# Equation of line parallel to another passing through a point

• Kernul
Oh, it's because they are parallel? So only the coefficient of the z term changes, right?In summary, the parametric equations for a line parallel to a given line and passing through a given point can be found by using the directional vector of the given line and the given point. The resulting equations can be converted into Cartesian form by eliminating the parameter. The coefficients of the Cartesian equations will be the same as those in the original equations, except for the coefficient of the z term, which will be the opposite.

## Homework Statement

Write the parametric and cartesian equations of the line passing through ##P = (-\frac{3}{10}, 0, \frac{1}{10})## and parallel to
$$r : \begin{cases} 10x + 4y - 3 = 0 \\ x + z = 0 \end{cases}$$

## The Attempt at a Solution

Since the two lines have to be parallel, that means that the two directional vectors are linearly dependent. So I first calculate the directional vector of ##r## as follow:
##\vec v_r = \begin{vmatrix}
\hat i & \hat j & \hat k \\
10 & 4 & 0 \\
1 & 0 & 1
\end{vmatrix} = (4, -10, -4)##
Now that I have the directional vector, I know that the directional vector of the line we are searching for (let's call it ##u##) has to be linearly dependent to it, so something like this: ##\vec v_u = \alpha(4, -10, -4) = (4\alpha, -10\alpha, -4\alpha)##
With this I have already an overview of the parametric form of ##u## using ##\vec v_u## and ##P##, which is:
$$u : \begin{cases} x = 4\alpha - \frac{3}{10}t \\ y = -10\alpha \\ z = -4\alpha + \frac{1}{10}t \end{cases}$$
I was thinking about substituting this three equations into ##r## and find ##\alpha## but I stopped because I was doing it not knowing why. If it's the right way, why should I substitute them? If not, then, what should I do after arriving at this point?

Kernul said:

## Homework Statement

Write the parametric and cartesian equations of the line passing through ##P = (-\frac{3}{10}, 0, \frac{1}{10})## and parallel to
$$r : \begin{cases} 10x + 4y - 3 = 0 \\ x + z = 0 \end{cases}$$

## The Attempt at a Solution

Since the two lines have to be parallel, that means that the two directional vectors are linearly dependent. So I first calculate the directional vector of ##r## as follow:
##\vec v_r = \begin{vmatrix}
\hat i & \hat j & \hat k \\
10 & 4 & 0 \\
1 & 0 & 1
\end{vmatrix} = (4, -10, -4)##
Now that I have the directional vector, I know that the directional vector of the line we are searching for (let's call it ##u##) has to be linearly dependent to it, so something like this: ##\vec v_u = \alpha(4, -10, -4) = (4\alpha, -10\alpha, -4\alpha)##

You just need a single direction vector. You could divide out a ##2## to make it simpler: ##\vec D = \langle 2,-5,-2\rangle##. You don't need the ##\alpha##.

With this I have already an overview of the parametric form of ##u## using ##\vec v_u## and ##P##, which is:
$$u : \begin{cases} x = 4\alpha - \frac{3}{10}t \\ y = -10\alpha \\ z = -4\alpha + \frac{1}{10}t \end{cases}$$
I was thinking about substituting this three equations into ##r## and find ##\alpha## but I stopped because I was doing it not knowing why. If it's the right way, why should I substitute them? If not, then, what should I do after arriving at this point?

Remember the form for a straight line: ##\langle x,y,z\rangle = \vec P_0 + t\vec D##, where ##\vec P_0## is the position vector to a point on the line and ##\vec D## is any direction vector. Recheck your equation.

LCKurtz said:
You just need a single direction vector. You could divide out a ##2## to make it simpler: ##\vec D = \langle 2,-5,-2\rangle##. You don't need the ##\alpha##.

Remember the form for a straight line: ##\langle x,y,z\rangle = \vec P_0 + t\vec D##, where ##\vec P_0## is the position vector to a point on the line and ##\vec D## is any direction vector. Recheck your equation.
Oh sorry, I messed up the equation.
So it would be like:
$$u : \begin{cases} x = -\frac{3}{10} + 2t \\ y = -5t \\ z = \frac{1}{10} -2t \end{cases}$$

Kernul said:
Oh sorry, I messed up the equation.
So it would be like:
$$u : \begin{cases} x = -\frac{3}{10} + 2t \\ y = -5t \\ z = \frac{1}{10} -2t \end{cases}$$
You can (and should) check it yourself. Does the line you found have the right direction (i.e., is it parallel to the given line)? Is there a value of t so that it goes through the given point? If you can answer yes to both questions, then you know you have the right equation.

• Kernul
Mark44 said:
You can (and should) check it yourself. Does the line you found have the right direction (i.e., is it parallel to the given line)? Is there a value of t so that it goes through the given point? If you can answer yes to both questions, then you know you have the right equation.
The fact that has the right direction can be already seen from the beginning, since it's where I started. They are linearly dependent so they are surely parallel.
About the value of ##t##, I can see that only ##t = 0## would give me the point again. Is it correct?
By the way, the Cartesian form would be:
$$\begin{cases} x + \frac{2}{5}y + \frac{3}{10} = 0 \\ - \frac{2}{5}y + z - \frac{1}{10} = 0 \end{cases}$$

Kernul said:
The fact that has the right direction can be already seen from the beginning, since it's where I started. They are linearly dependent so they are surely parallel.
About the value of ##t##, I can see that only ##t = 0## would give me the point again. Is it correct?
By the way, the Cartesian form would be:
$$\begin{cases} x + \frac{2}{5}y + \frac{3}{10} = 0 \\ - \frac{2}{5}y + z - \frac{1}{10} = 0 \end{cases}$$
If you express your Cartesian solution as equations in ##x, y## and ##x, z## then would expect the coefficients to be the same as in the original equations?

PeroK said:
If you express your Cartesian solution as equations in ##x, y## and ##x, z## then would expect the coefficients to be the same as in the original equations?
I put ##t = \frac{y}{-5}## so I guess it's normal that the coefficients remained the same. But in general I actually don't know if the coefficients can be the same as in the parametric form or not.

The first equation you were given was ##10x +4y -3 =0##

The first equation you got for the parallel line after multiplication was ##10x +4y +3=0##

Is that a surprise or coincidence?

PeroK said:
The first equation you were given was ##10x +4y -3 =0##

The first equation you got for the parallel line after multiplication was ##10x +4y +3=0##

Is that a surprise or coincidence?
Oh, it's because they are parallel? So only the coefficient changed?

Kernul said:
Oh, it's because they are parallel? So only the coefficient changed?
If the lines are parallel, then ##x,y,z## change in the same proportion to each other, so only the constant term changes.

For the other equation you should get ##x+z=## some constant.

Kernul said:
They are linearly dependent so they are surely parallel.
If two vectors are linearly dependent, then each one is a nonzero multiple of the other. The situation is not as obvious if you have three linearly dependent vectors, where none of them is necessarily a multiple of any of the other two.

PeroK said:
If the lines are parallel, then ##x,y,z## change in the same proportion to each other, so only the constant term changes.

For the other equation you should get ##x+z=## some constant.
So I did do it wrong. Then I have to put ##t = \frac{x}{2} + \frac{3}{20}##. Substituting this I would get:
$$\begin{cases} \frac{5}{2}x + y +\frac{3}{4} = 0 \\ x + z + \frac{1}{5} = 0 \end{cases}$$
Which has the same equations but with different constants.

Mark44 said:
If two vectors are linearly dependent, then each one is a nonzero multiple of the other. The situation is not as obvious if you have three linearly dependent vectors, where none of them is necessarily a multiple of any of the other two.
You mean in case there were three parallel lines? Shouldn't they all be linearly dependent in order for them to be parallel? Like, having only one parameter as a difference between them. For example: ##\vec v = (a, b, c)##, ##\vec u = \alpha(a, b, c)##, and ##\vec w = \beta(a, b, c)##

Mark44 said:
If two vectors are linearly dependent, then each one is a nonzero multiple of the other. The situation is not as obvious if you have three linearly dependent vectors, where none of them is necessarily a multiple of any of the other two.

Kernul said:
You mean in case there were three parallel lines?
Well, I wasn't talking about three parallel lines. If two (or more) vectors are parallel, then it's pretty obvious that they are linearly dependent.
Kernul said:
Shouldn't they all be linearly dependent in order for them to be parallel? Like, having only one parameter as a difference between them. For example: ##\vec v = (a, b, c)##, ##\vec u = \alpha(a, b, c)##, and ##\vec w = \beta(a, b, c)##
It's possible for three vectors to be linearly dependent where none is a multiple of any of the others. All they have to do is lie in the same plane. For example, u = <1, 0>, v = <0, 1>, w = <1, 1>.

Mark44 said:
Well, I wasn't talking about three parallel lines. If two (or more) vectors are parallel, then it's pretty obvious that they are linearly dependent.
It's possible for three vectors to be linearly dependent where none is a multiple of any of the others. All they have to do is lie in the same plane. For example, u = <1, 0>, v = <0, 1>, w = <1, 1>.
Yes, yes, I already knew that. Putting them that way, they would still be parallel though, right?

Kernul said:
Yes, yes, I already knew that. Putting them that way, they would still be parallel though, right?
No, they are not parallel. Why would you think that? All three point in different directions.

If two vectors are parallel, each one is a nonzero multiple of the other, which means that they are necessarily linearly dependent. My point was that if you have three vectors, they can be linearly dependent, but no two of them have to be parallel.

• Kernul
Mark44 said:
No, they are not parallel. Why would you think that? All three point in different directions.

If two vectors are parallel, each one is a nonzero multiple of the other, which means that they are necessarily linearly dependent. My point was that if you have three vectors, they can be linearly dependent, but no two of them have to be parallel.
Oh okay. Because I was getting confused about what I had said before.