AP EAMCET · Maths · Vector Algebra
\(P\) is the circumcentre of \(\triangle A B C\). If the position vectors of \(A, B, C\) and \(P\) are \(\bar{a}, \bar{b}, \bar{c}, \frac{\bar{a}+\bar{b}+\bar{c}}{4}\) respectively, then the position vector of the orthocentre of this triangle is
- A \(\bar{a}+\bar{b}+\bar{c}\)
- B \(\frac{\bar{a}+\bar{b}+\bar{c}}{2}\)
- C \(-\left(\frac{\bar{a}+\bar{b}+\bar{c}}{2}\right)\)
- D \(\overline{0}\)
Answer & Solution
Correct Answer
(B) \(\frac{\bar{a}+\bar{b}+\bar{c}}{2}\)
Step-by-step Solution
Detailed explanation
\( \bar{g} = \frac{2\bar{p} + \bar{h}}{3} \) \( \frac{\bar{a}+\bar{b}+\bar{c}}{3} = \frac{2\left(\frac{\bar{a}+\bar{b}+\bar{c}}{4}\right) + \bar{h}}{3} \) \( \bar{a}+\bar{b}+\bar{c} = \frac{\bar{a}+\bar{b}+\bar{c}}{2} + \bar{h} \)…
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