Let's go through how to solve this RI/RJC Prelim Vectors Question.

(i)

First, we can note that we can let $\mu=0$ and $\lambda=1$ for convenience since the vector equation holds for all values of $\lambda , \mu$. Then we have $\textbf{r}=\left( \begin{array}{c}6\\t\\2\end{array}\right)$. Putting this into the given equation in (i), we have $12+3t+2=5$, hence $t=-3$.

(ii) Let M be the midpoint of AB. Note that M will lie on the plane $\pi_1$! By the Midpoint Theorem,

$\vec{OM}=\frac{1}{2}[\vec{OA}+\vec{OB}]=\left( \begin{array}{c}4\\-6\\7+\frac{1}{2}s\end{array}\right)$.

Since M lies on the plane, $\vec{OM}\cdot \left(\begin{array}{c}2\\3\\1\end{array}\right)=5$. Expanding the dot product, we get $s=16$.

(iii) We need to show that the normals of the two planes are perpendicular.

$\mathbf{n_1}=\left( \begin{array}{c}2\\3\\1\end{array}\right)$ First we note that B also lies on $\pi_2$. We can obtain a normal of $\pi_2$ by taking the cross product $\mathbf{n_2}=\vec{AB}\times\left( \begin{array}{c}3\\-2\\0\end{array}\right)=\left( \begin{array}{c}4\\6\\-26\end{array}\right)$

We then calculate that $\mathbf{n_1}\cdot\mathbf{n_2}=0$. (shown)

If we draw a diagram, we can see that the shortest distance from A to the line of intersection is actually $|\vec{AM}|$, where M is the midpoint found earlier. $\vec{AM}=\vec{OM}-\vec{OA}=\left(\begin{array}{c}2\\3\\1\end{array}\right)$. Thus $|\vec{AM}|=\sqrt{2^2+3^2+1^2}=\sqrt{14}$.

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