Find Dot Product Between Vector CD & Vector K

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Discussion Overview

The discussion revolves around finding the dot product between vector CD, derived from points C(2;6) and D(0;10), and an unknown vector K. The context includes conditions that K must be perpendicular to vector A with components (-3, 2), have the same norm as A, and possess a negative x-component. The discussion involves theoretical reasoning and mathematical calculations.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant calculates the norm of vector CD as \(\sqrt{40}\) and expresses uncertainty about how to proceed without a calculator.
  • Another participant proposes vector K as \(\left<-2,-3\right>\) and vector CD as \(\left<-2,4\right>\), asking for the dot product.
  • A participant inquires about the method used to determine vector K.
  • Two methods are provided for finding vector K:
    • Using the slope of vector A to derive the slope of vector K, leading to the conclusion that K must be \(\left<-2,-3\right>\).
    • Setting up the dot product equation \(\vec{A} \cdot \vec{K} = 0\) and solving for K's components, resulting in the same conclusion.
  • Another participant notes that a vector perpendicular to (a, b) with the same norm can be expressed as (-b, a) or (b, -a), identifying the valid perpendicular vectors as (-2, 4) and (2, -3), with the first having a negative x-component.

Areas of Agreement / Disagreement

Participants present multiple methods for determining vector K, leading to a consensus on its components being \(\left<-2,-3\right>\), but there is no agreement on the final dot product calculation or its implications.

Contextual Notes

Participants rely on geometric interpretations and algebraic manipulations to derive vector K, but the discussion does not resolve the dot product calculation or any potential errors in reasoning.

sp3
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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 can't have a calculator).

Thank you for your help!
 
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vector K would be $\left<-2,-3\right>$

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

can you find the dot product?
 
Thanks for the reply, how did you find vector K?
 
sp3 said:
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 $\vec{K} = \left<x,y\right>$

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

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

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