Q99 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

$A(1,2,3), B(2,3,1)$ and $C(3,1,2)$ are three points. If the point $\mathrm{P}$ divides $\mathrm{AB}$ in the ratio $1: 2$ and the point $\mathrm{Q}$ divides $\mathrm{BC}$ in the ratio $-2: 3$, then the distance between $P$ and $Q$ is

  • A
    $\sqrt{312}$
  • B
    13
  • Correct answer
    C
    $\frac{2}{3} \sqrt{78}$
  • D
    25

Solution

$$ \begin{aligned} & x=\frac{2+2}{3}, y=\frac{3+4}{3}, z=\frac{1+6}{3} \\ & P(x, y, z)=\left(\frac{4}{3}, \frac{7}{3}, \frac{7}{3}\right) \\ & \text {. } \begin{array}{ll} & -2,1 \\ B & Q \end{array} \\ & (2,3,1) \quad(\alpha, \beta, \gamma) \quad(3,1,2) \\ & \alpha=\frac{-6+6}{1}, \beta=\frac{-2+9}{1}, \gamma=\frac{-4+3}{1} \\ & Q(\alpha, \beta, \gamma)=(0,7,-1) \\ & \therefore P Q=\sqrt{\left(\frac{4}{3}-0\right)^2+\left(\frac{7}{3}-7\right)^2+\left(\frac{7}{3}+1\right)^2} \\ & =\sqrt{\frac{16}{9}+\frac{196}{9}+\frac{100}{9}}=\frac{\sqrt{312}}{3}=\frac{2}{3} \sqrt{78} . \\ & \end{aligned} $$

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

Relevant Questions