If $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are the position vectors of the points $A, B, C, D$ respectively…

If $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are the position vectors of the points $A, B, C, D$ respectively such that $3 \bar{a}-\bar{b}+2 \bar{c}-4 \bar{d}=\overline{0}$, then the position vector of the point of intersection of the line segments $A C$ and $B D$ is
  1. $\frac{\bar{b}+3 \bar{d}}{4}$
  2. $\frac{3 \bar{a}+\bar{c}}{4}$
  3. $\frac{\bar{a}+\bar{c}}{2}$
  4. $\frac{\bar{b}+4 \bar{d}}{5}$

Solution

Given $3 \bar{a}-\bar{b}+2 \bar{c}-4 \bar{d}$ $\begin{aligned} 3 \bar{a}+2 \bar{c} &=\bar{b}+4 \bar{d} \\ \therefore & \frac{3 \bar{a}+2 \bar{c}}{3+2} \end{aligned}=\frac{\bar{b}+4 \bar{d}}{1+4}$ $\therefore \frac{3 \bar{a}+2 \bar{c}}{5}=\frac{\bar{b}+4 \bar{d}}{5}$

Asked in: MHT CET 2020 (12 Oct Shift 2)

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