- Thread starter
- #1

Thank you for your help!

- Thread starter sp3
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- Thread starter
- #1

Thank you for your help!

vector CD would be $\left<-2,4\right>$

can you find the dot product?

- Thread starter
- #3

two ways ...Thanks for the reply, how did you find vector K?

1. $\vec{A} = \left<-3,2 \right>$ has slope $m = -\dfrac{2}{3} \implies \vec{K}$ has slope $m_{\perp} = \dfrac{3}{2}$.

since $\vec{K}$ has a negative x component, then so does its y-component ... same magnitude means $\vec{K} = \left<x,y \right> = \left<-2,-3 \right>$

2. let $\vec{K} = \left<x,y\right>$

$\vec{A} \cdot \vec{K} = -3x + 2y = 0 \implies y = \dfrac{3}{2} x$

$|\vec{A}| = |\vec{K}| \implies \sqrt{(-3)^2 + 2^2} = \sqrt{x^2+y^2} \implies x^2+y^2 = 13 \implies x^2 + \dfrac{9}{4} x^2 = 13 \implies \dfrac{13}{4} x^2 = 13 \implies x = \pm 2$

$x < 0 \implies x = -2 \implies y = -3$

- Jan 29, 2012

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