1a: Take cross products of the normal to get the direction of the line:
Cross
@8
2,

1, 1
<
,
8
1, 0,

3
<D
8
3, 7, 1
<
Then parameterize the line using the point and the direction to get:
x
=
4
+
3
t
,
y
= 
7
t
,
z
= 
3
+
t
1b: The desired plane contains the normals of the two specifed plane, so its normal is the same cross product as in 1a. That gives
an equation of the plane as 3
H
x

4
L
+
7
H
y
+
2
L
+
H
z

3
L
=
0
2a: the gradient is:
f
=
x
2
y
+
y
2
z
x
2
y
+
y
2
z
8
D
@
f, x
D
, D
@
f, y
D
, D
@
f, z
D<
9
2 x y, x
2
+
2 y z, y
2
=
2b: the gradient will give a normal vector, so evaluating at the point gives:
9
2 x y, x
2
+
2 y z, y
2
= ê
.
8
x
Ø
2, y
Ø 
1, z
Ø
3
<
8

4,

2, 1
<
So an equation of the tangent plane is

4
H
x

2
L
+ 
2
H
y
+
1
L
+
H
z

3
L
=
0
2c: we just want
D
v
f
= “
f
ÿ
v
to be negative, so there are many correct answers. For example, moving in the direction given by
i gives:
8

4,

2, 1
<
.
8
1, 0, 0
<

4
3: First, sketch the triangular region to reverse the order, giving an integral which goes easily by substitution:
‡
0
1
‡
0
x
1
+
x
2
„
y
„
x
1
3
K

1
+
2
2
O
4a: Absolutely convergent by the ratio test
4b: Convergent by the alternating series test, and then absolutely convergent by the integral test: