MHT CET · Maths · Vector Algebra
If \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are lengths of the sides \(\mathrm{BC}, \mathrm{CA}, \mathrm{AB}\) respectively of \(\triangle \mathrm{ABC}\) and \(\mathrm{H}\) is any
point in the plane of \(\Delta \mathrm{ABC}\) such that a \(\overline{A H}+b \overline{B H}+c \overline{C H}=\overline{0}\), then \(\mathrm{H}\) is the
- A Circumcentre of \(\Delta \mathrm{ABC}\)
- B Incentre of \(\Delta \mathrm{ABC}\)
- C Centroid of \(\Delta \mathrm{ABC}\)
- D Orthocentre of \(\Delta \mathrm{ABC}\)
Answer & Solution
Correct Answer
(B) Incentre of \(\Delta \mathrm{ABC}\)
Step-by-step Solution
Detailed explanation

Consider \(\mathrm{H}\) to be origin.
Then position vector of the vertices \(A, B, C\) are \(\bar{a}, \bar{b}, \bar{c}\) respectively.
We have \(\mathrm{a} \overline{\mathrm{AH}}+\mathrm{b} \overline{\mathrm{BH}}+\mathrm{c} \overline{\mathrm{CH}}=\overline{0}\) i.e. \(\quad \mathrm{a} \overline{\mathrm{a}}+\mathrm{b} \overline{\mathrm{b}}+\mathrm{c} \overline{\mathrm{c}}=0\)
i.e. \(\frac{a \bar{a}+b \bar{b}+c \bar{c}}{a+b+c}=0\), which is position vector of incentre.
Hence \(\mathrm{H}\) is incentre of triangle.
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