Q79 / 150 · MHT CET 2025 (25 Apr Shift 2) · Mathematics

The altitude through vertex A of $\triangle \mathrm{ABC}$ with position vectors of points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ as $\bar{a}, \overline{\mathrm{~b}}, \overline{\mathrm{c}}$ respectively is

  • A
    $\frac{|\bar{b} \times \bar{c}|}{|\bar{c}-\bar{b}|}$
  • Correct answer
    B
    $\frac{|\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a}|}{|\bar{c}-\bar{b}|}$
  • C
    $\frac{|\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a}|}{|\bar{c} \times \bar{b}|}$
  • D
    $\frac{|\overline{\mathrm{b}} \times \overline{\mathrm{c}}|}{|\bar{a}|}$

Solution

Given vertices A, B, C with position vectors $\bar{a}, \bar{b}, \bar{c}$, the area of $\triangle \mathrm{ABC}$ is expressed as $\text{Area} = \frac{1}{2} |(\bar{b} - \bar{a}) \times (\bar{c} - \bar{a})|$.

Expanding, $(\bar{b} - \bar{a}) \times (\bar{c} - \bar{a}) = \bar{b} \times \bar{c} - \bar{b} \times \bar{a} - \bar{a} \times \bar{c} + \bar{a} \times \bar{a}$. Since $\bar{a} \times \bar{a} = 0$ and using antisymmetry, this simplifies to $\bar{a} \times \bar{b} + \bar{b} \times \bar{c} + \bar{c} \times \bar{a}$.


The area is also $\text{Area} = \frac{1}{2} |\bar{c} - \bar{b}| h$, where $h$ is the altitude from A to BC. Equating the two expressions: $\frac{1}{2} |\bar{c} - \bar{b}| h = \frac{1}{2} |\bar{a} \times \bar{b} + \bar{b} \times \bar{c} + \bar{c} \times \bar{a}|$.


Solving for $h$ yields $h = \frac{|\bar{a} \times \bar{b} + \bar{b} \times \bar{c} + \bar{c} \times \bar{a}|}{|\bar{c} - \bar{b}|}$, corresponding to option B.

Hence, the correct answer is option: 'B'.

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