MHT CET · Maths · Vector Algebra
A, B, C, D are four points in a plane with position vectors \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}, \overline{\mathrm{d}}\) respectively such that \((\overline{\mathrm{a}}-\overline{\mathrm{d}}) \cdot(\overline{\mathrm{b}}-\overline{\mathrm{c}})=(\overline{\mathrm{b}}-\overline{\mathrm{d}}) \cdot(\overline{\mathrm{c}}-\overline{\mathrm{a}})=0\). The point \(\mathrm{D}\), then is the of \(\triangle \mathrm{ABC}\)
- A centroid
- B circumcentre
- C incentre
- D orthocentre
Answer & Solution
Correct Answer
(D) orthocentre
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & (\overline{\mathrm{a}}-\overline{\mathrm{d}}) \cdot(\overline{\mathrm{b}}-\overline{\mathrm{c}})=(\overline{\mathrm{b}}-\overline{\mathrm{d}})(\overline{\mathrm{c}}-\overline{\mathrm{a}})=0 \\ & \overline{\mathrm{AD}} \cdot \overline{\mathrm{BC}}=\overline{\mathrm{BD}} \cdot \overline{\mathrm{CA}}=0 \\ & \Rightarrow \overline{\mathrm{AD}} \perp \overline{\mathrm{BC}} \text { and } \overline{\mathrm{BD}} \perp \overline{\mathrm{CA}}\end{aligned}\)

\(\Rightarrow D\) is the orthocentre of \(\triangle A B C\).

\(\Rightarrow D\) is the orthocentre of \(\triangle A B C\).
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