My exercise is a lot bigger but I really need help with this line equation$-$
x-2=4y=z+1
Because it has two equal signs it's throwing me off a bit. Now my question is can I write the equation like this? x-4y+z-1=0
Thanks in advance!
PS: I am asking this because I keep getting wrong answer in my exercise, which is to find the distance between two lines and I think that I might have problem with that equation.
Edit: Thank you very much :) Happy New Year too!
@rschwieb I have given two lines U:{ 2x+y-3z+5=0 and x-3y+5z-4=0 } and V: x-2=4y=z+1 and now i need to find distance between the lines U and V.
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$\begingroup$Ultimately, presenting the line in that format is giving you three equations: $$x-2=4y\\4y=z+1\\x-2=z+1$$ Any two of them will determine the line, and you cannot condense them down to a single equation, as that would determine a plane.
Added: As to the problem at hand (now that I have more information), I will assume that $U$ is intended to be the line which is the intersection of the two planes $2x+y-3z+5=0$ and $x-3y+5z-4=0.$
Before we deal with $U$ at all, let's find a more convenient way to describe $V.$ Using the first two of our equations for $V,$ we see that for any point $\langle x,y,z\rangle$ on $V,$ we have $x=4y+2$ and $z=4y-1.$ That is, the points on $V$ are exactly those of the form $\langle 4y+2,y,4y-1\rangle$ where $y$ is real. We can instead replace $y$ (which we'd like to mean something very specific) with the variable $t,$ and note that $$\begin{align}\langle 4t+2,t,4t-1\rangle &= \langle 4t,t,4t\rangle+ \langle 2,0,-1\rangle\\ &= \langle 4,1,4\rangle t+ \langle 2,0,-1\rangle,\end{align}$$ meaning that the points on $V$ are precisely the vectors obtained by starting at $\langle2,0,-1\rangle$ and then moving by a scalar multiple of $\langle4,1,4\rangle.$
Next, let's find a similar form for $U.$ There are several ways to go about this. Let's begin by finding a convenient "starting point"--the analog to $\langle2,0,-1\rangle$ for $V.$ If we start with the system $$\begin{cases}2x+y-3z+5=0\\x-3y+5z-4=0\end{cases}\tag{$\heartsuit$}$$ and make the substitution $z=0,$ we get the system $$\begin{cases}2x+y+5=0\\x-3y-4=0,\end{cases}$$ which we can solve to get $x=-\frac{11}7$ and $y=-\frac{13}7,$ so that our "starting point" is $\left\langle-\frac{11}7,-\frac{13}7,0\right\rangle.$ Now, we'd like for every point of $U$ to have the form $$\langle a,b,c\rangle s+\left\langle-\frac{11}7,-\frac{13}7,0\right\rangle$$ for some fixed constants $a,b,c,$ where $s$ varies over the reals. A few geometric observations will simplify the task of finding such $a,b,c.$ First, the direction vector $\langle 2,1,-3\rangle$ is orthogonal to the plane $2x+y-3z+5=0.$ (Why?) Likewise, $\langle 1,-3,5\rangle$ is orthogonal to the plane $x-3y+5z-4=0.$ Since $U$ lies in both of these planes, then $\langle a,b,c\rangle$ will be orthogonal to both $\langle2,1,-3\rangle$ and $\langle1,-3,5\rangle.$ (Why?) The simplest way to find such a vector is to take the cross product of the two vectors: $$\begin{align}\langle2,1,-3\rangle\times\langle1,-3,5\rangle &= \langle(1)(5)-(-3)(-3),(-3)(1)-(2)(5),(2)(-3)-(1)(1)\rangle\\ &= \langle-4,-13,-7\rangle.\end{align}$$ Any non-zero real multiple of this vector will do the trick. For simplicity, I will multiply by $-1,$ so our parametric form for $U$ is $$\langle 4,13,7\rangle s+\left\langle-\frac{11}7,-\frac{13}7,0\right\rangle.$$
Now, when we talk about the distance between two non-intersecting lines, we mean the length of a segment connecting them that is orthogonal to both lines (if any such segment exists). Once again, then, we can use the cross product to find a direction vector for that segment: $$\begin{align}\langle4,13,7\rangle\times\langle4,1,4\rangle &= \langle(13)(4)-(7)(1),(7)(4)-(4)(4),(4)(1)-(13)(4)\rangle\\ &= \langle45,12,-48\rangle.\end{align}$$ Again, any non-zero real multiple of this vector will do. I choose $\langle 15,4,-16\rangle.$
Almost there! Now we want the segment to start on $U$, have direction vector $\langle 15,4,-16\rangle,$ and end on $V,$ so we want some real $s,t,u$ such that $$\langle 15,4,-16\rangle u+\langle 4,13,7\rangle s+\left\langle-\frac{11}7,-\frac{13}7,0\right\rangle= \langle 4,1,4\rangle t+ \langle 2,0,-1\rangle\\\left\langle 4s+15u-\frac{11}7,13s+4u-\frac{13}7,7s-16u\right\rangle=\langle 4t+2,t,4t-1\rangle.$$ This gives us a system of equations $$\begin{cases} 4s+15u-\frac{11}7=4t+2\\ 13s+4u-\frac{13}7=t\\7s-16u=4t-1.\end{cases}$$ Solve for $u,$ then use distance formula to find the length of the vector $\langle 15u,4u,-16u\rangle,$ which is precisely the length of the segment in question. (Why?)
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