Q74 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics

If $|\vec{a}| = 3$, $|\vec{b}| = 4$, $|\vec{c}| = 5$ such that each vector is perpendicular to the sum of the other two, then $|\vec{a} + \vec{b} + \vec{c}|$ is equal to

  • A
    $5\sqrt{3}$
  • B
    $10\sqrt{2}$
  • Correct answer
    C
    $5\sqrt{2}$
  • D
    $4\sqrt{3}$

Solution

Given that each vector is perpendicular to the sum of the other two, we have:

$\vec{a} \cdot (\vec{b} + \vec{c}) = 0 \Rightarrow \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$

$\vec{b} \cdot (\vec{c} + \vec{a}) = 0 \Rightarrow \vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0$

$\vec{c} \cdot (\vec{a} + \vec{b}) = 0 \Rightarrow \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0$

Adding the above three equations, we get:

$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$

Using the identity for the square of the sum of three vectors:

$|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$

Substituting the given magnitudes $|\vec{a}| = 3$, $|\vec{b}| = 4$, and $|\vec{c}| = 5$:

$|\vec{a} + \vec{b} + \vec{c}|^2 = 3^2 + 4^2 + 5^2 + 0$

$|\vec{a} + \vec{b} + \vec{c}|^2 = 9 + 16 + 25 = 50$

$|\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2}$

Answer: $5\sqrt{2}$

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

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