y+9 = 0 (2).

Combined solution, x =- 18/ 17, y =-39/ 17.

∴ vector n = (- 18/ 17,-39/17,3) and vector a = (3,4,5).

Vector a, vector n = (3 3,4,5). (- 18/ 17,-39/ 17,3) =-54/ 17- 156/ 17.

| Vector A | = √ (3 2+4 2+5 2) = 5 √ 2.

| Vector n | = √ [(-18/17) 2+(-39/17) 2+3 2] = (√ 4446)/17.

Let the angle between vector a and vector n be

cos & lta，n & gt=a.n/|a|。 | n | =(45/ 17)/{(5√2)*[√4446)/ 17]} =(45√2223)/( 10 * 2223)。

∴cos <a，n & gt=9√2223/4446.