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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