If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are vectors of equal magnitude such that $(\mathbf{a},…

1. $\frac{3}{2}$
2. $-\frac{3}{2}$
3. $\frac{1}{2}$
4. $-\frac{1}{2}$

Solution

We have, $\cos \alpha=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}, \cos \beta=\frac{\mathbf{b} \cdot \mathbf{c}}{|\mathbf{b}||\mathbf{c}|}$ and $\cos \gamma=\frac{\mathbf{c} \cdot \mathbf{a}}{|\mathbf{c}||\mathbf{a}|}$ Also, it is given that $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=\lambda$ (say), where $\lambda>0$. Then, $\cos \alpha=\frac{\mathbf{a} \cdot \mathbf{b}}{\lambda^2} ; \cos \beta=\frac{\mathbf{b} \cdot \mathbf{c}}{\lambda^2}$ and $\cos \gamma=\frac{\mathbf{c} \cdot \mathbf{a}}{\lambda^2}$ $\Rightarrow \cos \alpha+\cos \beta+\cos \gamma=\frac{1}{\lambda^2}$
We know that, $ \begin{aligned} & |\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=(\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot(\mathbf{a}+\mathbf{b}+\mathbf{c}) \\ = & {\left[|\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})\right] } \\ \Rightarrow & {\left[\lambda^2+\lambda^2+\lambda^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})\right] \geq 0 } \\ & \quad\left[\because|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2 \geq 0\right] \\ \Rightarrow & 2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \geq-3 \lambda^2 \end{aligned} $
Now, from Eqs. (i) and (ii), we get $ \begin{aligned} \cos \alpha+\cos \beta+\cos \gamma & \geq \frac{1}{\lambda^2}\left(-\frac{3}{2} \lambda^2\right) \\ & =-\frac{3}{2} \end{aligned} $ Thus, the minimum value of $\cos \alpha+\cos \beta+\cos \gamma$ is $-\frac{3}{2}$

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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