2023-M2-10

Ans: (a) ${\displaystyle \frac{-2}{11}}\mathbf{i}-{\displaystyle \frac{5}{11}}\mathbf{j}-{\displaystyle \frac{1}{11}}\mathbf{k}$ (b) $-4\mathbf{i}-10\mathbf{j}-2\mathbf{k}$ (c) (i) $-\mathbf{i}+2\mathbf{k}$ (ii) $O$, $B$ and $G$ are collinear.

Sketch the graph according to the question.

1. Let $PR:RQ=1:\lambda$.

   In $\mathrm{\Delta }OPQ$, by section formula, we have

   $\begin{array}{rcl}\overrightarrow{OR}& =& {\displaystyle \frac{\lambda \overrightarrow{OP}+\overrightarrow{OQ}}{1+\lambda }}\\ \overrightarrow{OR}& =& {\displaystyle \frac{\lambda (-2\mathbf{i}-\mathbf{k})+(2\mathbf{i}-\mathbf{j}+\mathbf{k})}{1+\lambda }}\\ \overrightarrow{OR}& =& {\displaystyle \frac{2-2\lambda }{1+\lambda }}\mathbf{i}-{\displaystyle \frac{1}{1+\lambda }}\mathbf{j}+{\displaystyle \frac{1-\lambda }{1+\lambda }}\mathbf{k}\end{array}$

   Since $OR\perp OQ$, we have

   $\begin{array}{rcl}\overrightarrow{OR}\cdot \overrightarrow{OQ}& =& 0\\ {\displaystyle \frac{2-2\lambda }{1+\lambda }}\mathbf{i}-{\displaystyle \frac{1}{1+\lambda }}\mathbf{j}+{\displaystyle \frac{1-\lambda }{1+\lambda }}\mathbf{k}\cdot (2\mathbf{i}-\mathbf{j}+\mathbf{k})& =& 0\\ {\displaystyle \frac{2-2\lambda }{1+\lambda }}\times 2+{\displaystyle \frac{1}{1+\lambda }}+{\displaystyle \frac{1-\lambda }{1+\lambda }}& =& 0\\ 4-4\lambda +1+1-\lambda & =& \\ -5\lambda +6& =& 0\\ \lambda & =& {\displaystyle \frac{6}{5}}\end{array}$

   Hence, we have

   $\begin{array}{rcl}\overrightarrow{OR}& =& {\displaystyle \frac{2-2(\frac{6}{5})}{1+\frac{6}{5}}}\mathbf{i}-{\displaystyle \frac{1}{1+\frac{6}{5}}}\mathbf{j}+{\displaystyle \frac{1-\frac{6}{5}}{1+\frac{6}{5}}}\mathbf{k}\\ \overrightarrow{OR}& =& {\displaystyle \frac{-2}{11}}\mathbf{i}-{\displaystyle \frac{5}{11}}\mathbf{j}-{\displaystyle \frac{1}{11}}\mathbf{k}\end{array}$

2. Since $OR$ produced to $S$, then $O$, $R$ and $S$ are collinear. Hence, we have $\overrightarrow{OS}=\mu \overrightarrow{OR}$ for some real constant $\mu$.

   $\begin{array}{rcl}\overrightarrow{OS}& =& \mu \overrightarrow{OR}\\ \overrightarrow{OS}& =& {\displaystyle \frac{-2\mu }{11}}\mathbf{i}-{\displaystyle \frac{5\mu }{11}}\mathbf{j}-{\displaystyle \frac{\mu }{11}}\mathbf{k}\end{array}$

   Join $PQ$. By the theorem angle in the same segment, we have $\mathrm{\angle }QPS=\mathrm{\angle }QOS={90}^{\circ }$. Therefore, we have $PQ\perp PS$. Hence, we have

   $\begin{array}{rcl}\overrightarrow{PQ}\cdot \overrightarrow{PS}& =& 0\\ (\overrightarrow{OQ}-\overrightarrow{OP})\cdot (\overrightarrow{OS}-\overrightarrow{OP})& =& 0\\ (4\mathbf{i}-\mathbf{j}+2\mathbf{k})\cdot ((({\displaystyle \frac{-2\mu }{11}}+2)\mathbf{i}-{\displaystyle \frac{5\mu }{11}}\mathbf{j}+({\displaystyle \frac{-\mu }{11}}+1)\mathbf{k})& =& 0\\ 4({\displaystyle \frac{-2\mu }{11}}+2)+{\displaystyle \frac{5\mu }{11}}+2({\displaystyle \frac{-\mu }{11}}+1)& =& 0\\ -8\mu +88+5\mu -2\mu +22& =& 0\\ -5\mu +110& =& 0\\ \mu & =& 22\end{array}$

   Therefore, $\overrightarrow{OS}=-4\mathbf{i}-10\mathbf{j}-2\mathbf{k}$.

3. 1. A required vector

      $\begin{array}{cl}=& \overrightarrow{OP}\times \overrightarrow{OQ}\\ =& \left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -2& 0& -1\\ 2& -1& 1\end{array}\right|\\ =& \left|\begin{array}{cc}0& -1\\ -1& 1\end{array}\right|\mathbf{i}-\left|\begin{array}{cc}-2& -1\\ 2& 1\end{array}\right|\mathbf{j}+\left|\begin{array}{cc}-2& 0\\ 2& -1\end{array}\right|\mathbf{k}\\ =& -\mathbf{i}+2\mathbf{k}\end{array}$

   2. Sketch the graph according to the question.
      

      Since $\mathrm{\angle }QOS={90}^{\circ }$, then by the converse of angle in semi-circle, $QS$ is a diameter. Hence, we have

      $\begin{array}{rcl}\overrightarrow{OG}& =& {\displaystyle \frac{1}{2}}(\overrightarrow{OS}+\overrightarrow{OQ})\\ \overrightarrow{OG}& =& {\displaystyle \frac{1}{2}}(-4\mathbf{i}-10\mathbf{j}-2\mathbf{k}+2\mathbf{i}-\mathbf{j}+\mathbf{k})\\ \overrightarrow{OG}& =& -\mathbf{i}-{\displaystyle \frac{11}{2}}\mathbf{j}-{\displaystyle \frac{1}{2}}\mathbf{k}\end{array}$

      Since $\overrightarrow{BA}$ is perpendicular to the plane $\mathrm{\Pi }$, then $\overrightarrow{BA}\text{//}(\overrightarrow{OP}\times \overrightarrow{OQ})$. Then by the formula of vector projection, we have

      $\begin{array}{rcl}\overrightarrow{BA}& =& {\displaystyle \frac{\overrightarrow{OA}\cdot (\overrightarrow{OP}\times \overrightarrow{OQ})}{|\overrightarrow{OP}\times \overrightarrow{OQ}|}}\left({\displaystyle \frac{\overrightarrow{OP}\times \overrightarrow{OQ}}{|\overrightarrow{OP}\times \overrightarrow{OQ}|}}\right)\\ \overrightarrow{BA}& =& {\displaystyle \frac{(-6\mathbf{i}-22\mathbf{j}+2\mathbf{k})\cdot (-\mathbf{i}+2\mathbf{k})}{|-\mathbf{i}+2\mathbf{k}|}}\left({\displaystyle \frac{-\mathbf{i}+2\mathbf{k}}{|-\mathbf{i}+2\mathbf{k}|}}\right)\\ \overrightarrow{BA}& =& {\displaystyle \frac{(-6)(-1)+(-22)(0)+(2)(2)}{\sqrt{(-1{)}^2+(0{)}^2+(2{)}^2}}}\left({\displaystyle \frac{-\mathbf{i}+2\mathbf{k}}{\sqrt{(-1{)}^2+(0{)}^2+(2{)}^2}}}\right)\\ \overrightarrow{BA}& =& 2(-\mathbf{i}+2\mathbf{k})\\ \overrightarrow{BA}& =& -2\mathbf{i}+4\mathbf{k}\end{array}$

      Then, we have

      $\begin{array}{rcl}\overrightarrow{OB}& =& \overrightarrow{OA}-\overrightarrow{BA}\\ \overrightarrow{OB}& =& (-6\mathbf{i}-22\mathbf{j}+2\mathbf{k})-(-2\mathbf{i}+4\mathbf{k})\\ \overrightarrow{OB}& =& -4\mathbf{i}-22\mathbf{j}-2\mathbf{k}\\ \overrightarrow{OB}& =& 4(-\mathbf{i}-{\displaystyle \frac{11}{2}}\mathbf{j}-{\displaystyle \frac{1}{2}}\mathbf{k})\\ \overrightarrow{OB}& =& 4\overrightarrow{OG}\end{array}$

      Hence, $O$, $B$ and $G$ are collinear.