MHT CET · Maths · Vector Algebra
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
- A \(\frac{\bar{b}+3 \bar{d}}{4}\)
- B \(\frac{3 \bar{a}+\bar{c}}{4}\)
- C \(\frac{\bar{a}+\bar{c}}{2}\)
- D \(\frac{\bar{b}+4 \bar{d}}{5}\)
Answer & Solution
Correct Answer
(D) \(\frac{\bar{b}+4 \bar{d}}{5}\)
Step-by-step Solution
Detailed explanation
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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