# finding dot product

#### sp3

##### New member
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).

#### skeeter

##### Well-known member
MHB Math Helper
vector K would be $\left<-2,-3\right>$

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

can you find the dot product?

#### sp3

##### New member
Thanks for the reply, how did you find vector K?

#### skeeter

##### Well-known member
MHB Math Helper
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 $\vec{K} = \left<x,y\right>$

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

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

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

#### HallsofIvy

##### Well-known member
MHB Math Helper
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.