A First Course in Linear Algebra, 2017a

160 R n
⎡
Let ⃗q = ⎣
Then,
x
y
z
⎤
can be written as
⎦. Then, we can find ⃗p and ⃗p 0 by taking the position vectors of points P and P 0 respectively.
⎡
Here, the direction vector ⎣
⎦ as indicated above in Definition
4.19.
1
−6
6
⎤
⎡
⎣
x
y
z
⎤
⎦ = ⎣
⃗q = ⃗p 0 +t (⃗p − ⃗p 0 )
⎡
1
2
0
⎤
⎡
⎦ +t ⎣
1
−6
6
⎡
⎦ is obtained by ⃗p − ⃗p 0 = ⎣
⎤
⎦, t ∈ R
2
−4
6
⎤
⎡
⎦ − ⎣
1
2
0
⎤
Notice that in the above example we said that we found “a” vector equation for the line, not “the”
equation. The reason for this terminology is that there are infinitely many different vector equations for
the same line. To see this, replace t with another parameter, say 3s. Then you obtain a different vector
equation for the same line because the same set of points is obtained.
⎡ ⎤
1
In Example 4.21, the vector given by ⎣ −6 ⎦ is the direction vector defined in Definition 4.19. Ifwe
6
know the direction vector of a line, as well as a point on the line, we can find the vector equation.
Consider the following example.
Example 4.22: A Line From a Point and a Direction Vector
Find⎡a vector ⎤ equation for the line which contains the point P 0 =(1,2,0) and has direction vector
1
⃗d = ⎣ 2 ⎦
1
♠
Solution. We will use Definition 4.19 to write this line in the form ⃗p = ⃗p 0 + td, ⃗ t ∈ R. We are given the
⎡direction ⎤ vector ⃗d. In ⎡ order ⎤ to find ⃗p 0 , we can use the position vector of the point P 0 . Thisisgivenby
1
x
⎣ 2 ⎦.Letting⃗p = ⎣ y ⎦, the equation for the line is given by
0
z
⎡
⎣
x
y
z
⎤
⎡
⎦ = ⎣
1
2
0
⎤
⎡
⎦ +t ⎣
1
2
1
⎤
⎦, t ∈ R (4.6)
We sometimes elect to write a line such as the one given in 4.6 in the form
⎫
x = 1 +t ⎬
y = 2 + 2t where t ∈ R (4.7)
⎭
z = t
♠