Let the line L pass through the point (-3,5,2) and make equal angles with the positive coo…

Question

Let the line $L$ pass through the point $(-3,5,2)$ and make equal angles with the positive coordinate axes (direction $(1,1,1)$). If the distance of $L$ from the point $(-2,r,1)$ is $\sqrt{	frac{14}{3}}$, then the sum of all possible values of $r$ is:

Accepted Answer

$L:\ \dfrac{x+3}{1}=\dfrac{y-5}{1}=\dfrac{z-2}{1}=\lambda$, general point $R(\lambda-3,\ \lambda+5,\ \lambda+2)$.

Let $P=(-2,r,1)$. For $R$ to be the foot of perpendicular from $P$: $\vec{PR}\cdot(1,1,1)=0$:
$$(\lambda-1)+(\lambda+5-r)+(\lambda+1)=0 \implies 3\lambda+5-r=0 \implies \lambda=\frac{r-5}{3}.$$
So $R=\left(\dfrac{r-14}{3},\ \dfrac{r+10}{3},\ \dfrac{r+1}{3}\right)$. Setting $|PR|^2=\dfrac{14}{3}$:
$$\left(\frac{r-14}{3}+2\right)^2+\left(\frac{r+10}{3}-r\right)^2+\left(\frac{r+1}{3}-1\right)^2=\frac{14}{3}.$$
$$\frac{(r-8)^2+(10-2r)^2+(r-2)^2}{9}=\frac{14}{3} \implies (r-8)^2+(10-2r)^2+(r-2)^2=42.$$
$$6r^2-60r+168=42 \implies r^2-10r+21=0 \implies r=3,7.$$
Sum $=10$.

Solutions · 1