Question

Let \(\vec{a}, \vec{b}, \vec{c}\) be such that \(\vec{a} + \vec{b} + \vec{c} = 0\). If \(|\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 5\) then \(|\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}|\) is:

• A. \(20 \)
• B. \(24 \)
• C. \(25 \)
• D. \(30 \)
• E. \(35 \)

Hint:

Use $(\vec{a}+\vec{b}+\vec{c})^2=0$ directly to avoid lengthy calculations.

Answer 1

The Correct Option is C

Concept: If $\vec{a} + \vec{b} + \vec{c} = 0$, then: \[ |\vec{a} + \vec{b} + \vec{c}|^2 = 0 \] \[ \Rightarrow a^2 + b^2 + c^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0 \]

Step 1: Substitute magnitudes.
\[ 3^2 + 4^2 + 5^2 + 2S = 0 \] \[ 9 + 16 + 25 + 2S = 0 \Rightarrow 50 + 2S = 0 \Rightarrow S = -25 \]

Step 2: Take modulus.
\[ |S| = 25 \]