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:

The expression \( 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \) frequently appears in unit vector problems. Relate it back to the expansion of \( |\vec{a} + \vec{b} + \vec{c}|^2 \geq 0 \) to find the bounds for the sum.

Answer 1

Concept:
• A unit vector has a magnitude of 1, i.e., \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \).
• The squared magnitude of a vector difference is \( |\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y}) \).
• For any vector \( \vec{v} \), the squared magnitude \( |\vec{v}|^2 \geq 0 \).

Step 1: Expand each squared term using vector properties
Given that \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are unit vectors, we have \( |\vec{a}|^2 = 1 \), \( |\vec{b}|^2 = 1 \), and \( |\vec{c}|^2 = 1 \).
Expanding the terms individually:
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) = 1 + 1 - 2(\vec{a} \cdot \vec{b}) = 2 - 2(\vec{a} \cdot \vec{b}) \]
\[ |\vec{b} - \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 - 2(\vec{b} \cdot \vec{c}) = 1 + 1 - 2(\vec{b} \cdot \vec{c}) = 2 - 2(\vec{b} \cdot \vec{c}) \]
\[ |\vec{c} - \vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{c} \cdot \vec{a}) = 1 + 1 - 2(\vec{c} \cdot \vec{a}) = 2 - 2(\vec{c} \cdot \vec{a}) \]

Step 2: Sum the expanded terms
Let \( S = |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 \).
Substituting the expressions from
Step 1:
\[ S = [2 - 2(\vec{a} \cdot \vec{b})] + [2 - 2(\vec{b} \cdot \vec{c})] + [2 - 2(\vec{c} \cdot \vec{a})] \]
\[ S = 6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]

Step 3: Use the property of the sum of vectors to establish the inequality
Consider the vector \( \vec{a} + \vec{b} + \vec{c} \). Its squared magnitude must be non-negative:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 \geq 0 \]
Expanding the expression:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0 \]
Substitute the unit vector values (1+1+1=3):
\[ 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0 \]
Rearranging the terms:
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq -3 \]

Step 4: Apply the inequality to the sum \( S \)
Multiply the inequality from Step 3 by \( -1 \), noting that the inequality sign reverses:
\[ -2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \leq 3 \]
Add 6 to both sides of the inequality:
\[ 6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \leq 6 + 3 \]
\[ S \leq 9 \]
Hence, \( |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 \leq 9 \).