Question

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

Accepted Answer

Given, $\vec{a}, \vec{b}, \vec{c}$ are unit vectors.

$∴ |\vec{a}| = |\vec{b}| = |\vec{c}| = 1$

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

$\Rightarrow (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) + (\vec{b} - \vec{c}) \cdot (\vec{b} - \vec{c}) + (\vec{c} - \vec{a}) \cdot (\vec{c} - \vec{a}) = 9$

$\Rightarrow ({|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \vec{a} \cdot \vec{b}) + ({|\vec{b}|}^{2} + {|\vec{c}|}^{2} - 2 \vec{b} \cdot \vec{c}) + ({|\vec{c}|}^{2} + {|\vec{a}|}^{2} - 2 \vec{c} \cdot \vec{a}) = 9$

$\Rightarrow 6 - 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$

$\Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = \frac{- 3}{2} \dots (i)$

Let, $\vec{P} = \vec{a} + \vec{b} + \vec{c}$

$\Rightarrow |\vec{P}| = |\vec{a} + \vec{b} + \vec{c}|$

$\Rightarrow {|\vec{P}|}^{2} = {|\vec{a} + \vec{b} + \vec{c}|}^{2}$

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

Put, the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ from equation $(i)$ into the above equation

$\Rightarrow {|\vec{P}|}^{2} = 1 + 1 + 1 + 2 (\frac{- 3}{2})$

$\Rightarrow {|\vec{P}|}^{2} = 0$

$\Rightarrow \vec{P} = \vec{0}$

$\Rightarrow \vec{a} + \vec{b} + \vec{c} = \vec{0} \dots (ii)$

From equations $(i)$ and $(ii)$ , it is clear that $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = \frac{- 1}{2}$

Let, $λ = |2 \vec{a} + 5 \vec{b} + 5 \vec{c}|$

$\Rightarrow λ^{2} = {|2 \vec{a} + 5 \vec{b} + 5 \vec{c}|}^{2}$

$\Rightarrow λ^{2} = (2 \vec{a} + 5 \vec{b} + 5 \vec{c}) \cdot (2 \vec{a} + 5 \vec{b} + 5 \vec{c})$

$\Rightarrow λ^{2} = 4 {|\vec{a}|}^{2} + 25 {|\vec{b}|}^{2} + 25 {|\vec{c}|}^{2} + 20 \vec{a} \cdot \vec{b} + 50 \vec{b} \cdot \vec{c} + 20 \vec{c} \cdot \vec{a}$

$\Rightarrow λ^{2} = 54 + 20 (\frac{- 1}{2}) + 50 (\frac{- 1}{2}) + 20 (\frac{- 1}{2})$

$\Rightarrow λ^{2} = 9$

$\Rightarrow λ = 3$

Hence, $|2 \vec{a} + 5 \vec{b} + 5 \vec{c}| = 3$ .

Exercise-3

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Questions from Exercise-3 with Hints & Solutions

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