## anonymous 5 years ago Find the equation of a plane that contains the lines (x - 1)/6= y/8= (z + 2)/2 and (x + 1)/3= (y - 2)/4= z+ 5.

1. anonymous

$\frac{x-1}{6}=\frac{y}{8}=\frac{z+2}{2}$ and $\frac{x+1}{3}=\frac{y-2}{4}=z+5$ just to make it easier to look at

2. anonymous

the direction vector of the first line call it $\vec{v}=(6,8,2)$ and the corresponding direction vector of the second line call it $\vec{u}=(3,4,1)$

3. anonymous

Now calculate the cross-product $\vec{v}\times\vec{u}$ and to make calculations a little easier note the scalar value 2 is common to the components of $\vec{v}$

4. anonymous

on closer inspection those lines are parallel

5. anonymous

so form a vector from a point on the first line to a point on the second, the cross that vector with either of the direction vectors, that vector will be the normal vector for the plane that contains both lines, then just plug the components of your new vector , and a point from either line, into the standard equation for a plan and there you have it