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I have an idea that in order to get the equation given only the lines, it requires to do cross product and solving for the coordinates that satisfies the given two lines.

The problem is that only intersecting lines of planes are given. I have no problem for the cross product but I am confused for finding the coordinates since the answer would be infinite solution. What could be another way to find the points?

The original problem goes this way.

Determine an equation for each of the lines described below.

line of intersection of the planes 2x - y + z = -1 and x + 4y - z = 2.

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  • $\begingroup$ What do you mean only two lines are given? The question reads line of intersection of two planes. $\endgroup$
    – Math Lover
    Mar 14, 2021 at 11:19
  • $\begingroup$ You're given a system of two (linear, inhomogeneous) equations in three unknowns. Do you know how to solve such a system? The solution can be written as an equation of a line in 3-space. $\endgroup$ Mar 14, 2021 at 11:36
  • $\begingroup$ Yes I have an idea at least on how to solve such equations in three unknowns. However, it will result into equations, not coordinates. I don't know if that result can be applied in determining the equation. $\endgroup$
    – Fubuki
    Mar 14, 2021 at 12:45
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    $\begingroup$ "I don't know if that result can be applied in determining the equation." Well, if you don't know, you should try it, and see for yourself! $\endgroup$ Mar 14, 2021 at 22:03

2 Answers 2

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Crossing the normal vectors to the plane, you’ll get the direction vector of the line, i.e. $$(2,-1,1) \times (1,4,-1) =(-3,3,9) $$

Now, you need any one point that lies on the line. This will be any point that lies on both the planes. One variable is arbitrary. Let’s take $z=0$. Then you get $$2x-y=-1 \\ x+4y =2 \\ \implies x =-\frac 29, y= \frac 59$$ and thus, the equation of the line is $$\vec r = \left( -\frac 29, \frac 59, 0 \right) +\lambda(-3,3,9)$$

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  • $\begingroup$ Based on your solution, it shows that I can input any value on either of the variables x, y, or z since one variable is arbitrary and I only need one point. Am I right? $\endgroup$
    – Fubuki
    Mar 14, 2021 at 12:41
  • $\begingroup$ @Aiden That’s right. $\endgroup$
    – Tavish
    Mar 14, 2021 at 12:52
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Crossing the normal vectors to the plane, you’ll get the direction vector of the line, i.e. $$(2,-1,1) \times (1,4,-1) =(-3,3,9) $$

Now, you need any one point that lies on the line. This will be any point that lies on both the planes. One variable is arbitrary. Let’s take $z=0$. Then you get $$2x-y=-1 \\ x+4y =2 \\ \implies x =-\frac 29, y= \frac 59$$ and thus, the equation of the line is $$\vec r = \left( -\frac 29, \frac 59, 0 \right) +\lambda(-3,3,9)$$

Crossing the normal vectors to the plane, you’ll get the direction vector of the line, i.e. $$(2,-1,1) \times (1,4,-1) =(-3,3,9) $$

Now, you need any one point that lies on the line. This will be any point that lies on both the planes. One variable is arbitrary. Let’s take $z=0$. Then you get $$2x-y=-1 \\ x+4y =2 \\ \implies x =-\frac 29, y= \frac 59$$ and thus, the equation of the line is $$\vec r = \left( -\frac 29, \frac 59, 0 \right) +\lambda(-3,3,9)$$

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  • $\begingroup$ Why have you posted two copies of another user's answer? $\endgroup$ Mar 14, 2021 at 22:01
  • $\begingroup$ Based on the time you posted the answer, I think you have only copied Tavish's answer. $\endgroup$
    – Fubuki
    Mar 15, 2021 at 0:00

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