Given-a-b-c-and-d-are-reals-numbers-such-that-a-2-b-2-10-c-2-d-2-10-ab-cd-0-Find-ac-bd-

Question Number 212499 by efronzo1 last updated on 15/Oct/24

Given a, b, c and d are reals

numbers such that

{\begin{cases} a^{2} + b^{2} = 10 \\ c^{2} + d^{2} = 10 \\ ab + cd = 0 \end{cases}

Find ac + bd .

Answered by ajfour last updated on 15/Oct/24

{(a + b)}^{2} = 10 + 2 a b

{(c + d)}^{2} = 10 + 2 c d

l e t \frac{d}{b} = - \frac{c}{a} = k

c^{2} + d^{2} = k^{2} (a^{2} + b^{2}) = 10

\Rightarrow k^{2} = 1 a s (a^{2} + b^{2}) = 10

s a y k = 1 \Rightarrow

d = b, c = - a

a c + b d = - a^{2} + b^{2} = 10 - 2 a^{2} = 2 b^{2} - 10

i t h i n k i t d e p e n d s o n c h o i c e o f

a^{2}, b^{2} w i t h t h e c o n d i t i o n a^{2} + b^{2} = 10

Answered by Ghisom last updated on 16/Oct/24

a b + c d = 0 \Rightarrow d = - \frac{a b}{c}

c^{2} + d^{2} = 10 \Rightarrow b^{2} = \frac{c^{2} (10 - c^{2})}{a^{2}}

a^{2} + b^{2} = 10 \Rightarrow a^{4} - 10 a^{2} - c^{4} + 10 c^{2} = 0

\Rightarrow

c = \pm a \lor c = \pm \sqrt{10 - a^{2}}

\Rightarrow a c + b d = \pm 2 (a^{2} - 5) \lor a c + b d = 0

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