Vectors 1 Question 32
33. If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are unit vectors satisfying $|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|^{2}+|\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}|^{2}+|\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}|^{2}=9$, then
$|2 \overrightarrow{\mathbf{a}}+5 \overrightarrow{\mathbf{b}}+5 \overrightarrow{\mathbf{c}}|$ is equal to
(2012)
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Answer:
Correct Answer: 33. (3)
Solution:
- PLAN If $a, b, c$ are any three vectors
Then $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}|^{2} \geq 0$
$\Rightarrow|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}+|\overrightarrow{\mathbf{c}}|^{2}+2(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \geq 0$
$\therefore \quad \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}} \geq \frac{-1}{2}\left(|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}+|\overrightarrow{\mathbf{c}}|^{2}\right)$
Given, $|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|^{2}+|\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}|^{2}+|\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}|^{2}=9$
$\Rightarrow|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}-2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+|\overrightarrow{\mathbf{b}}|^{2}+|\overrightarrow{\mathbf{c}}|^{2}-2 \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+|\overrightarrow{\mathbf{c}}|^{2}+|\overrightarrow{\mathbf{a}}|^{2}$
$ -2 \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}=9 $
$ \begin{array}{ll} \Rightarrow 6-2(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}})=9 \quad[\because|\overrightarrow{\mathbf{a}}|=|\overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{c}}|=1] \\ \Rightarrow \quad \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}=-\frac{3}{2} \end{array} $
Also, $\quad \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}} \geq \frac{-1}{2}\left(|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}+|\overrightarrow{\mathbf{c}}|^{2}\right)$
$ \geq-\frac{3}{2} $
From Eqs. (i) and (ii), $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}|=0$
as $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}$ is minimum when $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}|=0$
$ \begin{array}{lc} \Rightarrow & \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=0 \\ \therefore & |2 \mathbf{a}+5 \mathbf{b}+5 \mathbf{c}|=|2 \mathbf{a}+5(\mathbf{b}+\mathbf{c})|=|2 \mathbf{a}-5 \mathbf{a}|=3 \end{array} $
