Question

If \( \vec{a}, \vec{b} \) and \( \vec{c} \) are unit vectors, then prove that \( |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 \leq 9 \).

Hint:

Problems involving sums of vector differences squared often rely on the expansion of \( |\Sigma \vec{v}|^2 \geq 0 \).
Remember that for unit vectors, the dot product \( \vec{u}\cdot\vec{v} = \cos \theta \), which ranges from -1 to 1.

Answer 1

Concept:
• Unit vectors: \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \).
• Expansion property: \( |\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2\vec{u}\cdot\vec{v} \).
• Sum squared property: \( |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \).
• Magnitudes are non-negative: \( |\vec{v}|^2 \geq 0 \).

Step 1: Expand the given expression using the dot product property
Let \( S = |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 \). Using the property \( |\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2\vec{u}\cdot\vec{v} \): \[ S = (|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b}) + (|\vec{b}|^2 + |\vec{c}|^2 - 2\vec{b}\cdot\vec{c}) + (|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c}\cdot\vec{a}) \] Since they are unit vectors, \( |\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1 \): \[ S = (1 + 1 - 2\vec{a}\cdot\vec{b}) + (1 + 1 - 2\vec{b}\cdot\vec{c}) + (1 + 1 - 2\vec{c}\cdot\vec{a}) \] \[ S = 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \]

Step 2: Find an inequality for the dot product sum
Consider the square of the sum of the three unit vectors: \[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \] Since \( |\vec{a} + \vec{b} + \vec{c}|^2 \geq 0 \): \[ 1 + 1 + 1 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \geq 0 \] \[ 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \geq -3 \] Multiplying by \(-1\) (reverses the inequality): \[ -2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \leq 3 \]

Step 3: Substitute back into the expression for \( S \)
From Step 1, \( S = 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \). Using the result from
Step 2: \[ S \leq 6 + 3 \] \[ S \leq 9 \] Hence proved.