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(C) Given that $a, b, c$ 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 \cd

Vedclass · Accepted Answer

$9$

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(C) Given that $a, b, c$ 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,we get:$|a-b|^2+|b-c|^2+|c-a|^2 = 6 - 2(a \cdot b + b \cdot c + c \cdot a) \quad \dots (i)$To maximize this expression,we need to minimize $(a \cdot b + b \cdot c + c \cdot a)$.We know that $|a+b+c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) \ge 0$.Since $|a|=|b|=|c|=1$,we have $1 + 1 + 1 + 2(a \cdot b + b \cdot c + c \cdot a) \ge 0$,which implies $2(a \cdot b + b \cdot c + c \cdot a) \ge -3$.The minimum value of $2(a \cdot b + b \cdot c + c \cdot a)$ is $-3$.Substituting this into equation $(i)$:$k = 6 - (-3) = 9$.Now,we need to calculate $k(2|a|^2+3|b|^2-4|c|^2)$.Since $|a|=|b|=|c|=1$,we have $|a|^2=|b|^2=|c|^2=1$.Thus,$k(2(1)+3(1)-4(1)) = 9(2+3-4) = 9(1) = 9$.

If $a, b, c$ are unit vectors and the maximum value of $|a-b|^2+|b-c|^2+|c-a|^2$ is $k$,then $k(2|a|^2+3|b|^2-4|c|^2)=$

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