Equation of a straight line in vector form

yonski · 2006-10-26 03:27:10

I've read through the section of my text book on this but i don't seem to understand it.

I'm looking for the equation of the line which goes through point A, with position vector a, and is parallel to the given vector b, right?

Since AP is parallel to b, AP = tb, where t is a scalar. Okay, i get that.

The vector b is the direction vector of the line. Right, okay.

Therefore the position vector r can be written as a+tb. Yep i see that.

Then it makes out as if r is the answer. But how is that the equation of the line which goes through point A and is parallel to b? It does neither!

Can anyone help explain it to me? Thanks!

Last edited by yonski (2006-10-26 03:30:24)

gnitsuk · 2006-10-26 22:07:26

Hi,

The vector equation of the line you seek is:

Where conventionally

is the unit vector in the direction of AP.

So, maybe the book is unclear here but the answer as to what is the equation of the line is not:

but is:

yonski · 2006-10-27 08:59:41

The book says r = a + tb.

But i thought i was looking for the equation of the line which goes through point A and is parallel to b? But r does neither of these? That's what i'm confused about.

gnitsuk · 2006-10-29 22:19:16

In the Cartesian system, a straight line has an equation of the form y = mx + c where m and c are contants.

So in other words, if you are given a point (x1,y1) and you can show that
y1 = m * x1 + c then the point (x1,y1) must be on the line. Further, if x1 and y1 cannot be related by the expression y1 = m * x1 + c then the point (x1,y1) is not on the line. So the equation y1 = m * x1 + c defines a propertiy which all points on the infinite line share and all points off of the infinite line do not share. This is why the expression is a useful one for defining a straight line in Cartesian space.

So , say we were asked to find the Cartesian equation of the line which passes through the point A(2,3) in the direction of the vector i+j.

This is pretty staight forward:

(2,3) is on the line as is (3,4). We know (3,4) is on the line as we can move to point (2,3) and travel in the direction i+j (one unit in the x direction and one unit in the y direction) to arrive at point (3,4).

Now the m in our Cartesian equation is the gradient and this will be (4 - 3) / (3 - 2) = 1 so the equation we seek is y = x + c. To find c just substitute in a known point on the line (say (2,3)) to give 3 = 2 + c therefore c = 1 so the equation we seek is y = x + 1. In other words, y is one more than x. So, (10,14) is not on this line as 14 is not one more than 10. But (-4,-3) is on the line as -3 is one more than -4.

Now, what if we were asked to find the vector equation of the line which passes through the point A(2,3) in the direction of the vector i+j.

Well, the direction of the line is i+j, let's call this vector b. So b = i + j. Now, just to be conventional let's find the unit vector in this direction. Well, that's easy enough. It's:

Now, the position vector of A is a = 2i + 3j. So we need to define a property which is true for all points on the desired line and not for all points off of the desired line.

So we could write r = a + tb' where t is any scalar number.

So we are saying that for all points on the line, r (remember, r is the position vector of any point on the line and is analogous to the (x,y) of any point on the line in Cartesian space) = a + tb'.

So, if a point is on the line, it's position can be written as a + tb' but if it is off of the line then it's position cannot. So, how about (-4,-3) is that on the line?

Well, its position vector is -4i-3j. and we can write that as:

So we have shown that the point (-4,-3) is on the line defined by the vector equation r = a + tb' as it's poition r can be expressed as the vector a plus minus-six-root-two lots of b'.

A point such as (10,14) with position vector 10i+14j cannot be so expressed and so is not on the line.

Incidentally, one need not use the unit vector (here b') in the direction of the line, but could use any vector in that direction. Obviously this would simply mean that each point on the line would have a different value for t. This is analogous to y = 2x + 5 and 3y = 6x + 15 being two expressions for that same line. they are the same equation.

Last edited by gnitsuk (2006-10-29 22:43:30)

Math Is Fun Forum

#1 2006-10-26 03:27:10

Equation of a straight line in vector form

#2 2006-10-26 22:07:26

Re: Equation of a straight line in vector form

#3 2006-10-27 08:59:41

Re: Equation of a straight line in vector form

#4 2006-10-29 22:19:16

Re: Equation of a straight line in vector form

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