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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
Options:
$3 \bar{a}-\bar{b}+2 \bar{c}-4 \bar{d}=\overline{0}$, then the position vector of the point of intersection of the line
segments $A C$ and $B D$ is
Solution:
1419 Upvotes
Verified Answer
The correct answer is:
$\frac{\bar{b}+4 \bar{d}}{5}$
Given $3 \bar{a}-\bar{b}+2 \bar{c}-4 \bar{d}$
$\begin{aligned} 3 \bar{a}+2 \bar{c} &=\bar{b}+4 \bar{d} \\ \therefore & \frac{3 \bar{a}+2 \bar{c}}{3+2} \end{aligned}=\frac{\bar{b}+4 \bar{d}}{1+4}$
$\therefore \frac{3 \bar{a}+2 \bar{c}}{5}=\frac{\bar{b}+4 \bar{d}}{5}$
$\begin{aligned} 3 \bar{a}+2 \bar{c} &=\bar{b}+4 \bar{d} \\ \therefore & \frac{3 \bar{a}+2 \bar{c}}{3+2} \end{aligned}=\frac{\bar{b}+4 \bar{d}}{1+4}$
$\therefore \frac{3 \bar{a}+2 \bar{c}}{5}=\frac{\bar{b}+4 \bar{d}}{5}$
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