# Thread: finding dot product

1. Hi! I'm given 2 points C(2;6) and D(0;10), a vector A with its components = (-3, 2). I'm asked to find the dot product between vector CD and an unknown vector K, knowing that K is perpendicular to A, same norm as A and with a negative x-component. I know that perpendicular means the dot product=0 and vector CD has a norm \sqrt{40} if i calculate it, but I have no clue how to solve it (we cant have a calculator).

Thank you for your help!

2.

3. vector K would be $\left<-2,-3\right>$

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

can you find the dot product?

4. Thread Author
Thanks for the reply, how did you find vector K?

5. Originally Posted by sp3
Thanks for the reply, how did you find vector K?
two ways ...

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

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

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

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

6. Equivalently, one vector perpendicular perpendicular to (a, b) with the same norm is (-b, a), another is (b, -a). Here, K= (-3, 2) so those two perpendicular vector are (-2, 4) and (2, -3). The first has x component negative.

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