Vector Algebra Question - CBSE

Question

Vectors $\vec{a}, \vec{b}, \vec{c}$ are such that $|\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 5$ and each of them is perpendicular to the sum of the other two. Find $|\vec{a} + \vec{b} + \vec{c}|$.

Accepted Answer

From perpendicularity: $\vec{a}\cdot(\vec{b}+\vec{c}) = \vec{b}\cdot(\vec{c}+\vec{a}) = \vec{c}\cdot(\vec{a}+\vec{b}) = 0$. [1.5 Marks]
Sum of dot products $= 0 \Rightarrow 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 0$. [1.5 Marks]
$|\vec{a}+\vec{b}+\vec{c}|^2 = 9 + 16 + 25 = 50$. [1.0 Mark]
$|\vec{a}+\vec{b}+\vec{c}| = 5\sqrt{2}$. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Setting up perpendicularity equations: 1.5 Marks
Evaluating sum of pairwise dot products: 1.5 Marks
Expanding magnitude square of 3 vectors sum: 1.0 Mark
Evaluating final magnitude $= 5\sqrt{2}$: 1.0 Mark

Question:

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