Vector of a line
λb
a
b
r
Vector equation:
r  a  b
Where r, a, and b are vectors and  is a scalar
(parameter) and it is the same as t used in the book.
 b1 
 x   a1 



b 
 y a 
   2
 2
Hint: Think of the a as a fixed point and
b as the gradient
Parametric form
x  a1  b1
y  a2  b2
Cartesian form
x  a1 y  a2

b1
b2
Example 1: Find the vector equation of a line L, which passes through A
(2,5) and parallel to vector 3i  4 j .
Solution: Since the lines are parallel the vector can be used for b
r  2i  5 j    3i  4 j or r   2  3  i   5  4  j
Or
 2  3 
r    t 
 5   4 
Note we can convert this into a standard equation form by using Cartesian
form:
x2 y 5

, so  4 x  8  3 y  15
3
4
4
23
4 x  3 y  23, or y   x 
3
3
Example 2: Find the vector equation of the line L, passing through A (1,4)
and B (5,8). Give both the parametric and Cartesian form of L.
A
P
B
O
 1   5  4
 4 8  4
1
1  1  1
Then OP: OP  OA  OP     4       t  
 4
1  4  1 
 1
 1
First we need to find: AB  OA  OB            4  
Should read OP = OA + AP
Therefore the parametric form would be: x  1  t , and y  4  t
The Cartesian form would be:
x 1 y  4

or y  x  3
1
1
Example 3: Find the angle between the lines:
x  2 y 1
x2 y4


and
4
3
1
2
First write the equations in vector equation or parametric form:
x  2 y 1

4
3
 x  2 
 4



 y   1
 3
   
 
 x   2 
 1



 y  4 
2
   
 
 4
 1 
The first line is parallel to   and the second is parallel to  
 3
2
x2 y4

1
2
To find the angle we need to use the scalar product equation:
a1b1  a2b2  a b cos
4  1  3  2  5  5 cos
2
 cos so   79.7
5 5
Lines in 3-D
 x   a1   b1 
r   y    a2   t  b2  , R (x, y, z) is any point on the line.
 z  a  b 
   3  3
x  a1  tb1 

y  a2  tb2  are the parametric equations.
z  a3  tb3 
Example 4: Find the angle between the lines:
 1  1 
0 1
L1 : r1   1   t  2  , L2 : r2   2   t  2 
2  3
 1  3 
   
   
a1b1  a2b2  a3b3
 cos
r1 r2
1 4  9
 1  cos
14 14
The lines are parallel so the angle is 0.
Relationships between lines:
Parallel
Intersecting
Coincident
Skew: do not intersect, not parallel and are not coplanar. They can still be
considered to form an angle by translating one of the lines.
Example 5: Are the lines parallel? If not what is the angle between them.
Line 1: x  2  3s, y  3  1s and z  4  2s
Line 2: x  4  6s, y  2s and z  1  4s
Line 3: x  1  2s, y  1  2s and z  4s
Solution:
 6 
3
 
 
Lines 1 and 2 are parallel since 2  2 1
 
 
 4 
2
 
 
Angle between lines 1 and 3:
77.4, 102.6 degrees
Example 6: Classify the following line pairs as either parallel, intersecting or skew, and
each case find the measure of the acute angle between them:
x  1  2t , y  2  12t and z  4  12t
x  4s  3, y  3s  2 and z  s  1
2
4
Is (−12) = 𝑘 ( 3 ) ?
12
−1
No so the lines are not parallel.
2
1
Finding the angle: (−12) = 2 (−6)
12
6
4 − 18 − 6
√1 + 36 + 36√16 + 9 + 1
= 𝑐𝑜𝑠𝜃
−20
= 𝑐𝑜𝑠𝜃
√73√26
𝜃 = 117.3273 𝑜𝑟 62.67268
The angle is 62.7
Do the lines intersect?
For this we need to make the equations equal and see if they work.
For x: −1 + 2𝑡 = 4𝑠 − 3 so 𝑡 = 2𝑠 − 1
For y: 2 − 12𝑡 = 3𝑠 + 2 so −4𝑡 = 𝑠
−1
4
Therefore if we solve this we find that 𝑡 = 9 , 𝑠 = 9.
Does this work?
We need to check this in the third set (for z): 4 + 12𝑡 = −𝑠 − 1
−1
4
When we plug this in they are not equal since 4 + 12 ( 9 ) ≠ − 9 − 1.
The lines do not intersect, and are not parallel therefore they are skew.
Example 7: Find the point of intersection L1 r1  2i  3j    2i  j and
L2 r2  5i  2 j    i  2 j
If they intersect then we can find the values of  when r1  r2
 12 16 
, 
5
5

Solution: 
Extra Work:
10.
a)
i. ⃗⃗⃗⃗⃗
𝑂𝐹 = 2𝑖 + 5𝑗 + 3𝑘
ii. ⃗⃗⃗⃗⃗
𝐴𝐺 = −2𝑖 + 5𝑗 + 3𝑘
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ | = √4 + 25 + 9 = √38
b) |𝑂𝐹 | = √4 + 25 + 9 = √38, |𝐴𝐺
⃗⃗⃗⃗⃗ ∙ 𝐴𝐺
⃗⃗⃗⃗⃗ = −4 + 25 + 9 = 30
iii. 𝑂𝐹
c)
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
𝑂𝐹∙𝐴𝐺
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ |
|𝑂𝐹 ||𝐴𝐺
=
30
√38√38
= cos 𝜃, 𝑠𝑜 𝜃 = 37.9
should be -2+3s=-20-4t
Note the intersection point is (5, -8, 15)
60 degrees