Question

Find the equation of a line passing through the points $A(0,6,-9)$ and $B(-3,-6,-3)$ $D$ is the foot of paspendicular drawn from a point $C(7,4,-1)$ on the line $AB$, then find the co-ordinates of the point $D$ and the equation of the line $CD$.

Answer 1

$A(0,6,-9)$ and $B(-3,-6,-3)$

$\therefore$ Equation of line through $A$ & $B$

$\frac{x}{3}=\frac{y-6}{12}=\frac{z+9}{-6}$

Let $(3r,12r+6,-6r-9)$ be the foot of $\perp$ from $C(7,4,-1)$.

$\therefore (3r-7,12r+2,-6r-8)$ is $\perp$ to $AB$.

$\therefore (3r-7)3+12(12r+2)-6(-6r-8)=0\Rightarrow r=\frac{-17}{63}$

$\therefore$ Co-ordinates of point $D=(\frac{-17}{21},\frac{53}{21},\frac{-155}{21})$

$\therefore$ Equation of line $CD$,

$\frac{x-7}{7+\frac{17}{21}}=\frac{y-4}{4-\frac{53}{21}}=\frac{z+1}{-1+\frac{155}{21}}\Rightarrow \frac{x-7}{164}=\frac{y-4}{31}=\frac{z+1}{134}$