A First Course in Linear Algebra, 2017a

246 R n
Now, we need to write ⃗y as the sum of a vector in W and a vector in W ⊥ . This can easily be done as
follows:
⃗y =⃗z +(⃗y −⃗z)
since⃗z is in W and as we have seen⃗y −⃗z is in W ⊥ .
The vector⃗y −⃗z is given by
⎡ ⎤ ⎡
1
⃗y −⃗z = ⎢ 2
⎥
⎣ 3 ⎦ − ⎢
⎣
4
Therefore, we can write⃗y as
⎡
⎢
⎣
1
2
3
4
⎤
⎡
⎥
⎦ = ⎢
⎣
2
2
2
4
2
2
2
4
⎤
⎤
⎡
⎥
⎦ = ⎢
⎡
⎥
⎦ + ⎢
⎣
⎣
−1
0
1
0
−1
0
1
0
⎤
⎥
⎦
⎤
⎥
⎦
♠
Example 4.149: Point in a Plane Closest to a Given Point
Find the point Z in the plane 3x + y − 2z = 0 that is closest to the point Y =(1,1,1).
Solution. The solution will proceed as follows.
1. Find a basis X of the subspace W of R 3 defined by the equation 3x + y − 2z = 0.
2. Orthogonalize the basis X to get an orthogonal basis B of W.
3. Find the projection on W of the position vector of the point Y .
We now begin the solution.
1. 3x + y − 2z = 0 is a system of one equation in three variables. Putting the augmented matrix in
reduced row-echelon form:
[
3 1 −2 0
]
→
[
1
1
3
− 2 3
0 ]
gives general solution x = − 1 3 s + 2 3t, y = s, z = t for any s,t ∈ R. Then
⎧⎡
⎨
W = span ⎣ − 1 ⎤ ⎡ ⎤⎫
2
3 3
⎬
1 ⎦, ⎣ 0 ⎦
⎩
⎭
0 1
⎧⎡
⎨
Let X = ⎣
⎩
W.
−1
3
0
⎤
⎡
⎦, ⎣
2
0
3
⎤⎫
⎬
⎦ .ThenX is linearly independent and span(X)=W, soX is a basis of
⎭