Homework due 26th January
Solutions to compulsory questions
Problem (
§
1.3, Exercise 24).
Find an equation for the plane that passes through the point
(2
,

1
,
3) and is perpendicular to the line
v
= (1
,

2
,
2) +
t
(3
,

2
,
4).
Solution.
The plane has a normal vector 3
i

2
j
+ 4
k
, and so has equation given by
(
(
x
i
+
y
j
+
z
k
)

(2
i

j
+ 3
k
)
)
·
(3
i

2
j
+ 4
k
) = 0
.
Equivalently,
3
x

2
y
+ 4
z
= 20
.
Problem (
§
1.3 Exercise 32).
Given vectors
a
and
b
, do the equations
x
×
a
=
b
and
x
·
a
=
k
a
k
determine a unique vector
x
? Argue geometrically and analytically.
There are quite a few ways to approach this problem. Here are a few.
Solution 1.
Firstly, if
a
=
0
then any vector
x
satisﬁes the equations, and so the solution is not
unique in this case. Suppose now that
a
6
=
0
.
If there are two solutions,
x
and
y
say, then
x
·
a
=
k
a
k
y
·
a
=
k
a
k
and
x
×
a
=
b
y
×
a
=
b
.
Subtracting gives
(
x

y
)
·
a
= 0
and
(
x

y
)
×
a
=
0
,
and so
x

y
is both perpendicular and parallel to
a
. Therefore
x

y
=
0
, and so
x
=
y
.
Therefore the solution, if it exists, is unique provided
a
6
=
0
. (The question did not actually ask
you to prove the existence of a solution).
Solution 2.
A more geometric argument proceeds by determining the length and direction of
x
.
If
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 Spring '08
 PARKINSON
 Math, Multivariable Calculus, 2k, 0k, 0j, 3j, 1 3 j

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