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Finding the parametric and vector forms of the line is perpendicular to lines $(4t,1+2t,3t)$ and $(−1+s,−7+2s,−12+3s)$

And passes through the point of the intersection of two lines

A vector perpendicular to these lines is $$v = (4, 2, 3) \times (1, 2, 3) = [0, -9, 6]$$

How would I write the vector / parametric form?

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HINT

We have found the direction vector $\vec v$ for the perpendicular line, now we need the intersection point $P_0$ to determine the parametric equation

$$P(t)=P_0+t\vec v$$

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  • $\begingroup$ $[x, y, z] = [4,2,3] + t[0,-9,6]$? $\endgroup$ – Tree Garen Sep 28 '18 at 12:36
  • $\begingroup$ There is a problem with the intersection point. Indeed for $t = 1$ we obtain $(4,3,3)$ and for $s=5$ we have $(4,3,3)$. $\endgroup$ – user Sep 28 '18 at 12:41

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