If a,b,and c are unit vectors,then |a - b|^2 | vedclass

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(B) Given that $a$,$b$,and $c$ are unit vectors,so $|a| = |b| = |c| = 1$.We know that $|a - b|^2 = |a|^2 + |b|^2 - 2(a \cdot b) = 1 + 1 - 2(a \cdot b)

Vedclass · Accepted Answer

$9$

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(B) Given that $a$,$b$,and $c$ are unit vectors,so $|a| = |b| = |c| = 1$.We know that $|a - b|^2 = |a|^2 + |b|^2 - 2(a \cdot b) = 1 + 1 - 2(a \cdot b) = 2 - 2(a \cdot b)$.Similarly,$|b - c|^2 = 2 - 2(b \cdot c)$ and $|c - a|^2 = 2 - 2(c \cdot a)$.Adding these three expressions:$|a - b|^2 + |b - c|^2 + |c - a|^2 = 6 - 2(a \cdot b + b \cdot c + c \cdot a)$.We also know that $|a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) = 3 + 2(a \cdot b + b \cdot c + c \cdot a)$.Therefore,$2(a \cdot b + b \cdot c + c \cdot a) = |a + b + c|^2 - 3$.Substituting this into the sum:$|a - b|^2 + |b - c|^2 + |c - a|^2 = 6 - (|a + b + c|^2 - 3) = 9 - |a + b + c|^2$.Since $|a + b + c|^2 \ge 0$,the maximum value of the expression is $9 - 0 = 9$.Thus,the expression does not exceed $9$.

If $a$,$b$,and $c$ are unit vectors,then $|a - b|^2 + |b - c|^2 + |c - a|^2$ does not exceed

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