Let a- and b- are two non-zero- non-collinear vectors -a-1 such that vectors 3a-b- and 2b-a-b- a- represents two sides of a triangle- If area of triangle is 3-4-b-4-4 then the value of -b-2 is

V_1 = 3(a×b) v_2 = 2 (b (a. b)a) v_1. v_2 = 0 |v_1| = 3 |a||b| sinθ |v_2| = 2 √((b (a. b) a)^2) = 2 √(|b|^2 2 (a. b)^2 + (a . b)^2) = 2 √(|b|^2 (a. b)^2 ...

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Byju's Answer

Standard XII

Mathematics

Complex Numbers

Let a⃗ and b⃗...

Question

Let $\to a$ and $\to b$ are two non-zero, non-collinear vectors $(| \to a | = 1)$ such that vectors $3 (\to a \times \to b)$ and $2 (\to b - (\to a . \to b) \to a)$ represents two sides of a triangle. If area of triangle is $\frac{3}{4} (| \to b |^{4} + 4)$ then the value of $| \to b |^{2}$ is

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Solution

${\to v}_{1} = 3 (\to a \times \to b)$
${\to v}_{2} = 2 (\to b - (\to a . \to b) \to a)$
${\to v}_{1} . {\to v}_{2} = 0$
$| {\to v}_{1} | = 3 | \to a | | \to b | sin θ$
$| {\to v}_{2} | = 2 \sqrt{(\to b - (\to a . \to b) \to a)^{2}}$
$= 2 \sqrt{| \to b |^{2} - 2 (\to a . \to b)^{2} + (\to a . \to b)^{2}}$
$= 2 \sqrt{| b |^{2} - (\to a . \to b)^{2}}$
$= 2 | \to b | sin θ$
$A r e a = \frac{1}{2} . 3 | \to b | sin θ . 2 | \to b | sin θ$
$= \frac{3}{4} (| \to b |^{4} + 4 |)$
$\Rightarrow | \to b |^{2} {sin}^{2} θ = \frac{1}{4} (| \to b |^{4} + 4)$
$4 {sin}^{2} θ = | \to b |^{2} + \frac{4}{| \to b |^{2}} \geq 4$
using AM $\geq$ GM
So equality holds because ${sin}^{2} θ = 1$
$| \to b |^{4} = 4$
$| \to b |^{2} = 2$

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Let →a and →b are two non-zero, non-collinear vectors (|→a|=1) such that vectors 3(→a×→b) and 2(→b−(→a.→b)→a) represents two sides of a triangle. If area of triangle is 34(|→b|4+4) then the value of |→b|2 is

Let $\to a$ and $\to b$ are two non-zero, non-collinear vectors $(| \to a | = 1)$ such that vectors $3 (\to a \times \to b)$ and $2 (\to b - (\to a . \to b) \to a)$ represents two sides of a triangle. If area of triangle is $\frac{3}{4} (| \to b |^{4} + 4)$ then the value of $| \to b |^{2}$ is