21-256 Homework 2 (solutions)
Due Friday 23rd May 2014
1. Find the equation of a sphere if one of its diameters has end-points (2, 1, 4) and (4, 3, 10).
The center of the sphere is the midpoint of the given points, i.e. (3, 2, 7). The radius is the
distance between the center and one of the end-points, thus
p
√
√
r = (4 − 3)2 + (3 − 2)2 + (10 − 7)2 = 1 + 1 + 9 = 11
Hence the equation of the sphere is
(x − 3)2 + (y − 2)2 + (z − 7)2 = 11
2. Describe in words the region of R3 represented by the equation x = z.
It is a plane formed by taking the line x = z in the xz-plane and extending it infinitely
along the y-axis.
−3
5
.
and w =
3. Find v + w, 2v + 3w, kvk and kv − wk when v =
−6
−12
2
5−3
=
v+w =
−18
−12 − 6
1
10 − 9
2·5
+ 3 · (−3)
=
=
2v + 3w =
−42
−24 − 18
2 · (−12) + 3 · (−6)
p
√
√
kvk = 52 + (−12)2 = 25 + 144 = 169 = 13
5 − (−3) 8 p
√
√
=
= 82 + (−6)2 = 64 + 36 = 100 = 10
kv − wk = −12 − (−6) −6 –
–
–
–
4. Find a + b, 2a + 3b, kak and ka − bk when a = 2i − 4j + 4k and b = 2j − k.
– a + b = 2i + (−4 + 2)j + (4 − 1)k = 2i − 2j + 3k
– 2a+3b = (2·2)i+(2·(−4)+3·2)j+(2·4+3·(−1))k = 4i+(−8+6)j+(8−3)k = 4i−2j+5k
p
√
√
– kak = 22 + (−4)2 + 42 = 4 + 16 + 16 = 36 = 6
p
– √
ka − bk = k2i +√((−4) − 2)j + (4 − (−1))kk = k2i − 6j + 5kk = 22 + (−6)2 + 52 =
4 + 36 + 25 = 65
5. Find (2i + j) · (i − j + k).
(2i + j) · (i − j + k) = 2 · 1 + 1 · (−1) + 0 · 1 = 2 − 1 + 0 = 1
 
 
4
2
6. Find the acute angle between the vectors 0 and −1.
2
0
Denote the angle by θ. Then
     
4
2
 0 · −1 




2
0

−1 
θ = cos      
 4 2 


0 −1 

2 0 4 · 2 + 0 · (−1) + 2 · 0
p
= cos−1 √
42 + 02 + 22 · 22 + (−1)2 + 02
8+0+0
√
= cos−1 √
16 + 0 + 4 · 4 + 1 + 0
8
−1
√
√
= cos
20 · 5
8
−1
√
= cos
100
4
= cos−1
5
!


 
3
1
7. Find the scalar and vector projections of  6  onto 2.
−2
3
 
 
1
3



6
and w = 2. Then
Write v =
3
−2
v·w
3 · 1 + 6 · 2 + (−2) · 3
projw (v) =
w=
w·w
12 + 2 2 + 3 2
 
1
3 + 12 − 6
9  
2
=
w=
1+4+9
14
3
v·w
9
compw (v) =
=√
kwk
14
w
8. Find the scalar and vector projections of i + j + k onto i − j + k.
Write v = i + j + k and w = i − j + k. Then
1 · 1 + 1 · (−1) + 1 · 1
w
w=
w·w
12 + 1 2 + 1 2
1
1−1+1
w = (i − j + k)
=
1+1+1
3
v·w
1
compw (v) =
=√
kwk
3
projw (v) =
v·w
9. Compute (j+7k)×(2i−j+4k) and verify that it is orthogonal to both j+7k and 2i−j+4k.
Write v = j + 7k and w = 2i − j + 4k. Then

 

1·4
− (−1) · 7
4+7
−
4 · 0  = 14 − 0 = 11i + 14j − 2k
v×w = 7·2
0 · (−1) −
2·1
0−2
So
v · (v × w) = 0 · 11 + 1 · 14 + 7 · (−2) = 0 + 14 − 14 = 0
w · (v × w) = 2 · 11 + (−1) · 14 + 4 · (−2) = 22 − 14 − 8 = 0
and hence v and w are both orthogonal to v × w.
10. Find the volume of the parallelepiped with adjacent edges P Q, P R and P S, where the
points P, Q, R, S are given by
P = (−2, 1, 0),
Q = (2, 3, 2),
R = (1, 4, −1),
S = (3, 6, 1)
The adjacent edges are given by

  

  

  
2 − (−2)
4
1 − (−2)
3
3 − (−2)
5
−−→ 
−→
−→
3 − 1  = 2 , P R =  4 − 1  =  3  , P S =  6 − 1  = 5
PQ =
2−0
2
(−1) − 0
−1
1−0
1
So the volume V is given by
Now
−−→ −→ −→
V = |P Q · (P R × P S)|

 
  
3·1
− 5 · (−1)
3+5
8
−→ −→ 




1·3
P R × P S = (−1) · 5 −
= −5 − 3 = −8
3·5
−
3·5
15 − 15
0
So it follows that
V = |4 · 8 + 2 · (−8) + 2 · 0| = |32 − 16| = 16
Extra credit problems
E1. Prove that ka + bk 6 kak + kbk for all n-dimensional vectors a, b, and give a geometric
interpretation. Under what conditions is it true that ka + bk = kak + kbk?
Proof. First note that a · b 6 |a · b| 6 kakkbk. Indeed, let
f (t) = ka − tbk2 = (a − tb) · (a − tb) = kak2 − 2t(a · b) + t2 kbk2
Note that f (t) > 0 for all t. Putting t =
kak2 − 2
a·b
therefore gives
kbk2
(a · b)2 (a · b)2
+
>0
kbk2
kbk2
Rearranging gives (a · b)2 6 kak2 kbk2 . Taking square roots gives |a · b| 6 kakkbk, as
desired.
It follows that
ka + bk2 = (a + b) · (a + b)
=a·a+a·b+b·a+b·b
= kak2 + 2a · b + kbk2
6 kak2 + 2kakkbk + kbk2
= (kak+kbk)2
Since both quantities are positive, we can take their square root and deduce that
ka + bk 6 kak+kbk
Equality holds if and only if a and b are parallel.
E2. Prove that a × (b × c) = (a · c)b − (a · b)c for all 3-dimensional vectors a, b, c, and use this
result to prove that
a × (b × c) + b × (c × a) + c × (a × b) = 0
Proof. This amounts to writing out the formulae to both sides and is a very tedious exercise in
algebraic manipulation. Well done if you did it!