# Finding the implicit equation for a line

I'm trying to find the equation of a line that passes through (1, 2, -1) and creates equivalent angles with the three positive axes. I'm at a loss, though I'd like just a small hint, not the full answer. Thanks!

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Wait, does $L : \lambda(1,1,1)+(1,2,-1)$ satisfy what's being asked? – F M Mar 11 '11 at 14:37
The idea of $\lambda(1,1,1)$ is a good one-it certainly makes a direction with the same angle to each axis. However with the point $(1,2,-1)$ the line does not actually hit any of the positive axes. From that point of view there is no solution as the only line through $(1,2,-1)$ that touches all the axes goes through the origin and does not make the same angle with each one. – Ross Millikan Mar 11 '11 at 15:42

Let $a$ be the angle between $\textbf{v}$ and $\textbf{x}$, $b$ be the angle between $\textbf{v}$ and $\textbf{y}$ and $c$ be the angle between $\textbf{v}$ and $\textbf{z}$. Then you are looking at the direction cosines. In other words, you are looking at the following: $$\alpha := \cos a := \frac{\textbf{v} \cdot \hat{\textbf{x}}}{|\textbf{v}|}$$ $$\beta := \cos b := \frac{\textbf{v} \cdot \hat{\textbf{y}}}{|\textbf{v}|}$$ and $$\gamma := \cos c := \frac{\textbf{v} \cdot \hat{\textbf{z}}}{|\textbf{v}|}$$

Then you can solve for $a$, $b$, and $c$.

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Some hints:

If you have two points $x,y$ on a line, you can parameterize the line. How?

The vector $x-y$ is parallel to the line. How do you compute the angle between two vectors?

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