A First Course in Linear Algebra, 2017a

176 R n
Solution. By Definition 4.42, P is a point in the plane if it satisfies the equation
⃗n • ( −→ 0P − −→ 0P 0 )=0
Given the above⃗n, P 0 ,andP, this equation becomes
⎡ ⎤ ⎛⎡
⎤ ⎡ ⎤⎞
1 5 2
⎣ 2 ⎦ • ⎝⎣
4 ⎦ − ⎣ 1 ⎦⎠ =
3 1 4
⎡
⎣
1
2
3
⎤
⎛⎡
⎦ • ⎝⎣
= 3 + 6 − 9 = 0
3
3
−3
⎤⎞
⎦⎠
Therefore P =(5,4,1) is contained in the plane.
⎡
Suppose⃗n = ⎣
Then
a
b
c
⎤
⎦, P =(x,y,z) and P 0 =(x 0 ,y 0 ,z 0 ).
⎡
⎣
a
b
c
⎤
⎛⎡
⎦ • ⎝⎣
⎡
⎣
x
y
z
a
b
c
⃗n • ( −→ 0P − −→ 0P 0 ) = 0
⎤ ⎡ ⎤⎞
x 0
⎦ − ⎣ y 0
⎦⎠ = 0
z 0
⎤
⎡
⎦ • ⎣
⎤
x − x 0
y − y 0
⎦ = 0
z − z 0
a(x − x 0 )+b(y − y 0 )+c(z − z 0 ) = 0
♠
We can also write this equation as
ax + by + cz = ax 0 + by 0 + cz 0
Notice that since P 0 is given, ax 0 + by 0 + cz 0 is a known scalar, which we can call d. This equation
becomes
ax + by + cz = d
Definition 4.44: Scalar Equation of a Plane
⎡ ⎤
a
Let ⃗n = ⎣ b ⎦ be the normal vector for a plane which contains the point P 0 =(x 0 ,y 0 ,z 0 ).Then if
c
P =(x,y,z) is an arbitrary point on the plane, the scalar equation of the plane is given by
where a,b,c,d ∈ R and d = ax 0 + by 0 + cz 0 .
ax + by + cz = d