Q71 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics

If $ABC$ is a right-angled triangle in which $BC$ is the longest side and the position vector of $B$ and $C$ are respectively $3\hat{i} - 2\hat{j} + \hat{k}$ and $5\hat{i} + \hat{j} - 3\hat{k}$, then the value of $\overline{AB} \cdot \overline{AC} + \overline{BA} \cdot \overline{BC} + \overline{CA} \cdot \overline{CB}$ is

  • A
    $25$
  • B
    $27$
  • Correct answer
    C
    $29$
  • D
    $31$

Solution

Since $ABC$ is a right-angled triangle and $BC$ is the longest side, the right angle must be at $A$.

Therefore, $\overline{AB} \cdot \overline{AC} = 0$.

The given expression is $\overline{AB} \cdot \overline{AC} + \overline{BA} \cdot \overline{BC} + \overline{CA} \cdot \overline{CB}$.

Substituting $\overline{AB} \cdot \overline{AC} = 0$, we get:

$0 + \overline{BA} \cdot (\overline{BA} + \overline{AC}) + \overline{CA} \cdot (\overline{CA} + \overline{AB})$

$= |\overline{BA}|^2 + \overline{BA} \cdot \overline{AC} + |\overline{CA}|^2 + \overline{CA} \cdot \overline{AB}$

Since $\overline{BA} \cdot \overline{AC} = 0$ and $\overline{CA} \cdot \overline{AB} = 0$, the expression simplifies to:

$|\overline{BA}|^2 + |\overline{CA}|^2$

By Pythagoras theorem, $|\overline{BA}|^2 + |\overline{CA}|^2 = |\overline{BC}|^2$.

Given the position vectors of $B$ and $C$:

$\vec{B} = 3\hat{i} - 2\hat{j} + \hat{k}$

$\vec{C} = 5\hat{i} + \hat{j} - 3\hat{k}$

$\overline{BC} = \vec{C} - \vec{B} = (5 - 3)\hat{i} + (1 - (-2))\hat{j} + (-3 - 1)\hat{k} = 2\hat{i} + 3\hat{j} - 4\hat{k}$

$|\overline{BC}|^2 = (2)^2 + (3)^2 + (-4)^2 = 4 + 9 + 16 = 29$

Thus, the value of the given expression is $29$.

Answer: $29$

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

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