Lecture 1b

Vector Equation of a Line in R2

(pages 5-6)

The equation x2= (0.5)x1+ 1 deﬁnes the x2component of a point on the line

in terms the x1component. But when we graph the line, we think less of this

equation, and more of the following two facts: the x2-intercept is 1 and the

slope is 0.5. And so, when graphing, we may start at the point (0,1), and then

move “up one, right two” to ﬁnd a second point on the line (the point (2,2)),

and then connecting those points gives us our line. In fact, though, we could

have used any multiple of “up one, right two” to ﬁnd a second point on the

line. So, we can think of the line x2= (0.5)x1+ 1 as starting at the point (0,1),

and adding all scalar multiples of the directions “up one, right two”. Well, if

we decide to think of vectors as directions instead of points, then the direction

“up one, right two” corresponds to the vector 2

1. Then we can turn the

equation x2= (0.5)x1+ 1 into the equation x1

x2=0

1+t2

1for all

t∈R. Notice that picking t= 0 gives us our starting point of 0

1.

Deﬁnition: A line through ~p with direction vector ~

dis the set

{~p +t~

d|t∈R}

which has the vector equation

~x =~p +t~

d , t ∈R

Note that in the case when ~p =~

0, we end up with the equation ~x =t~

d, which

is simply all scalar multiples of ~

d. This ends up being the equation of the line

through the origin that goes through ~

d. Another thing to note is that since the

direction vector ~

dcorresponds to the slope of a line, we see that two lines are

parallel if and only if their direction vectors are NON-ZERO MULTIPLES of

each other.

Sometimes it is useful to write the components of a vector equation separately.

For example, the equation x1

x2=0

1+t2

1is the same as saying

x1= 2tand x2= 1 + t(for the same t). In this case, tis called a ”parameter”,

and the equations are known as parametric equations.

1

Deﬁnition The parametric equation of the line ~x =~p +t~

dis the collection of

equations

x1=p1+td1

x2=p2+td2

t∈R

Now, if we solve both of these equations for t, we get

t=x1−p1

d, t =x2−p2

d2

But since t=t, we get

x1−p1

d=x2−p2

d2

and then solving for x2in terms of x1, we get equation x2=p2+d2

d1(x1−p1).

This is the type of equation we started with!

Deﬁnition: The scalar form of the equation of a line is x2=p2+d2

d1(x1−p1),

where p1

p2is a point on the line, and d1

d2is a direction vector for the line.

EXAMPLE The vector equation of a line passing through the point P(−2,2)

with direction vector ~

d=2

3has vector equation

~x =−2

2+t2

3t∈R

It has parametric equations

x1=−2+2t

x2= 2 + 3tt∈R

Solving these parametric equations for tgives us

t=x1+ 2

2=x2−2

3

Solving for x2gives us the scalar form of the line: x2=3

2(x1+ 2) + 2, or

x2=3

2x1+ 5.

2

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###### Document Summary

Vector equation of a line in r2 (pages 5-6) The equation x2 = (0. 5)x1 + 1 de nes the x2 component of a point on the line in terms the x1 component. But when we graph the line, we think less of this equation, and more of the following two facts: the x2-intercept is 1 and the slope is 0. 5. In fact, though, we could have used any multiple of up one, right two to nd a second point on the line. So, we can think of the line x2 = (0. 5)x1 + 1 as starting at the point (0, 1), and adding all scalar multiples of the directions up one, right two . Well, if we decide to think of vectors as directions instead of points, then the direction. Up one, right two corresponds to the vector. Then we can turn the (cid:21) (cid:20) 2 (cid:20) x1 (cid:21) 1 (cid:20) 0 (cid:21) (cid:21) (cid:20) 2 (cid:21)

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