2
S.
18.02 SOLUTIONS TO EXERCISES
1B.
Dot Product
1B2
A
.
B
=
c

4; therefore (a) orthogonal if c
=
4,
c4
b) cos 0
=
dm&'
the angle 0 is acute if cose
>
0, i.e., if c
>
4.
1B3
Place the cube in the first octant so the origin is at one corner
P, and i, j
,
k are
three edges. The longest diagonal PQ
=
i
+
j
+
k;
a face diagonal PR
=
i
+
j.
1B4
QP
=
(a, 0, 2),
QR
=
(a, 2,2), therefore
a) QP
.
=
a2

4;
therefore PQR is a right angle if a2

4
=
0, i.e., if a
=
f
2.
4
b) cose
=
drndrn'
the angle is acute if cos0
>
0, i.e., if a2

4
>
0, or
la1
>
2, i.e., a
>
2 or a
<
2.
1B5
a)
F
.
u
=
I/& b) u
=
dir A
=
A/7, so
F

u
=
417
1B6
After dividing by IOPI, the equation says cos0
=
c, where 6 is the angle between OP
and u; call its solution Oo
=
cos' c. Then the locus is the nappe of a right circular cone
with axis in the direction u and vertex angle 200.
In particular this cone is
a) a plane if Oo
=
n/2, i.e., if c
=
0
b) a ray if Bo
=
0, n, i.e., if c
=
f
1
c) nonexistent if Bo is nonexistent, i.e.,
c
>
1or c
<
1.
1B7
I
i'l
=
I
j'l
=

Jz
=
1;a picture shows the system is righthanded.
b) A.
il=
1/a; A.
jl=
5/a;

if

5j'
since they are perpendicular unit vectors, A
=
i'

j'
.
it+
j'.
C) Solving, i
=

J=
Jz
'
'
2(i1j')

3(i1+j')il
thus A=2i 3j
=
Jz
Jz

,
as before.
1B8
a) Check that each has length
1,and the three dot products
.
j', i'
.
k',
.
k'
are 0; make a sketch to check righthandedness.
b) A.il=&,
A.jl=O, A.kl=&,
therefore, A=fii1+&k'.
1B9
Let u
=
dir A, then the vector ucomponent of
B
is
(B
.u)u. Subtracting it off gives
a vector perpendicular to
u (and therefore also to A); thus