Equations of Lines — Question 9

Question 9 – Solution

Strategy The two traces determine the line. Parameterize from $A$ toward $B$, then impose $y=0$ for the $xz$-plane.

Line equations A direction is $B-A=\left\langle -2,4,4\right\rangle=2\left\langle -1,2,2\right\rangle$. Thus $$\boxed{\vec r=\left\langle 2,-1,0\right\rangle+t\left\langle -1,2,2\right\rangle},$$ $$x=2-t,\qquad y=-1+2t,\qquad z=2t,$$ and $$\boxed{\frac{x-2}{-1}=\frac{y+1}{2}=\frac z2}.$$

$xz$-plane trace Set $y=0$. Then $-1+2t=0$, so $t=1/2$ and $$\boxed{C=\left(\tfrac 32,0,1\right)}.$$

Division ratio In this reduced parameterization, $B$ occurs at $t=2$. Thus $C$ lies one-half unit from $A$ in parameter and three-halves units from $B$, giving $$\boxed{AC:CB=1:3}.$$

Verification $C$ has $y=0$, and substituting its coordinates into the symmetric equation gives the common value $1/2$.