The point of intersection C of the plane 8x+y+2z=0 and the line joining the points A(-3,-6…

Question

The point of intersection $C$ of the plane $8x+y+2z=0$ and the line joining the points $A(-3,-6,1)$ and $B(2,4,-3)$ divides the line segment $AB$ internally in the ratio $k:1$. If $a,b,c$ (with $|a|,|b|,|c|$ coprime) are the direction ratios of the perpendicular from the point $C$ on the line $\dfrac{1-x}{1}=\dfrac{y+4}{2}=\dfrac{z+2}{3}$, then $|a+b+c|$ is equal to:

Answer: 10

Accepted Answer

Line $AB$: $\dfrac{x-2}{5}=\dfrac{y-4}{10}=\dfrac{z+3}{-4}=\lambda$; point $(5\lambda+2,\ 10\lambda+4,\ -4\lambda-3)$. On the plane:
$$8(5\lambda+2)+10\lambda+4-8\lambda-6=0 \implies \lambda=-\frac13.$$
$$C=\left(\frac13,\frac23,-\frac53\right).$$
With $L:\ \dfrac{x-1}{-1}=\dfrac{y+4}{2}=\dfrac{z+2}{3}=\mu$, foot of perpendicular $D$ from $C$: solving $\vec{CD}\cdot(-1,2,3)=0$ gives $\mu=\dfrac{11}{14}$, and
$$\vec{CD}=\left(-\frac{5}{42},-\frac{130}{42},\frac{85}{42}\right)\ \propto\ (-1,-26,17).$$
$|a+b+c|=|-1-26+17|=10.$

Solutions · 1