Question #d62c9

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Question #d62c9

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Answer 1 · 2017-03-30T10:15:56

$b = \pm \sqrt{15}$ $b=+-sqrt15$

Explanation:

$the magnitude of < 2, 6 > is$ $" the magnitude of " <2,6>" is"$

$\sqrt{2^{2} + 6^{2}} = \sqrt{40}$ $sqrt(2^2+6^2)=sqrt40$

$require < - 5, b > to have the same magnitude$ $"require " <-5,b>" to have the same magnitude"$

$\Rightarrow \sqrt{{(- 5)}^{2} + b^{2}} = \sqrt{40}$ $rArrsqrt((-5)^2+b^2)=sqrt40$

$\Rightarrow \sqrt{25 + b^{2}} = \sqrt{40}$ $rArrsqrt(25+b^2)=sqrt40$

$square both sides$ $color(blue)" square both sides"$

$\Rightarrow 25 + b^{2} = 40 \to b^{2} = 15$ $rArr25+b^2=40tob^2=15$

$\Rightarrow b = \pm \sqrt{15}$ $rArrb=+-sqrt15$

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Question #d62c9

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