Lines and points in 3d

Problem

The points \(A(-2|1|4)\) and \(B(-4|0|6)\) are given.

1. Determine the coordinates of the point \(C\) such that \(\vec{CA} = 2\cdot \vec{AB}\).

2. The straight line \(g\) passes through points \(A\) and \(B\). Consider straight lines fulfilling the conditions I and II:

   1. Each of the lines intersects the line \(g\) orthogonally.

   2. The distance of each of these lines from point \(A\) is \(3\).

   Determine an equation for one of those lines.

Solution of part a

We start by plugging \(\vec{CA}=\vec A - \vec C\) and \(\vec{AB}=\vec B - \vec A\) into the equation

\[\vec A - \vec C = 2 \cdot \left(\vec B - \vec A\right)\,,\]

and, subsequently, solve for \(\vec C\):

\[\begin{split}\vec C = \vec A - 2 \cdot \left(\vec B - \vec A\right) = 3\cdot\vec A - 2\cdot\vec B = 3 \cdot\begin{pmatrix} -2\\1\\4\end{pmatrix} - 2\cdot\begin{pmatrix} -4\\0\\6\end{pmatrix} = \begin{pmatrix} 2\\3\\0\end{pmatrix}\end{split}\]

This calculation can quickly be checked with Sage:

The coordinates of \(C\) are thus \((2|3|0)\).

Solution of part b

First, we observe that the distance from \(B\) to \(A\) is \(3\) because the line \([AB]\) has the length

\[\begin{split}|\vec{AB}| = \left|\begin{pmatrix}-2\\-1\\2\end{pmatrix}\right| = \sqrt{(-2)^2+(-1)^2+2^2} = 3\,.\end{split}\]

Sage can also confirm this:

Therefore, we choose \(B\) as the point on the line which has the required closest distance to \(A\). We then are left with determining the direction vector \(\vec G\) of the line. It has to be perpendicular to \(\vec{AB}\) and thus has to fulfill the condition

\[\vec{AB}\cdot\vec G = 0\,.\]

One can then simply guess a solution, for example

\[\begin{split}\vec G = \begin{pmatrix}1\\0\\1\end{pmatrix}\,.\end{split}\]

The equation of the line \(G\) thus is:

\[\begin{split}g: \vec X = \begin{pmatrix}-4\\0\\6\end{pmatrix} + \lambda \begin{pmatrix}1\\0\\1\end{pmatrix},\quad\lambda\in\mathbb{R}\,.\end{split}\]

We can determine the direction vector with Sage choosing for example the \(x_1\)- and \(x_2\)-coordinates at discretion. To check the above result, we set them to \(1\) and \(0\), respectively.