UBC Mathematics 152 Midterm I Answer to Version B Q.B3
Solution:
(a) In the parametric formula for L, the direction is the vector multiplied by the parameter
 
1
~lL = 5
2
(b) Similarly, the direction of line M is
~lM
 
2
= 10 = 2~lL .
4
Thus, ~lL and ~lM are constant multiples of each other ⇒ L, M are parallel.
Alternatively,
i
~lL × ~lM = 1
2
j
5
10
k
2 = i(20 − 20) − j(4 − 4) + k(10 − 10) = ~0
4
⇒
They are parallel!
(c) Answer 1: It is impossible to turn p~L = (1, 0, 0) into p~M = (2, 2, 2) by adding a scalar
multiple of (1, 5, 2).
Answer 2: If identical, d~ = p~M − p~L = u~lL , but d~ = p~M − p~L = (1, 2, 2) 6= u(1, 5, 2).
Answer 3: If identical, p~M = (2, 2, 2) should satisfy the equation for L, thus
 
   

  
1
1
2
1+s
2
0 + s 5 = 2 ⇒  5s  = 2 ⇒ No solutions!
0
2
2
2s
2
Answer 4: If identical, L and M must have infinitely many points of intersection. By equating the two equations for the two lines,
  

 
   
 
 
   
s − 2t
1
2
1
2
1
1
2
1
0 + s 5 = 2 + t 10 ⇒ s 5 − t 10 = 2 ⇒ 5s − 10t = 2
2
2s − 4t
2
4
2
4
0
2
2




1 −2 | 1
1 −2 | 1
(2)=(2)−5(1)
⇒ 5 −10 | 2 −−−−−−−−→ 0 0 | −3 ⇒ No solution = no point of intersection!
2 −4 | 2 (3)=(3)−2(1) 0 0 | 0
~ the two
Answer 5: If the projection of d~ = p~M − p~L = (1, 2, 2) onto ~lL is identical to d,
lines are identical. If not, they are different.
 
1
~lL
15
1
~
~
d~|| = Proj~lL d~ = (~lL · d)
=
lL = 5 6= d!
30
2
||~lL ||2
2
(d) Answer 1: It is obvious the length of d~⊥ is the distance between the two lines
1
 
 
 
1
1
1
1
1
d~⊥ = d~ − d~|| = 2 − 5 = −1
2
2
2
2
2
r
⇒
||d~⊥ || =
1 1
+ +1=
4 4
r
√
6
6
=
.
4
2
Alternatively, using Pythagorean theorem
r
r
√
q
6
6
1 25
2
2
~
~
+ 1) =
=
.
d = ||d|| − ||d|| || = 9 − ( +
4
4
4
2
Answer 2: Let ~n be a vector connecting p~M = (2, 2, 2) to any point on L:
 
    

1
1
2
s−1
~n = 0 + s 5 − 2 = 5s − 2 .
0
2
2
2s − 2
Now, we slide the point on L until ~n ⊥ ~lL . To find the value of s at this point, we require
~n · ~lL = 0
⇒
s − 1 + 5(5s − 2) + 2(2s − 2) = 0
⇒
30s − 15 = 0
⇒
s=
1
.
2
Plug the value into the expression for ~n
1
~n =
2
5
2
2
2
  1
−1
−2
− 2 =  21 
−1
−2
s
⇒
d = ||~n|| =
1
−
2
2
r
r
√
2
1
6
6
1
1
+
+ 12 =
+ +1=
=
.
2
4 4
4
2
(2,2,2)
(2,2,2)
l
dl
d
M
M
dll
(1,0,0)
(1,0,0)
L
n (s=1/2)
L
n (s) = s(1,5,2)−(1,2,2)