If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are unit vectors and the maximum value of…

If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are unit vectors and the maximum value of $|\mathbf{a}-\mathbf{b}|^2+|\mathbf{b}-\mathbf{c}|^2+|\mathbf{c}-\mathbf{a}|^2$ is $k$, then $k\left(2 \mathbf{a}^2+3 \mathbf{b}^2-4 \mathbf{c}^2\right)=$
  1. 6
  2. 8
  3. 9
  4. 12

Solution

$\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are unit vectors. $ \begin{gathered} |\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=1 \\ \text { Now, }|\mathbf{a}-\mathbf{b}|^2+|\mathbf{b}-\mathbf{c}|^2+|\mathbf{c}-\mathbf{a}|^2 \\ =2\left(\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2\right)-2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \\ =2(1+1+1)-2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \end{gathered} $
$ |\mathbf{a}-\mathbf{b}|^2+|\mathbf{b}-\mathbf{c}|^2+|\mathbf{c}-\mathbf{a}|^2 $ is maximum when $2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})$ is minimum using. $ |\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) $ We know that, $ \begin{aligned} & \mathbf{a}+\mathbf{b}+\mathbf{c}=0, \text { then } \mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a} \text { is minimum } \\ & 0=1+1+1+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \\ & 2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})=-3 \\ & \text { Substitute in Eq. (i) } \\ & \qquad=6-(-3)=9 \end{aligned} $ Substitute in Eq. (i) Hence, $k=9$ $ =6-(-3)=9 $ Now, $k\left(2 \mathbf{a}^2+3 \mathbf{b}^2-4 \mathbf{c}^2\right)=9(2+3-4)=9(1)=9$

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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