Equal distances in a line to two other given lines I know that it is likely to be a very elementary question, but appreciate so much your help.
The exercise asks to find points in $r$ such that the distance from other two given lines ($s_1$ and $s_2$) is the same.
$r:x-1=2y=z$
$s_1:\begin{cases}
x=2 \\ 
y=\gamma \\
z=0
\end{cases}$
$s_2:\begin{cases}
x=0 \\
y=0 \\
z=\delta 
\end{cases}$
My attempt:
I know that the distance between two skewed lines is the projection of the vector determined by any two given points in each line in the cross product of director vectors of them both.
The parametric equation of $r$ is$$r:\begin{cases}
x=\lambda +1 \\ 
y=\frac{\lambda}{2} \\
z=\lambda
\end{cases}$$so a direction vector for $r$ is $v= \left (1,\dfrac{1}{2},1\right )$, a direction vector for $s1$ is $(0,1,0)$ and for $s2$ is $(0,0,1)$.$$\frac{(v\times (0,1,0))\cdot \left (\lambda +1-2,\frac{\lambda}{2}-\gamma ,\lambda \right )}{|v\times (0,1,0)|}= 
\frac{(v\times (0,0,1))\cdot \left (\lambda +1,\frac{\lambda}{2}-\delta ,\lambda \right )}{|v\times (0,0,1)|}.$$
The answer are the points $(1,0,0)$ and $(19/3,8/3,16/3)$, but I really don't see how to get rid off all those variables $(\lambda ,\delta ,\gamma )$ to reach those two points. Can someone clarify it? Thanks in advance.
 A: We can parameterize $r$ using $x$, i.e., all points on $r$ look as $(2t+1, t, 2t)$,
or $(\lambda+1, \frac{\lambda}{2}, \lambda)$.
Given a line $\mathbf p + t \mathbf e$, where $\mathbf e$ is a unit vector, and a point $\mathbf q$ the
squared distance between them is given by
$$ d^2 = |\mathbf q-\mathbf p|^2 - ((\mathbf q-\mathbf p)\cdot \mathbf e)^2$$
For $s_1$ (where $\mathbf p = (2,0,0), \mathbf e = (0,1,0)$ and $\mathbf q = (x,y,z)$:
$$
d_1^2 = (x-2)^2 + y^2 + z^2 - y^2
\;\;\;=x^2 + z^2 -4x+4
$$
For $s_2$ (where $\mathbf p = (0,0,0), \mathbf e = (0,0,1)$ :
$$
d_2^2 = x^2+y^2+z^2-z^2
\;\;\;=x^2+y^2
$$
This means that for the point $(x,y,z)$ to be equidistant from $s_1$ and $s_2$ we need to
have
$$x^2+z^2-4x+4 = x^2 + y^2$$
or
$$z^2-y^2-4x+4 = 0$$
Using parametrization of $r$ we get:
$$
\lambda^2 - \frac{\lambda^2}{4} - 4 \lambda -4+ 4 = 0
$$
or
$$3\lambda^2 - 16\lambda =0$$
The solutions are $\lambda=0$ and $\lambda=\frac{16}{3}$, and the corresponding points on
$R$ are thus $(1,0,0)$ and $(\frac{19}{3}, \frac{8}{3}, \frac{16}{3})$
A: By applying distance between 2 skew lines formula, we get that the distance between $r$ and $s_1$ is $0.71$ while between $r$ and $s_2$ is $0.45$. Even if they were the same distance apart, it would not prove that the same point is equidistant to both lines. So this is not the right approach. We need to instead use the formula between a line and a point.
The distance between a point $P(x_0, y_0, z_0)$, associated with the direction vector $\vec{p}$, and the line $\vec{r} = \vec{a} + t\vec{b}$ is
$$d = \frac{|\vec{b} \times (\vec{p} - \vec{a})|}{|\vec{b}|}$$
$s_1 = (2, 0, 0) + \gamma (0, 1, 0)$. $s_2 = (0,0,0) + \delta(0, 0, 1)$. As you have mentioned, the desired points on $r$ can be expressed by $(\lambda + 1, 0.5\lambda, \lambda)$. Plugging these in we get
$$d = \frac{|(0,1,0) \times (\lambda - 1, 0.5\lambda, \lambda)|}{|(0, 1, 0)|} = \frac{|(0,0,1) \times (\lambda + 1, 0.5\lambda, \lambda)|}{|(0, 0, 1)|}$$
Expand this we obtain
$$d = |(\lambda, 0, 1 - \lambda)| = |(- 0.5\lambda, \lambda + 1, 0)|$$
Expand the norm, and solve for lambda, we get
$$d^2 = \lambda^2 + 1 - 2\lambda + \lambda^2 = 0.25\lambda^2 + \lambda^2 + 2\lambda + 1$$
$$\lambda = 0, \frac{16}{3}$$
This translates into the points $(1, 0, 0)$ and $(\frac{19}{3}, \frac{8}{3}, \frac{16}{3})$.
