Question

If \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors, then \( |\vec{a}-\vec{b}|^2 + |\vec{b}-\vec{c}|^2 + |\vec{c}-\vec{a}|^2 \) does not exceed

• A. 4
• B. 9
• C. 8
• D. 6

Hint:

Vector Identities. Use \(|\vec{x-\vec{y|^2 = |\vec{x|^2 + |\vec{y|^2 - 2\vec{x\cdot\vec{y\) and \(|\vec{x+\vec{y+\vec{z|^2 = |\vec{x|^2 + |\vec{y|^2 + |\vec{z|^2 + 2(\vec{x\cdot\vec{y + \vec{y\cdot\vec{z + \vec{z\cdot\vec{x) \ge 0\). For unit vectors, \(|\vec{x|=1\).

Answer 1

The Correct Option is B

We use the property \( |\vec{x}|^2 = \vec{x} \cdot \vec{x} \).
$$ |\vec{a}-\vec{b}|^2 = (\vec{a}-\vec{b}) \cdot (\vec{a}-\vec{b}) = \vec{a}\cdot\vec{a} - 2\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{b} $$ Since \(\vec{a}, \vec{b}, \vec{c}\) are unit vectors, \(|\vec{a}|=|\vec{b}|=|\vec{c}|=1\), and \(\vec{a}\cdot\vec{a} = |\vec{a}|^2 = 1\), etc.
$$ |\vec{a}-\vec{b}|^2 = 1 - 2\vec{a}\cdot\vec{b} + 1 = 2 - 2\vec{a}\cdot\vec{b} $$ Similarly, $$ |\vec{b}-\vec{c}|^2 = 2 - 2\vec{b}\cdot\vec{c} $$ $$ |\vec{c}-\vec{a}|^2 = 2 - 2\vec{c}\cdot\vec{a} $$ Summing these three expressions: $$ S = |\vec{a}-\vec{b}|^2 + |\vec{b}-\vec{c}|^2 + |\vec{c}-\vec{a}|^2 = (2 - 2\vec{a}\cdot\vec{b}) + (2 - 2\vec{b}\cdot\vec{c}) + (2 - 2\vec{c}\cdot\vec{a}) $$ $$ S = 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) $$ Now consider the magnitude squared of the sum of the vectors: $$ |\vec{a}+\vec{b}+\vec{c}|^2 = (\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) $$ $$ = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) $$ $$ |\vec{a}+\vec{b}+\vec{c}|^2 = 1 + 1 + 1 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) $$ $$ |\vec{a}+\vec{b}+\vec{c}|^2 = 3 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) $$ Since the square of a magnitude must be non-negative: $$ |\vec{a}+\vec{b}+\vec{c}|^2 \ge 0 $$ $$ 3 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \ge 0 $$ $$ 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \ge -3 $$ Now substitute this back into the expression for S: $$ S = 6 - [2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})] $$ Since \(2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \ge -3\), then \(-[2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})] \le 3\).
Therefore, $$ S \le 6 + 3 = 9 $$ The sum does not exceed 9.