If $\hat{\mathbf{a}}, \hat{\mathbf{b}}$ and $\hat{\mathbf{c}}$ are vectors with magnitudes 2,3 and 4 respectively, then the best upper bound of $|\hat{\mathbf{a}}-\hat{\mathbf{b}}|^2+|\hat{\mathbf{b}}-\hat{\mathbf{c}}|^2+|\hat{\mathbf{c}}-\hat{\mathbf{a}}|^2$ among the given values is

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Aicharya · Accepted Answer

Given, $|\mathrm{a}|=2,|\mathrm{~b}|=3$ and $|\mathrm{c}|=4$ $ \begin{aligned} & \begin{aligned} & 	herefore|a-b|^2+|b-c|^2+|c-a|^2 \ &= a^2+b^2-2 a \cdot b+b^2+c^2-2 b \cdot c \ & \quad+c^2+a^2-2 a \cdot c \ &= 2\left(a^2+b^2+c^2ight)-2(a \cdot b+b \cdot c+c \cdot a) \ & \begin{array}{rl} 2 & 2(a \cdot b+b \cdot c+c \cdot a) \end{array} \ &=2\left(a^2+b^2+c^2ight) \ & \quad-\left(|a-b|^2+|b-c|^2+|c-a|^2ight) \end{aligned} \end{aligned} $ We know, $(a+b+c)^2 \geq 0$ $ \begin{aligned} & 	herefore \quad a^2+b^2+c^2+2(a \cdot b+b \cdot c+c \cdot a) \geq 0 \ & \Rightarrow \quad a^2+b^2+c^2+2\left(a^2+b^2+c^2ight) \ & -\left(|a-b|^2+|b-c|^2+|c-a|^2ight) \geq 0 \ & \Rightarrow|\mathrm{a}-\mathrm{b}|^2+|\mathrm{b}-\mathrm{c}|^2+|\mathrm{c}-\mathrm{a}|^2 \ & \leq 3\left(a^2+b^2+c^2ight) \ & \Rightarrow|a-b|^2+|b-c|^2+|c-a|^2 \ & \leq 3\left(2^2+3^2+4^2ight) \ & \Rightarrow|a-b|^2+|b-c|^2+|c-a|^2 \ & \leq 3(4+9+16) \ & \Rightarrow|a-b|^2+|b-c|^2+|c-a|^2 \leq 87 \ & \end{aligned} $

If $\hat{\mathbf{a}}, \hat{\mathbf{b}}$ and $\hat{\mathbf{c}}$ are vectors with magnitudes 2,3 and 4…

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