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Find the line by a given point $(1, 2, 3)$, the line has common points with $\frac{x}{2}=\frac{y+1}{-2}=\frac{z-2}{1}$ and $\frac{x}{4}=\frac{y+2}{0}=\frac{z}{3}$

What am I doing wrong?

I decide to give a line as an intersection of two planes. A plane can be defined by a line and a point that is not on that line.

Consider a plane defined by $(1, 2, 3)$ and $\frac{x}{2}=\frac{y+1}{-2}=\frac{z-2}{1}$. If $\frac{x}{2}=\frac{y+1}{-2}=\frac{z-2}{1}$ lies in the plane, then the plane's normal vector can be any vector that is orthogonal to $\{2, -2, 1\}$; for example, $\{2, 1, -2\}=\{A, B, C\}$. Using $Ax+By+Cz+D=0$, we get $2x+y-2z+2=0$.

Consider a plane defined by $(1, 2, 3)$ and $\frac{x}{4}=\frac{y+2}{0}=\frac{z}{3}$. Analogically, we get $3x+y-4z+7=0$.

The answer in the book is $\begin{cases} 5x+y-8z+17=0 \\ 12x+9y-16z+18=0 \end{cases}$ , which didn't confuse me (an infinite number of intersecting planes can give a single line) until I saw that the direction vectors of my answer and the book's answer don't match. In fact, the direction vector of my answer is $\{2, -2, 1\}$, which is just the first line from the exercise.

I suspect I went astray with the second line, but maybe the entire reasoning is wrong.

Thank you.

fragileradius
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  • What is $$\frac{y+2}{0}$$? – Dr. Sonnhard Graubner Nov 03 '18 at 12:32
  • @Dr.SonnhardGraubner It tells you the line has direction vector ${4, 0, 3}$ and all points that belong to the line are $(x, -2, z)$ (I may remember incorrectly though). – fragileradius Nov 03 '18 at 12:34
  • @Dr.SonnhardGraubner $\frac{x}{4}=\frac{y+2}{0}=\frac{z}{3}$ means that $0\cdot x=4\cdot (y+2)$ and so on. It is not a division. It is a way of represent the line that contains the point $(0,-2,0)$ and has the direction $(4,0,3).$ – mfl Nov 03 '18 at 12:42

1 Answers1

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Consider a plane defined by $(1, 2, 3)$ and $\frac{x}{2}=\frac{y+1}{-2}=\frac{z-2}{1}$. If $\frac{x}{2}=\frac{y+1}{-2}=\frac{z-2}{1}$ lies in the plane, then the plane's normal vector can be any vector that is orthogonal to $\{2, -2, 1\}$; for example, $\{2, 1, -2\}=\{A, B, C\}$. Using $Ax+By+Cz+D=0$, we get $2x+y-2z+2=0$.

This is not true. Since the line is in the plane the normal vector is orthogonal to $(2,-2,1).$ But it can't be any orthogonal vector. Since $(1,2,3)$ and $(0,-1,2)$ are points of the plane the normal vector is also perpendicular to $(1,3,1).$ So, the normal vector is

$$(2,-2,1)\times (1,3,1)=(-5,-1,8).$$

mfl
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