Math 23
B. Dodson
Week 3 Homework:
12.5 Lines, Planes
12.6 Quadratic Surfaces
13.1, 13.2 Vector functions, derivatives
Problem 12.5.3: Give vector and (scalar) parametric
equations for the line through the point
(2,4,10) parallel to the vector
<
3
,
1
,
8
> .
Solution: The vector equation
±
OP
=
±
OP
0
+
t
±
d,
for when the position vector of the point
P
(
x, y, z
)
puts
P
on the line through
P
0
(
x
0
, y
0
, z
0
) with
direction vector
±
d
=
< a, b, c >
gives
< x, y, z >
=
< x
0
, y
0
, z
0
>
+
t < a, b, c >
=
<

2
,
4
,
10
>
+
t <
3
,
1
,
8
>,
which we can view as a “pointslope” equation,
where
P
0
is the point, and
±
d
gives
the direction of the line. (Here in the
position vector
±
OP, O
=
O
(0
,
0
,
0)
is the Origin.)
To get the scalar equations, we use scalar mult.
and vector add to write the vector equation
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.
as
< x, y, z >
=
<

2 + 3
t,
4 +
t,
10 + 8
t >,
and simply readoﬀ
x
=

2 + 3
t, y
= 4 +
t, z
= 10 + 8
t.
Here, each value of the parameter
t
gives a
point on the line.
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 Spring '06
 YUKICH
 Calculus, Equations, Derivative, Scalar, Vector Space, Parametric Equations, Parametric equation

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