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Logarithmic Inequalities

Let a,b and c...

Question

Let $\vec{a}$ , $\vec{b}$ and $\vec{c}$ be three unit vectors such that ${|\vec{a} - \vec{b}|}^{2} + {|\vec{a} - \vec{c}|}^{2} = 8$ , then ${|\vec{a} + 2 \vec{b}|}^{2} + {|\vec{a} + 2 \vec{c}|}^{2}$ is equal to

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Solution

Step 1: Solving the given equation
$a$ , $b$ , $c$ be three unit vectors.
$\Rightarrow |a| = |b| = |c| = 1$
${|\vec{a} - \vec{b}|}^{2} + {|\vec{a} - \vec{c}|}^{2} = 8 \Rightarrow (\vec{a} - \vec{b}) . (\vec{a} - \vec{b}) + (\vec{a} - \vec{c}) . (\vec{a} - \vec{c}) = 8 [∵ \vec{a} . \vec{a} = {|a|}^{2}] \Rightarrow \vec{a} . \vec{a} + (\vec{a} . - \vec{b}) + (- \vec{b} . \overset{⇀}{a}) + (- \vec{b} . - \vec{b}) + \vec{a} . \vec{a} + (\vec{a} . - \vec{c}) + (- \vec{c} . \vec{a}) + (- \vec{c} . - \vec{c}) = 8 \Rightarrow {|\vec{a}|}^{2} - 2 (\vec{a} . \vec{b}) + {|\vec{b}|}^{2} + {|\overset{⇀}{a}|}^{2} - 2 (\vec{a} . \vec{c}) + {|\vec{c}|}^{2} = 8 \Rightarrow 1 - 2 (\vec{a} . \vec{b}) + 1 + 1 - 2 (\vec{a} . \vec{c}) + 1 = 8 \Rightarrow 4 - 2 [(\vec{a} . \vec{b}) + (\vec{a} . \vec{c})] = 8 \Rightarrow - 2 [(\vec{a} . \vec{b}) + (\vec{a} . \vec{c})] = 8 - 4 \Rightarrow [(\vec{a} . \vec{b}) + (\vec{a} . \vec{c})] = - 2$
Step 2: Finding ${|\vec{a} + 2 \vec{b}|}^{2} + {|\vec{a} + 2 \vec{c}|}^{2}$
${|\vec{a} + 2 \vec{b}|}^{2} + {|\vec{a} + 2 \vec{c}|}^{2} = {|a|}^{2} + 4 (\vec{a} . \vec{b}) + 4 {|\vec{b}|}^{2} + {|a|}^{2} + 4 (\vec{a} . \vec{c}) + 4 {|\vec{c}|}^{2} [∵ {(a + b)}^{2} = a^{2} + 2 a b + b^{2}] = 1 + 4 (\vec{a} . \vec{b}) + 4.1 + 1 + 4 (\vec{a} . \vec{c}) + 4.1 [∵ |a| = 1] = 10 + 4 [(\vec{a} . \vec{b}) + (\vec{a} . \vec{c})] = 10 - 8 [∵ (\vec{a} . \vec{b}) + (\vec{a} . \vec{c}) = - 2] = 2$
Hence, the value of ${|\vec{a} + 2 \vec{b}|}^{2} + {|\vec{a} + 2 \vec{c}|}^{2}$ is equal to $2$ .

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