3 Intersecting Planes

Finding $\Pi _{1}$

Express $x$ , $y$ and $z$ in terms of $\lambda$ by equating $\lambda$ to each component of $l$

$x=2+5\lambda$ , $y=4-2\lambda$ , $z=-3-\lambda$

Hence, the vector equation of $l$ is

$r=\begin{pmatrix} 2\\ 4\\ -3 \end{pmatrix} +\lambda \begin{pmatrix} 5\\ -2\\ -1 \end{pmatrix}$

Let $X$ be the point on $l$ such that $\vec{AX}$ is perpendicular to $l$

$\vec{AX}=X-A=\vec{n}$

$=\begin{pmatrix} 2+5\lambda \\ 4-2\lambda \\ -3-\lambda \end{pmatrix}-\begin{pmatrix} 8\\ 2\\ 1 \end{pmatrix}$

$=\begin{pmatrix} -6+5\lambda \\ 2-2\lambda \\ -4-\lambda \end{pmatrix}$

Using the fact that $\vec{n}\cdot \vec{l}=0$

$\begin{pmatrix} -6+5\lambda \\ 2-2\lambda \\ -4-\lambda \end{pmatrix}\cdot \begin{pmatrix} 5\\ -2\\ -1 \end{pmatrix}=0$

$-30+25\lambda -4+4\lambda +4+\lambda =0$

$\lambda =1$

$\vec{n}=\begin{pmatrix} -1\\ 0\\ -5 \end{pmatrix}$

Express $\Pi _{1}$ in Cartesian form

$\begin{pmatrix} x\\ y\\ z \end{pmatrix}\cdot \begin{pmatrix} -1\\ 0\\ -5 \end{pmatrix}=\begin{pmatrix} 8\\ 2\\ 1 \end{pmatrix}\cdot \begin{pmatrix} -1\\ 0\\ -5 \end{pmatrix}=-8-5=-13$

$x+5z=13$

Find the general equation for the intersection point of $\Pi _{1}$ and $\Pi _{2}$

$\left\{\begin{matrix} x+5z=13\\ -3x+20y-15z=1 \end{matrix}\right.$

Adding the equations gives $20y=40$ so $y=2$

$x=13-5z$ and since $z$ can be any number, it can be replaced by variable $t$

Summarising the results

$x=13-5t$ , $y=2$ , $z=t$ where $t \in \mathbb{R}$

Substitute $x$ , $y$ and $z$ into $\Pi _{3}$ to find $t$

$3(13-5t)+2-t=9$

$t=2$

The coordinates of the common point are

$\left ( 13-5t, 2, t \right )=\left ( 3, 2, 2 \right )$

Sum of coordinates

$3+2+2=\textbf{7}$