C4 vectors question

What does it mean to " write the equation of a straight line in vector form " ?

Does it have anything to do with y=mx+c ?

I don't really get the explanation in my C4 Edexcel book. They talk about finding the equation of a line that passes through the point A ( which has a position vector a ) and is parallel to a given vector b.
But they then end up showing how to get some other vector r... :confused:

Reply 1

mintmocha

Oh which exercise and question is it?
I have the book.

Reply 2

TimmonaPortella

Original post by confuzzled92

I don't really get the explanation in my C4 Edexcel book. They talk about finding the equation of a line that passes through the point A ( which has a position vector a ) and is parallel to a given vector b.
But they then end up showing how to get some other vector r... :confused:

Do you understand the bold?
You have a position vector and a direction vector with a scalar. Since you start from the point given by the position vector, and can go any distance along the direction vector due to the scalar, any point on the line can be given through the equation.

What vector r? You'll need to give the textbook's example I think.

edit: I think your lambda/mu/whatever is referred to as a scalar. Anyway, you have some amount of the direction vector, so can reach any point on the line through the equation depending on the value of your lambda/mu/whatever.

(edited 13 years ago)

Reply 3

soutioirsim

The way my teacher explained to me was that if you have a vector equation for a line in the form:

y = a + tb (Where a and b are vectors)

Then the vector a is to "get to" the line.

The vector b is the actual line.

"t" is a scalar quantity which represents how far along the lin you want to go.

I don't know if that will help you but I find it useful to think of it this way.

Reply 4

dknt

Original post by confuzzled92

The general equation for a vector line is given by:

r=a +sm

Now comare this to y=mx + c

Notice that a corresponds to the c i.e. just a random point on the line.
m represents the direction vector, or the gradient of the line. If two vectors are parallel, their direction vectors are the same (i.e. same gradient)

As soutioirsim said, s or t or

\mu

\lambda

represents a scalar quantity, to give you the position vector of a vector on that line. Note that if this =0, then you are left with r=a, which is why a is just any point on the line.

(edited 13 years ago)

Reply 5

Farhan.Hanif93

Original post by soutioirsim

Two little niggles with that definition, it's not entirely correct.

Firstly, it's a bit strange to define a vector line as

y=...

as you haven't indicated that y is a vector when it actually is. It also allows for confusion with cartesian coordinates to occur. It's conventional to write a vector line as

\mathbf{r}=...

.

Secondly, saying that "

\mathbf{b}

is actually the line" is a little misleading.

\mathbf{b}

is just a vector parallel to the line and it is able to move freely in vector space i.e. a vector in the same direction as the line but not necessarily bounded by having to go through any particular point. Whereas

\mathbf{r}

is a vector in the direction of

\mathbf{b}

that passes through the point with position vector

\mathbf{a}

, which means that it is not free to move about freely in vector space and it must pass through a specific set of points.

Reply 6

confuzzled92

Original post by mintmocha

Oh which exercise and question is it?
I have the book.

Pg 75, the bottom of the page

Reply 7

mintmocha

Original post by confuzzled92

Pg 75, the bottom of the page

Oh, it's a general statement.
Okay so say you have the vector AO=a.
Now, the vector of AR is not given to you but you know that the vector b is parallel to it. The condition is that if vectors are parallel then one vector is the multiple of another. In this case you are given the vector b so b has to be multiplied by something to give AR. The scalar here to assumed to be 't' so AR=tb.
Hence, the position vector of r=AO+AR
which is r=a+tb
This 'r' is thus the vector equation of a straight line passing through a point A and parallel to vector b.

ETA: you're basically using the triangle law

(edited 13 years ago)

Reply 8

confuzzled92

Original post by mintmocha

I understand why vector r is defined as it is. But I don't really get what vector r has to do with the equation in vector form of AR (Btw, is the book talking about finding the equation in vector form of AR? Or of the entire line passing through AR? :confused:

)

Reply 9

soutioirsim

Original post by Farhan.Hanif93

Two little niggles with that definition, it's not entirely correct.

Firstly, it's a bit strange to define a vector line as

y=...

as you haven't indicated that y is a vector when it actually is. It also allows for confusion with cartesian coordinates to occur. It's conventional to write a vector line as