Let $\mathbf{O A}=\mathbf{a}, \mathbf{O B}=\mathbf{b}$ be two non collinear vectors, $\mathbf{O P}=x_1 \mathbf{a}+y_1 \mathbf{b}, \mathbf{O Q}=x_2 \mathbf{a}+y_2 \mathbf{b}$ and $\mathbf{A}^{\prime} \mathbf{O}=\mathbf{O A}, \mathbf{B}^{\prime} \mathbf{O}=\mathbf{O B}$. If $x_1=\frac{-3}{4}, x_2=\frac{1}{3}$, $y_1=\frac{7}{4}, y_2=\frac{5}{3}$, then

• Correct answer
  A

  $P$ lies inside the $\triangle A^{\prime} O B$ and $Q$ lies outside the $\triangle A O B$
• B

  $P$ lies outside the $\triangle A O B^{\prime}$ and $Q$ lies on the $\triangle A^{\prime} O B^{\prime}$
• C

  $P$ lies inside the $\triangle A O B$ and $Q$ lies outside the $\triangle A O B^{\prime}$
• D

  $P$ lies on the $\triangle A^{\prime} O B$ and $Q$ lies outside the $\triangle A O B$

Solution

We have, $\begin{aligned} & \mathbf{O A}=\mathbf{a}, \mathbf{O B}=\mathbf{b} \\ & \mathbf{O P}=x_1 \mathbf{a}+y_1 \mathbf{b}, \mathbf{O Q}=x_2 \mathbf{a}+y_2 \mathbf{b} \\ & \text { where, } x_1=\frac{-3}{4}, x_2=\frac{1}{3}, y_1=\frac{7}{4}, y_2=\frac{5}{3} \\ & \therefore \quad \mathbf{O P}=\frac{-3}{4} \mathbf{a}+\frac{7}{4} \mathbf{b} \\ & \text { OQ }=\frac{1}{3} \mathbf{a}+\frac{5}{3} \mathbf{b} \\ & \text { OP }=\frac{7 \mathbf{b}-3 \mathbf{a}}{7-3} \end{aligned}$ $\therefore$ OP divides $\mathbf{A B}$ externally in ratio $7: 3$. Also, $\mathbf{A}^{\prime} \mathbf{O}=\mathbf{O A}=\mathbf{a}$ $\begin{aligned} & \therefore \quad \mathbf{O A}^{\prime}=-\mathbf{a} \\ & \mathbf{B}^{\prime} \mathbf{O}=\mathbf{O B}=\mathbf{b} \\ & \mathbf{O B}^{\prime}=-\mathbf{b} \\ & \end{aligned}$
Clearly $p$ lies inside the $\triangle A^{\prime} O B$ and $Q$ lies outside the $\triangle A O B$.

Hence, the correct answer is option: 'A'.