If a-b and c are unit vectors satisfying -a-b-2-b-c-2-c-a-2-9- then -2a-5b-5c-

Given : |a b|^2+|b c|^2+|c a|^2=9 ⇒ 2(|a|^2+|b|^2+|c|^2) 2(a·b+b·c+c·a)=9 nbsp; Since a,b and c are unit vectors ⇒ 6 2(a·b+b·c+c·a)=9 ⇒ nbsp;a·b+b·c+c·a= 32 ...

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Standard IX

Mathematics

Triangle Inequality

If a,b and c ...

Question

If $\to a, \to b$ and $\to c$ are unit vectors satisfying $| \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2} = 9,$ then $| 2 \to a + 5 \to b + 5 \to c | =$

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Solution

Given :
$| \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2} = 9 \Rightarrow 2 (| \to a |^{2} + | \to b |^{2} + | \to c |^{2}) - 2 (\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a) = 9$
Since $\to a, \to b$ and $\to c$ are unit vectors
$\Rightarrow 6 - 2 (\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a) = 9 \Rightarrow \to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a = - \frac{3}{2} \dots (i)$
As, $| \to a + \to b + \to c |^{2} \geq 0$
$\Rightarrow | \to a |^{2} + | \to b |^{2} + | \to c |^{2} + 2 (\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a) \geq 0 \Rightarrow \to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a \geq - \frac{3}{2}$
Since, $\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a = - \frac{3}{2}$
So, $| \to a + \to b + \to c | = 0$
$\Rightarrow \to a + \to b + \to c = \to 0 \dots (i i)$
$\Rightarrow | 2 \to a + 5 \to b + 5 \to c | = | 2 \to a + 5 (- \to a) | = 3 | \to a | = 3$

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Triangle Inequality

Standard IX Mathematics

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If →a,→b and →c are unit vectors satisfying |→a−→b|2+|→b−→c|2+|→c−→a|2=9, then |2→a+5→b+5→c|=

If $\to a, \to b$ and $\to c$ are unit vectors satisfying $| \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2} = 9,$ then $| 2 \to a + 5 \to b + 5 \to c | =$