Why this isn't working? Find the points of the line $r$ that has the distance $\sqrt{\frac{14}{3}}$ from line $s$.

Question

I have the line $$r:\\x+y=2\\x=y+z$$

and $$s:x=y=z+1$$

I need to find the points of $r$ that has distance $\sqrt{\frac{14}{3}}$ from $s$.

What I tried:

By using the formula for distance of a point to a line:

$$d(P,l) = \frac{||\vec{AP}\times\vec v||}{||\vec v||}$$ where $\vec v$ is a direction vector to the line. I've choose the point $(1,1,0)$ from $s$, and a generic point from $r$ has the form:

$$x = \lambda\\y = 2-\lambda\\z = -2+2\lambda$$ (I don't know if this is rigth, I just choose $z=\lambda$ in the equations for $r$ and got this). So the vector $\vec{AP}=(1-\lambda, -1+\lambda,2-2\lambda)$ $$d(P,r) = \frac{||(1-\lambda, -1+\lambda,2-2\lambda)\times(1,1,0)||}{\sqrt{2}} = \sqrt{\frac{14}{3}}$$

There is no solution for $\lambda$ because the cross product gives a vector not related to $\lambda$.

Why this isn't working, but the answer given below Works?

Answer 1

Let $P(t, 2-t, 2t-2)$ be a general point on r.

A vector parallel to s is given by $\vec{b}=\langle1,1,1\rangle$, and a vector parallel to r is given by $\vec{a}=\vec{QP}=\langle t-1,1-t,2t-2\rangle$ where $Q=(1,1,0)$ is on both r and s.

Then $\vec{a}\times\vec{b}=\langle 3-3t,t-1,2t-2\rangle$, and the distance between r and s is given by

$d=\displaystyle\frac{\left|\vec{a}\times\vec{b}\right|}{\left|\vec{b}\right|}=\frac{\sqrt{14(t-1)^2}}{\sqrt{3}}=\frac{\sqrt{14}}{\sqrt{3}}\left|t-1\right|$,

so $\displaystyle d=\frac{\sqrt{14}}{\sqrt{3}}\iff t=0$ or $t=2$ $\iff P=(0,2,-2)$ or $P=(2,0,2)$.

---

Alternate solution:

Let $P(t, 2-t, 2t-2)$ be a general point on r and let $\vec{b}=\langle1,1,1\rangle$ be a direction vector for s as above,

and let $\vec{a}=\vec{QP}=\langle t, 2-t, 2t-1\rangle$ where $Q=(0,0,-1)$ is a point on s.

By the Pythagorean Theorem,

$\displaystyle d^2=|\vec{a}|^2-(\text{comp}_{\vec{b}}\vec{a})^2=\vec{a}\cdot\vec{a}-\frac{(\vec{a}\cdot\vec{b})^2}{\vec{b}\cdot\vec{b}}=t^2+(2-t)^2+(2t-1)^2-\frac{(2t+1)^2}{3}$, so

$\displaystyle d^2=\frac{14}{3}\iff 3(6t^2-8t+5)-(4t^2+4t+1)=14\iff 14t^2-28t=0$

$\iff14t(t-2)=0\iff t=0$ or $t=2$ $\iff P=(0,2,-2)$ or $P=(2,0,2)$.