# Finding the parametric and vector forms of the line is perpendicular to two lines

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?

## 1 Answer

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

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

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