Thread: Equation of a line

I don't get the explanation.

Given that line k passes through point A(4,-5,3) and has direction vector (-2,3,0), the equation of line k can be expressed as:

(1)

If only the denominator is 0, then equation 1 is meaningless. (What does that mean)

If when the denominator is 0, we assume that the numerator is also 0 (why?) and the fraction can be any value (why?), then equation 1 can be expressed as:

(2)

z=3 (how do you get that?)

Equations 1 and 2 mean the same thing. Equation 2 is usually used.

The usual way to write this is .
The z-value is constant. There is no change in the z-direction.

You can also write that as "parametric equations":
x= 4- 2t, y= -5+ 3t, z= 3+ 0t= 3

Solving the first two equations for t, t= (4- x)/2 and t= (5+y)/3. Set those equal: (4-x)/2= (5+y)/2. Since there is no t in the z equation, you cannot solve for t- but you still have z= 3.

Do you mind answering the questions I wrote in my first post? There is a reason I wrote them.

Originally Posted by chengbin

Do you mind answering the questions I wrote in my first post? There is a reason I wrote them.

What questions do you think that the two of us did not answer?
Because of this question, I wonder if you even understand the question.

Ahh!

If when the denominator is 0, we assume that the numerator is also 0 (why?) and the fraction can be any value (why?)
z=3 (how do you get that?)

is the same as a= 0(x). But 0 times any number is 0 so that only makes sense if a= 0- that is, if "the numerator is also 0". but in that case 0= 0(x) for any x. That is why "the fraction can be any value". So for to make any sense, the numerator must be 0: z- 3= 0 so z= 3.

That's abusing the notation slightly. In fact, you can't divide by 0. All of this should be done in terms of limits to be valid.

Originally Posted by HallsofIvy

Ahh!

is the same as a= 0(x). But 0 times any number is 0 so that only makes sense if a= 0- that is, if "the numerator is also 0". but in that case 0= 0(x) for any x. That is why "the fraction can be any value". So for to make any sense, the numerator must be 0: z- 3= 0 so z= 3.

That's abusing the notation slightly. In fact, you can't divide by 0. All of this should be done in terms of limits to be valid.

Thanks a lot for clearing this up. Just what I needed to understand this.