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I'm having problems with this:

Find the parametric equation of the line that passes through the point $(-1, 4, 5)$ and is perpendicular to the line: $$x = -2 + t$$ $$y = 1 - t$$ $$z = 1 + 2t$$

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    $\begingroup$ wouldn't any line in the plane $x-y+2z = 5$ do? $\endgroup$ – abel Jun 15 '15 at 20:59
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Let $X = (-2+t, 1-t, 1+2t)$ be a point on the line, and $x_0 = (-1,4,5)$ the given point. We want $(X - x_0) \cdot (1, -1, 2) = 0$:

$$(-2 + t + 1)(1) + (1 - t - 4)(-1) + (1+2t -5)(2) = 0$$ $$6t - 6= 0$$ $$t = 1$$

Pick $x_1$ to be the point on $X$ at $t=1$, so $(-1, 0, 3)$.

Now all we need is an equation connecting $x_0$ and $x_1$. One such is:

$$\begin{split} x &= -1\\ y &= -4t + 4\\ z &= -2t + 5\end{split}$$

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