3
Solutions
Solution of problem 1.1:
The point
P
on the curve corresponding to the
value of the parameter
t
= 0 has position vector
−→
r
(0) =
(
2
,
1
,
0
)
. In
other words
P
has coordinates
P
(2
,
1
,
0).
The tangent vector to the
curve at
P
is the vector
d
−→
r
dt
(0) =
parenleftbigg
d
dt
(big
t
2
+ 3
t
+ 2
, e
t
cos
t,
ln(
t
+ 1)
)big
parenrightbigg

t
=0
=
parenleftbigg(bigg
2
t
+ 3
, e
t
cos
t
−
e
t
sin
t,
1
t
+ 1
)biggparenrightbigg

t
=0
=
(
3
,
1
,
1
)
.
The tangent line
L
at
t
= 0 is the line passing through
P
and having
the tangent vector
d
−→
r /dt
(0) as a direction vector. Thus
L
is given by
the parametric equations
x
= 2 + 3
s,
y
= 1 +
s,
z
=
s.
To intersect the line
L
with the plane
x
+
y
+
z
= 8 we substitute the
parametric expressions for
x
,
y
, and
z
in the equation of the plane and
solve for
s
. We get (2+3
s
)+(1+
s
)+
s
= 8, i.e.
s
= 1. Thus the point
of intersection is the point on the line which corresponds to the value
of the parameter
s
= 1, i.e. the point
(5
,
2
,
1)
. The correct answer is
(D)
.
Solution of problem 1.2:
A vector is perpendicular to a plane if and only
if it is parallel to a normal vector for the plane.
To find a vector
−→
n
which is normal to the plane containing
P
(2
,
1
,
5),
Q
(
−
1
,
3
,
4), and
R
(3
,
0
,
6) we need to find two nonparallel vectors in
the plane and compute their cross products. The vectors
−→
PQ
=
(−
1
−
2
,
3
−
1
,
4
−
5
)
=
(−
3
,
2
,
−
1
)
−→
PR
=
(
3
−
2
,
0
−
1
,
6
−
5
)
=
(
1
,
−
1
,
1
)
9