Equation of line parallel to another passing through a point

Kernul

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?

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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.

Kernul
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:
$$u : \begin{cases} x = -\frac{3}{10} + 2t \\ y = -5t \\ z = \frac{1}{10} -2t \end{cases}$$

Mentor
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
Kernul
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:
$$\begin{cases} x + \frac{2}{5}y + \frac{3}{10} = 0 \\ - \frac{2}{5}y + z - \frac{1}{10} = 0 \end{cases}$$

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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?

Kernul
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.

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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?

Kernul
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?

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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.

Mentor
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.

Kernul
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.

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)##

Mentor
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?
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>.

Kernul
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?

Mentor
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
Kernul
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.