Question

Find the area of the parallelogram with vertices k(1, 2, 1), l(1, 3, 4), m(5, 8, 4), and n(5, 7, 1).

Answer 1

The answer is:
A
=
√
265
.

There are two ways, the first one ie VERY LONG and complicate, the second one VERY SHORT and easy, but we have to use the vectorial product.

The first one:

First of all, let's check if the shape is really a parallelogram:

KL=√(xK−xL)2+(yK−yL)2+%(xK
−
z
L
)
2
=

=
√
(
1
−
1
)
2
+
(
2
−
3
)
2
+
(
3
−
6
)
2
=
√
0
+
1
+
9
=
√
10
.

M
N
=
√
(
3
−
3
)
2
+
(
8
−
7
)
2
+
(
6
−
3
)
2
=
√
0
+
1
+
9
=
√
10
.

So
K
L
=
M
N


The direction of
K
L
is the vector
→
v
such as:

→
v
=
(
x
K
−
x
L
,
y
K
−
y
L
,
z
K
−
z
L
)
=
(
0
,
1
,
3
)
.

The direction of
M
N
is the vector
→
w
such as:

→
w
=
(
x
M
−
x
N
,
y
M
−
y
N
,
z
M
−
z
N
)
=
(
0
,
1
,
3
)
.

So
→
v
is parallel to
→
w
.

So, since
K
L
=
M
N
and
K
L
is parallel to
M
N
, the shape is a parallelogram.

The area of a parallelogram is:
A
=
b
⋅
h
.

We can assume that the base
b
is
K
L
=
√
10
, but finding the height is more complicated, because it is the distance of the two line
r
, that contains
K
and
L
, and
s
, that contains
M
and
N
.

A plane, perpendicular to a line, can be written:

a
(
x
−
x
P
)
+
b
(
y
−
y
P
)
+
c
(
z
−
z
P
)
=
0
,

where
→
d
(
a
,
b
,
c
)
is a whatever vector perpendicular to the plan, and
P
is a whaterver point that lies on the plan.

To find
π
, that is a plan perpendicular to
r
, we can assume that
→
d
=
→
v
and
P
=
K
.

So:

π
:
0
(
x
−
1
)
+
1
(
y
−
2
)
+
3
(
z
−
3
)
=
0
⇒
y
+
3
z
−
11
=
0
.

A line can be written as the system of three equation in parametric form:

x
=
x
P
+
a
t

y
=
y
P
+
b
t

z
=
z
P
+
c
t


Where
P
is a whatever point of the line and
→
d
(
a
,
b
,
c
)
is a whatever vector, direction of the line.

To find
s
, we can assume that
P
=
M
, and
→
d
=
→
w
.

So
s
:

x
=
3
+
0
t

y
=
8
+
1
t

z
=
6
+
3
t


or:

x
=
3

y
=
8
+
t

z
=
6
+
3
t
.

Now, solving the system between
π
and
s
we can find
Q
, foot of the height conducted from
K
to
s
.

y
+
3
z
−
11
=
0

x
=
3

y
=
8
+
t

z
=
6
+
3
t


8
+
t
+
3
(
6
+
3
t
)
−
11
=
0
⇒
10
t
=
−
15
⇒
t
=
−
3
2
.

So, to find the point
Q
, it is necessary to put
t
=
−
3
2
in the equation of
s
.

x
=
3

y
=
8
−
3
2

z
=
6
+
3
(
−
3
2
)


So:

x
=
3


y
=
13
2


z
=
3
2


Now, to find
h
, we can use the formula of the distance of two points,
K
and
Q
, just seen before:

h
=
√
(
1
−
3
)
2
+
(
2
−
13
2
)
2
+
(
3
−
3
2
)
2
=
√
2
2
+
(
9
2
)
2
+
(
3
2
)
2
=
√
4
+
81
4
+
9
4
=
√
16
+
81
+
9
4
=
√
106
2
.

Finally the area is:

A
=
√
10
√
106
2
=
√
1060
2
=
√
4
⋅
265
2
=
√
265
.