In a triangle PQR- let a-QR- b-RP and c-PQ- If -a-3- and -b-4 and a-c-b-c-a-b-a-a-b- Then the value of -a-b-2 is-

Determining the value of |a×b|^2:Given: a=QR, b=RP and c=PQ. We know that, a+b+c=0, From this we get,a= (b+c)0ex0exc= (a+b)Therefore, ⇒a·(c b)/c·(a b)=|a/a+b|0 ...

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Byju's Answer

Standard XII

Mathematics

AM,GM,HM Inequality

In a triangle...

Question

In a triangle $P Q R$ , let $\vec{a} = \vec{Q R}$ , $\vec{b} = \vec{R P}$ and $\vec{c} = \vec{P Q}$ . If $|\vec{a}| = 3$ , and $|\vec{b}| = 4$ and $\frac{\vec{a} \cdot (\vec{c} - \vec{b})}{\vec{c} \cdot (\vec{a} - \vec{b})} = |\frac{\vec{a}}{\vec{a} + \vec{b}}|$ . Then the value of ${|\vec{a} \times \vec{b}|}^{2}$ is?

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Solution

Determining the value of ${|\vec{a} \times \vec{b}|}^{2}$ :
Given: $\vec{a} = \vec{Q R}$ , $\vec{b} = \vec{R P}$ and $\vec{c} = \vec{P Q}$ .
We know that, $\vec{a} + \vec{b} + \vec{c} = 0$ , From this we get,
$\vec{a} = - (\vec{b} + \vec{c}) \vec{c} = - (\vec{a} + \vec{b})$
Therefore,
$\Rightarrow \frac{\vec{a} \cdot (\vec{c} - \vec{b})}{\vec{c} \cdot (\vec{a} - \vec{b})} = |\frac{\vec{a}}{\vec{a} + \vec{b}}| \Rightarrow \frac{- (\vec{b} + \vec{c}) (\vec{c} - \vec{b})}{- (\vec{a} + \vec{b}) (\vec{a} - \vec{b})} = |\frac{\vec{a}}{\vec{a} + \vec{b}}| \Rightarrow \frac{{|\vec{c}|}^{2} - {|\vec{b}|}^{2}}{{|\vec{a}|}^{2} - {|\vec{b}|}^{2}} = \frac{3}{3 + 4} [∵ \vec{a} = 3, \vec{b} = 4] \Rightarrow \frac{{|\vec{c}|}^{2} - 4^{2}}{3^{2} - 4^{2}} = \frac{3}{7} \Rightarrow \frac{{|\vec{c}|}^{2} - 16}{9 - 16} = \frac{3}{7} \Rightarrow \frac{{|\vec{c}|}^{2} - 16}{- 7} = \frac{3}{7} \Rightarrow {|\vec{c}|}^{2} - 16 = \frac{3}{7} \times - 7 \Rightarrow {|\vec{c}|}^{2} - 16 = - 3 \Rightarrow {|\vec{c}|}^{2} = 16 - 3 \Rightarrow {|\vec{c}|}^{2} = 13$
Now we have $\vec{a} + \vec{b} = - \vec{c}$
$\Rightarrow {|\vec{a} + \vec{b}|}^{2} = {|- \vec{c}|}^{2} \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} \cdot \vec{b} = {|\vec{c}|}^{2} \Rightarrow 9 + 16 + 2 \vec{a} \cdot \vec{b} = 13 [∵ {|\vec{c}|}^{2} = 13] \Rightarrow 25 + 2 \vec{a} \cdot \vec{b} = 13 \Rightarrow 2 \vec{a} \cdot \vec{b} = 13 - 25 \Rightarrow 2 \vec{a} \cdot \vec{b} = - 12 \Rightarrow \vec{a} \cdot \vec{b} = - 6$
Now add the square of the formula
$\Rightarrow {|\vec{a} \cdot \vec{b}|}^{2} + {|\vec{a} \times \vec{b}|}^{2} = a^{2} b^{2} \sin^{2} θ + a^{2} b^{2} \cos^{2} θ \Rightarrow {|\vec{a} \cdot \vec{b}|}^{2} + {|\vec{a} \times \vec{b}|}^{2} = a^{2} b^{2} \Rightarrow {|\vec{a} \cdot \vec{b}|}^{2} + {|\vec{a} \times \vec{b}|}^{2} = {|a|}^{2} \cdot {|b|}^{2} \Rightarrow {|\vec{a} \times \vec{b}|}^{2} + 6^{2} = 3^{2} \times 4^{2} \Rightarrow {|\vec{a} \times \vec{b}|}^{2} + 36 = 9 \times 16 \Rightarrow {|\vec{a} \times \vec{b}|}^{2} = 144 - 36 \Rightarrow {|\vec{a} \times \vec{b}|}^{2} = 108$
Hence, the value of ${|\vec{a} \times \vec{b}|}^{2}$ is $108$ .

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In a triangle PQR, let a→=QR→, b→=RP→ and c→=PQ→. If a→=3, and b→=4 and a→·c→-b→c→·a→-b→=a→a→+b→. Then the value of a→×b→2 is?