21-256 Homework 3 (solutions)
Due Wednesday 28th May 2014
1. Use the scalar triple product to verify that following three vectors are coplanar:
u = i + 5j − 2k,
v = 3i − j,
w = 5i + 9j − 4k

       

1
3
5
1
(−1) · (−4) − 9 · 0
 5  · −1 ×  9  =  5  ·  0 · 5 − (−4) · 3 
−2
0
−4
−2
3 · 9 − 5 · (−1)
   
1
4
=  5  · 12
−2
32
= 1 · 4 + 5 · 12 − 2 · 32
= 4 + 60 − 64
=0
So the vectors are coplanar.
2. Find the acute angle between the lines 2x − y = 3 and 3x + y = 7 in R2 .
– Two points lying in the first line are A = (0, −3) and B = ( 23 , 0), so its direction is
3
−−→
AB = 2 . We can multiply by a scalar without changing the direction of the line;
3
1
2
.
multiplying by 3 gives
2
– Two points lying in the second line are C = (0, 7) and D = ( 37 , 0), so its direction is
7 −−→
1
3
3
CD =
.
, or equivalently by multiplying by 7 we get
−3
−7
So the angle is
 
1
1
·


2
−3
−1

cos−1 
 1 1  = cos
2 −3 −1
= cos
1 · 1 + 2 · (−3)
p
√
2
1 + 22 12 + (−3)2
−5
√ √
5 10
Depending on how you did this, you may have got
at an acute and an obtuse angle simultaneously.
π
4
−1
= cos
!
−5
√
5 2
−1
= cos
−1
√
2
=
3π
4
instead. This is because two lines meet
3. Find the two unit vectors (in R2 ) that are parallel to the tangent line to the curve y = x2
at the point (2, 4).
dy
dx
= 2x, so the tangent line when x = 2 has a gradient 2 · 2 = 4. Thus the
√
√
1
direction of the tangent line is
. This has magnitude 12 + 42 = 17, so the two unit
4
vectors are
−1 1
1
1
√
√
and
17 4
17 4
First note that
4. Find the vector equation of the line through the point (0, 14, −10) which is parallel to the
line r = (−1 + 2λ)i + (6 − 3λ)j + (3 + 9λ)k.
The direction vector of the lines are the same, namely 2i − 3j + 9k, so the line we seek has
equation


 
0
2



14 + λ −3
r=
−10
9
or equivalently r = 2λi + (14 − 3λ)j + (−10 + 9λ)k.
5. Find the vector equation of the line of intersection of the planes x + 2y + 3z = 1 and
x − y + z = 1.
– Point on line. By inspection we can see that (1, 0, 0) lies on both lines. (We could
have solved simultaneously, but that’s more work!)
– Direction of line. The direction of the line is the cross product of the normal vectors
of the planes; thus
  
    
5
2 · 1 − (−1) · 3
1
1
2 × −1 =  3 · 1 − 1 · 1  =  2 
−3
1 · (−1) − 1 · 2
1
3
So the line has equation
 
 
1
5
r = 0 + λ  2 
0
−3
6. Find the vector equation and the linear equation of the plane through the origin and the
points (2, −4, 6) and (5, 1, 3).
Call the points P and Q respectively. The normal vector to the plane is therefore
    
 

2
5
(−4) · 3 − 1 · 6
−18
−−→ −−→     
6 · 5 − 3 · 2  =  24 
OP × OQ = −4 × 1 =
6
3
2 · 1 − 5 · (−4)
22
The origin (0, 0, 0) lies on the plane, and therefore it has equation


−18
r ·  24  = 0
22
or equivalently
−18x + 24y + 22z = 0
7. Find the point at which the following lines intersect, and find an equation of the plane
containing both lines:
 
 
 
 
1
1
2
1
r = 1 + λ −1 , r = 0 + µ −1
0
2
2
0
Solving the corresponding system of equations gives


1 + λ = 2 + µ
1 − λ = −µ


2λ = 2
The third equation gives λ = 1. The second equation gives µ = 0. This is consistent with
the first equation, so the lines intersect. Their point of intersection is thus
(1 + 1 · 1, 1 + 1 · (−1), 0 + 1 · 2) = (2 + 0 · 1, 0 + 0 · (−1), 2 + 0 · 0) = (2, 0, 2)
1
3
3
.
and
as a linear combination of the two vectors
8. Express the vector
−3
5
−2
1
3
3
. That is, we need to solve the
+µ
We need to solve the equation
=λ
−3
5
−2
system
(
3λ + µ = 3
5λ − 3µ = −2
The first equation gives µ = 3 − 3λ. Substituting into the second gives
5λ − 9 + 9λ = −2
and therefore µ = 3 − 3 ·
1
2
⇒ 14λ = 7
⇒
= 32 . So
1 3
3 1
3
=
+
−2
2 5
2 −3
λ=
1
2
2
4
,
is linearly independent.
3
1
2
4
0
Suppose λ
+µ
=
. Then
3
1
0
(
2λ + 4µ = 0
3λ + µ = 0
9. Show that the set
The second equation gives µ = −3λ. The first equation thus gives 2λ−12λ = 0, so −10λ = 0
and hence λ = 0. So µ = −3 · 0 = 0. So the linear combination is trivial, and the vectors
are LI.
10. Show that the set {2i + 4j − 4k, −3i + 2j + k, −2i + 4j − k} is linearly dependent.
 
 
   
2
−3
−2
0
We need to solve λ1  4  + λ2  2  + λ3  4  = 0. That is
−4
1
−1
0


2λ1 − 3λ2 − 2λ3 = 0
4λ1 + 2λ2 + 4λ3 = 0


−4λ1 + λ2 − λ3 = 0
The third equation gives λ3 = −4λ1 + λ2 . Substituting into the other two equations gives
(
10λ1 − 5λ2 = 0
−12λ1 + 6λ2 = 0
and rearranging both of these quations yields 2λ1 = λ2 . Setting λ1 = 1 thus gives λ2 = 2
and λ3 = −4 · 1 + 2 = −2. So a nonzero solution exists; indeed
   
 
 
0
−2
−3
2
 4  + 2  2  − 2  4  = 0
1
−1
0
−4
so the zero vector is a nontrivial linear combination of the three vectors, so they’re LD.
Extra credit problems
E1. Two parallel planes have equations
ax + by + cz + d1 = 0
and ax + by + cz + d2 = 0
Prove that the distance between the two planes is √
|d1 − d2 |
.
a2 + b2 + c2
The distance from one plane to the other is precisely the distance from any one point on
the first plane to the second plane.
We may assume a 6= 0; otherwise relabel stuff (a, b, c can’t all be zero). Then the point
1
2
( −d
the first plane and ( −d
a , 0, 0) lies on
a , 0, 0) lies on the second plane. The vector between
 d1 −d2 
a
these points is  0 . Using the formula for the distance, we thus have that the distance
0
is
   d −d  1
2
a
a
b ·  0  d1 −d2
c
+
b
·
0
+
c
·
0
a
·
0
a
 
√
=
2 + b2 + c2
a a
 b 
c d1 − d2
= √
a2 + b2 + c2 |d1 − d2 |
=√
a2 + b2 + c2