Question:
Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are \(\vec{a},\ \vec{b},\ \vec{c}\) and \(\frac{\vec{a}+\vec{b}+\vec{c}}{4}\) respectively, then the position vector of the orthocentre of this triangle, is :
Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are \(\vec{a},\ \vec{b},\ \vec{c}\) and \(\frac{\vec{a}+\vec{b}+\vec{c}}{4}\) respectively, then the position vector of the orthocentre of this triangle, is :
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The centroid G divides the line from orthocentre H to circumcentre P in the ratio 2:1, so \(H=3G-2P\).
Updated On: Nov 11, 2026
- \(\vec{a}+\vec{b}+\vec{c}\)
- \(-\left(\frac{\vec{a}+\vec{b}+\vec{c}}{2}\right)\)
- \(\vec{0}\)
- \(\frac{\vec{a}+\vec{b}+\vec{c}}{2}\)
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The Correct Option is D
Solution and Explanation
Centroid \(\vec G=\frac{\vec a+\vec b+\vec c}{3}\). Circumcentre \(\vec P=\frac{\vec a+\vec b+\vec c}{4}\).
H, G, P are collinear with \(HG:GP=2:1\), so \(\vec G=\frac{\vec H+2\vec P}{3}\Rightarrow\vec H=3\vec G-2\vec P\).
\(\vec H=(\vec a+\vec b+\vec c)-\frac{\vec a+\vec b+\vec c}{2}=\frac{\vec a+\vec b+\vec c}{2}\).
Answer: (4) \(\frac{\vec{a}+\vec{b}+\vec{c}}{2}\)
H, G, P are collinear with \(HG:GP=2:1\), so \(\vec G=\frac{\vec H+2\vec P}{3}\Rightarrow\vec H=3\vec G-2\vec P\).
\(\vec H=(\vec a+\vec b+\vec c)-\frac{\vec a+\vec b+\vec c}{2}=\frac{\vec a+\vec b+\vec c}{2}\).
Answer: (4) \(\frac{\vec{a}+\vec{b}+\vec{c}}{2}\)
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