The points $A$ and $B$ have position vectors \(\left( \begin{matrix} -1 \ 2 \ 5 \ \end{matrix

Question

The points $A$ and $B$ have position vectors $\left( \begin{matrix} -1 \ 2 \ 5 \ \end{matrix} \right)$ and $\left( \begin{matrix} 1 \ -2 \ 8 \ \end{matrix} \right)$ respectively. The point $C$ is such that $\overrightarrow{BC}=2\overrightarrow{AB}$. Find the position vector of $C$.[2] The point $D$ has position vector $\left( \begin{matrix} 1 \ 2 \ d \ \end{matrix} \right)$ and is such that $\left| \overrightarrow{AD} \right|=\left| \overrightarrow{BD} \right|$. Find the value of $d$.[2] Use a scalar product to find angle $ADB$.[3] Find exactly the position vector of the point $P$, where $P$ is the center of the circle that passes through $A$,$B$ and $D$.[5]

Accepted Answer

(a) $\overrightarrow{OC}=\left( \begin{matrix} 5 \ -10 \ 14 \ \end{matrix} \right)$ (b) $\frac{17}{2}$ (c) ${{83.8}^{\circ }}$ (d) $\overrightarrow{OP}=\frac{1}{72}\left( \begin{matrix} 7 \ 14 \ 482 \ \end{matrix} \right)$

N2023 P2 Q3

N2023 P2 Q3

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