SATHEE: Vectors 1 Question 32

Vectors 1 Question 32

33. If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $| \vec{a} - \vec{b} |^{2} + | \vec{b} - \vec{c} |^{2} + | \vec{c} - \vec{a} |^{2} = 9$ , then

$| 2 \vec{a} + 5 \vec{b} + 5 \vec{c} |$ is equal to

(2012)

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Solution:

PLAN If $a, b, c$ are any three vectors

Then $| \vec{a} + \vec{b} + \vec{c} |^{2} \geq 0$

$\Rightarrow | \vec{a} |^{2} + | \vec{b} |^{2} + | \vec{c} |^{2} + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0$

$∴ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \geq \frac{- 1}{2} (| \vec{a} |^{2} + | \vec{b} |^{2} + | \vec{c} |^{2})$

Given, $| \vec{a} - \vec{b} |^{2} + | \vec{b} - \vec{c} |^{2} + | \vec{c} - \vec{a} |^{2} = 9$

$\Rightarrow | \vec{a} |^{2} + | \vec{b} |^{2} - 2 \vec{a} \cdot \vec{b} + | \vec{b} |^{2} + | \vec{c} |^{2} - 2 \vec{b} \cdot \vec{c} + | \vec{c} |^{2} + | \vec{a} |^{2}$

$- 2 \vec{c} \cdot \vec{a} = 9$

$\begin{array}{ll} \Rightarrow & 6 - 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9 [∵ | \vec{a} | = | \vec{b} | = | \vec{c} | = 1] \\ \Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = - \frac{3}{2} \end{array}$

Also, $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \geq \frac{- 1}{2} (| \vec{a} |^{2} + | \vec{b} |^{2} + | \vec{c} |^{2})$