A First Course in Linear Algebra, 2017a

172 R n
We will conclude this section with an important application of projections. Suppose a line L and a
point P are given such that P is not contained in L. Through the use of projections, we can determine the
shortest distance from P to L.
Example 4.40: Shortest Distance from a Point to a Line
Let P =(1,3,5) be a⎡point⎤
in R 3 ,andletL be the line which goes through point P 0 =(0,4,−2) with
direction vector d ⃗ 2
= ⎣ 1 ⎦. Find the shortest distance from P to the line L, and find the point Q on
2
L that is closest to P.
Solution. In order to determine the shortest distance from P to L, we will first find the vector −→ P 0 P and then
find the projection of this vector onto L. The vector −→ P 0 P is given by
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
1 0 1
⎣ 3 ⎦ − ⎣ 4 ⎦ = ⎣ −1 ⎦
5 −2 7
Then, if Q is the point on L closest to P, it follows that
−−→ −→
P0 Q = projd
⃗ P 0 P
( −→
P0 P • d
=
⃗ )
⃗d
‖⃗d‖ 2
⎡ ⎤
= 15 2
⎣ 1 ⎦
9
2
⎡ ⎤
= 5 2
⎣ 1 ⎦
3
2
Now, the distance from P to L is given by
‖ −→ QP‖ = ‖ −→ P 0 P − −−→ P 0 Q‖ = √ 26
The point Q is found by adding the vector −−→ P 0 Q to the position vector −→ 0P 0 for P 0 as follows
⎡ ⎤
⎡
⎣
0
4
−2
⎤
⎡
⎦ + 5 ⎣
3
2
1
2
⎤
⎦ =
Therefore, Q =( 10 3 , 17
3 , 4 3 ). ♠
⎢
⎣
10
3
17
3
4
3
⎥
⎦