MATH1920 HOMEWORK 5, SECTION 14.6
6
(a) The normal vector of the tangent plane is given by the gradient vector
∇
f

(1
,
1
,

1)
= (2
x

y,

x

2
y,

1)

(1
,
1
,

1)
= (1
,

3
,

1)
Equation of tangent plane: (
x

1)

3(
y

1)

(
z
+ 1) = 0, i.e.
x

3
y

z
+ 1 = 0
(b) The normal line is in the direction of the gradient vector (1
,

3
,

1). So the equation
of the normal line is
x
= 1 +
t
,
y
= 1

3
t
,
z
=

1

t
.
14
Let
f
(
x,y,z
) =
xyz
,
g
(
x,y,z
) =
x
2
+ 2
y
2
+ 3
z
2
. Note that
∇
f

(1
,
1
,
1)
= (
yz,xz,xy
)

(1
,
1
,
1)
= (1
,
1
,
1)
∇
g

(1
,
1
,
1)
= (2
x,
4
y,
6
z
)

(1
,
1
,
1)
= (2
,
4
,
6)
The line tangent to the curve of intersection of the level surfaces is perpendicular to both
normal vectors of the surfaces, which are the gradient vectors as above. So the tangent line
is in the direction of the vector
(
i
+
j
+
k
)
×
(2
i
+ 4
j
+ 6
k
) =
±
±
±
±
±
±
±
i
j k
1 1
1
2 4
6
±
±
±
±
±
±
±
= 2
i

4
j
+ 2
k
Factoring out 2, we can use
i

2
j
+
k
instead. The equation of the tangent line is