AP EAMCET · Maths · Vector Algebra
A point \(C\) with position vector \(\frac{3 \bar{a}+4 \bar{b}-5 \bar{c}}{3}\) (where \(\bar{a}, \bar{b}\) and \(\bar{c}\) are non coplanar vectors) divides the line joining \(A\) and \(B\) in the ratio \(2: 1\). If the position vector of \(A\) is \(\bar{a}-2 \bar{b}+3 \bar{c}\), then the position vector of \(B\) is
- A \(2 \bar{a}+3 \bar{b}-4 \bar{c}\)
- B \(2 \bar{a}-3 \bar{b}+4 \bar{c}\)
- C \(2 \bar{a}+3 \bar{b}+4 \bar{c}\)
- D \(\bar{a}+3 \bar{b}-4 \bar{c}\)
Answer & Solution
Correct Answer
(D) \(\bar{a}+3 \bar{b}-4 \bar{c}\)
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