Question

Find a unit vector orthogonal to the plane containing the points A= 1,0,0 , B= 3,−1,−3 , and C= 1,3,−2 .

Answer 1

Answer:

unit normal vector n will be n=(a,b,c) = (4/√171,11/√171,6/√171)

Step-by-step explanation:

There are several ways to solve this problem

1) build 2 vectors AB and BC such that the vectorial product ABxBC is the orthogonal vector to the plane , then find unit vector

2) since the 3 points belongs to the plane solve a linear system of 4 equation with 4 variables

for the second solution , the equation of the plane with normal vector n=(a,b,c) and containing the point (x₀,y₀,z₀) is

a*(x-x₀)+b*(y-y₀)+c*(z-z₀) =0

and

a²+b²+c² = 1 (unit vector)

then choosing A=(x₀,y₀,z₀)=(1,0,0)

a*(x-1)+b*(y-0)+c*(z-0) =0

for B

a*(3-1)+b*(-1-0)+c*(-3-0) =0

1) 2*a - b - 3*c =0

for C

a*(1-1)+b*(3-0)+c*(-2-0) =0

2) 3*b - 2*c=0 → b= 2/3*c

replacing in 1)

2*a -  2/3*c - 3*c =0

2*a-11/3*c=0 → a=11/6*c

thus

a²+b²+c² = 1

(11/6*c)²+(2/3*c)²+c² = 1

(121/36+4/9+1)*c² = 1

171/36*c²=1 → c= 6/√171

therefore

a=11/6*c = 11/6*6/√171= 11/√171

b=2/3*c= 2/3*6/√171= 4/√171

then the unit normal vector n will be

n=(a,b,c) = (4/√171,11/√171,6/√171)