Let xyz be a triangle and a-YZ-b-ZX-c-XY- If -a-6-c-14-a-b-30- then the values of -a-b-b-c-2100 is

Since a+b+c=XY+YZ+ZX ∴a+b+c=0 ⇒ |a+b|=|c| ⇒ 36+|b|^2+60=196 ⇒ |b|=10 and a·b=|a||b|cosθ ⇒ 30=60cosθ ⇒θ =60^∘ Now, |a×b+b×c|^2 =|a×b b× (a+b)|^2 =4|a×b|^2 =4|a|^ ...

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Byju's Answer

Standard XII

Mathematics

Condition for Three Vectors to Form a Triangle

Let xyz be a ...

Question

Let $△ x y z$ be a triangle and $\to a = - - \to Y Z, \to b = - - \to Z X, \to c = - - \to X Y$ . If $| \to a | = 6, | \to c | = 14, \to a \cdot \to b = 30,$ then the values of $\frac{| \to a \times \to b + \to b \times \to c |^{2}}{100}$ is

108.00

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108

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108.0

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Solution

Since $\to a + \to b + \to c = - - \to X Y + - - \to Y Z + - - \to Z X$
$∴ \to a + \to b + \to c = \to 0$
$\Rightarrow | \to a + \to b | = | \to c |$
$\Rightarrow 36 + | \to b |^{2} + 60 = 196$
$\Rightarrow | \to b | = 10$
and $\to a \cdot \to b = | \to a | | \to b | cos θ$
$\Rightarrow 30 = 60 cos θ$
$\Rightarrow θ = 60^{\circ}$
Now, $| \to a \times \to b + \to b \times \to c |^{2}$
$= | \to a \times \to b - \to b \times (\to a + \to b) |^{2}$
$= 4 | \to a \times \to b |^{2}$
$= 4 | \to a |^{2} | \to b |^{2} {sin}^{2} θ$
$= 4 \times 36 \times 100 \times \frac{3}{4}$
$= 10800$
Hence the value of $\frac{| \to a \times \to b + \to b \times \to c |^{2}}{100}$ is $108.$

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Let △xyz be a triangle and →a=−−→YZ,→b=−−→ZX,→c=−−→XY. If |→a|=6,|→c|=14,→a⋅→b=30, then the values of |→a×→b+→b×→c|2100 is

Let $△ x y z$ be a triangle and $\to a = - - \to Y Z, \to b = - - \to Z X, \to c = - - \to X Y$ . If $| \to a | = 6, | \to c | = 14, \to a \cdot \to b = 30,$ then the values of $\frac{| \to a \times \to b + \to b \times \to c |^{2}}{100}$ is