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Expansion of (a ± b ± c)²

If a, b, c ...

Question

If $a, b, c$ and $A, B, C \in R - (0}$ such that $a A + b B + c C + \sqrt{(a^{2} + b^{2} + c^{2}) (A^{2} + B^{2} + C^{2})} = 0$ , then value of $\frac{a B}{b A} + \frac{b C}{c B} + \frac{c A}{a C}$ is

A

3

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B

4

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C

5

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D

6

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Solution

The correct option is A $3$
$a A + b B + c C + \sqrt{(a^{2} + b^{2} + c^{2}) (A^{2} + B^{2} + C^{2})} = 0$

$\Rightarrow a A + b B + c C = - \sqrt{(a^{2} + b^{2} + c^{2}) (A^{2} + B^{2} + C^{2})}$

squaring both sides, we get
$\Rightarrow {(a A + b B + c C)}^{2} = {(- \sqrt{(a^{2} + b^{2} + c^{2}) (A^{2} + B^{2} + C^{2})})}^{2}$

$\Rightarrow {(a A + b B + c C)}^{2} = (a^{2} + b^{2} + c^{2}) (A^{2} + B^{2} + C^{2})$

$\Rightarrow a^{2} A^{2} + b^{2} B^{2} + c^{2} C^{2} + 2 A b B + 2 b B c C + 2 c C a A = a^{2} (A^{2} + B^{2} + C^{2}) + b^{2} (A^{2} + B^{2} + C^{2}) + c^{2}$
$(A^{2} + B^{2} + C^{2})$

$\Rightarrow a^{2} A^{2} + b^{2} B^{2} + c^{2} C^{2} + 2 A b B + 2 b B c C + 2 c C a A = a^{2} A^{2} + a^{2} B^{2} + a^{2} C^{2} + b^{2} A^{2} + b^{2} B^{2} + b^{2} C^{2} +$
$c^{2} A^{2} + c^{2} B^{2} + c^{2} C^{2}$

$\Rightarrow 2 A a b B + 2 b B c C + 2 c C a A = a^{2} B^{2} + a^{2} C^{2} + b^{2} A^{2} + b^{2} C^{2} + c^{2} A^{2} + c^{2} B^{2}$

$\Rightarrow 2 A a b B + 2 b B c C + 2 c C a A = a^{2} B^{2} + b^{2} A^{2} + a^{2} C^{2} + c^{2} A^{2} + b^{2} C^{2} + c^{2} B^{2}$

$\Rightarrow 0 = a^{2} B^{2} - 2 A a b B + b^{2} A^{2} + a^{2} C^{2} - 2 c C a A + c^{2} A^{2} + b^{2} C^{2} 2 b B c C + c^{2} B^{2}$

$\Rightarrow 0 = {(a B - b A)}^{2} + {(a C - c A)}^{2} + {(b C - B c)}^{2}$

$\Rightarrow {(a B - b A)}^{2} = 0, {(a C - c A)}^{2} = 0, {(b C - B c)}^{2} = 0$

$\Rightarrow a B - b A = 0, a C - c A = 0, b C - B c = 0$

$\Rightarrow a B = b A, a C = c A, b C = B c$

Substituting the values of $a B = b A, a C = c A, b C = B c$ in $\frac{a B}{b A} + \frac{b C}{c B} + \frac{c A}{a C}$ we get

The value of $\frac{a B}{b A} + \frac{b C}{c B} + \frac{c A}{a C} = \frac{b A}{b A} + \frac{c B}{c B} + \frac{c A}{c A} = 1 + 1 + 1 = 3$

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Similar questions

∣ ∣ ∣ ∣ \begin{matrix} b + c & c + a & a + b c + a & a + b & b + c a + b & b + c & c + a \end{matrix} ∣ ∣ ∣ ∣ == 2 (a + b + c) (a b + b c + c a - a^{2} - b^{2} - c^{2})

If a + b + c = 0, then prove the following

(a) (b + c) (b − c) + a(a + 2b) = 0

(b) a(a² − bc) + b(b² − ca) + c(c² − ab) = 0

(c) a(b² + c²) + b(c² + a²) + c(a² + b²) = −3abc

(d)

Q. Prove the following identity:

\frac{a {(b - c)}^{2}}{(c - a) (a - b)} + \frac{b {(c - a)}^{2}}{(a - b) (b - c)} + \frac{c {(a - b)}^{2}}{(b - c) (c - a)} = a + b + c

△_{1}, △_{2}

be the areas of two triangles with vertices

(b, c), (c, a), (a, b)

(a c - b^{2}, a b - c^{2}), (b a - c^{2}, b c - a^{2}), (c b - a^{2}, c a - b^{2})

\frac{△_{1}}{△_{2}} = (a + b + c)^{2}

Q. If a² + b² + c² − ab − bc − ca =0, then

(a) a + b + c

(b) b + c = a

(c) c + a = b

(d) a = b = c

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