Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The position vectors of the given points are $A(2,-3,0)$, $B(1,-4,-2)$, $C(4,6,8)$, and $D(7,0,10)$.

First, we find the vector $\overline{AB}$:
$\overline{AB}=(1-2)\hat{i}+(-4-(-3))\hat{j}+(-2-0)\hat{k}$
$\overline{AB}=-\hat{i}-\hat{j}-2\hat{k}$

Next, we find the vector $\overline{CD}$:
$\overline{CD}=(7-4)\hat{i}+(0-6)\hat{j}+(10-8)\hat{k}$
$\overline{CD}=3\hat{i}-6\hat{j}+2\hat{k}$

The magnitude of the projection of $\overline{AB}$ on $\overline{CD}$ is given by $\left|\frac{\overline{AB}\cdot \overline{CD}}{|\overline{CD}|}\right|$.

Calculating the dot product $\overline{AB}\cdot \overline{CD}$:
$\overline{AB}\cdot \overline{CD}=(-1)(3)+(-1)(-6)+(-2)(2)$
$\overline{AB}\cdot \overline{CD}=-3+6-4=-1$

Calculating the magnitude of $\overline{CD}$:
$|\overline{CD}|=\sqrt{3^2+(-6{)}^2+2^2}$
$|\overline{CD}|=\sqrt{9+36+4}=\sqrt{49}=7$

Therefore, the magnitude of the projection is:
$\left|\frac{-1}{7}\right|=\frac{1}{7}$

Answer: $\frac{1}{7}$