
#1
June 2nd, 2019,
16:49
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 xcomponent. 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!

June 2nd, 2019 16:49
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MHB Journeyman
#2
June 2nd, 2019,
17:52
vector K would be $\left<2,3\right>$
vector CD would be $\left<2,4\right>$
can you find the dot product?

#3
June 7th, 2019,
12:31
Thread Author
Thanks for the reply, how did you find vector K?

MHB Journeyman
#4
June 7th, 2019,
15:31
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 ycomponent ... same magnitude means $\vec{K} = \left<x,y \right> = \left<2,3 \right>$
2. let $\vec{K} = \left<x,y\right>$
$\vec{A} \cdot \vec{K} = 3x + 2y = 0 \implies y = \dfrac{3}{2} x$
$\vec{A} = \vec{K} \implies \sqrt{(3)^2 + 2^2} = \sqrt{x^2+y^2} \implies x^2+y^2 = 13 \implies x^2 + \dfrac{9}{4} x^2 = 13 \implies \dfrac{13}{4} x^2 = 13 \implies x = \pm 2$
$x < 0 \implies x = 2 \implies y = 3$

#5
June 15th, 2019,
07:51
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.