Answer to Question #209788 in Analytic Geometry for Tshego

Question

Determine proj_au the orthogonal projection of u and a and deduce ||proj_au|| for

a.) u=<-1 3>, a=<-1,-3>;

b.)u=<-2,1,-3>, a=<-2,1,2>.

Answer 1

$proj_{\vec a}\vec u=\dfrac{\vec u\cdot\vec a}{|\vec a|^2}\vec a$

(2.1)

$\vec u=\langle-1,3\rangle, \vec a=\langle-1,-3\rangle$

$\vec u\cdot\vec a=-1(-1)+3(-3)=-8$

$|\vec a|^2=(-1)^2+(-3)^2=10$

$proj_{\vec a}\vec u=\dfrac{\vec u\cdot\vec a}{|\vec a|^2}\vec a=\dfrac{-8}{10}\langle-1,-3\rangle$

$=\langle0.8,-0.3\rangle$ $|proj_{\vec a}\vec u|=\sqrt{(0.8)^2+(-0.3)^2}=\sqrt{0.73}$

(2.2)

$\vec u=\langle-2,1, -3\rangle, \vec a=\langle-2,1, 2\rangle$

$\vec u\cdot\vec a=-2(-2)+1(1)-3(2)=-1$

$|\vec a|^2=(-2)^2+(1)^2+(2)^2=9$

$proj_{\vec a}\vec u=\dfrac{\vec u\cdot\vec a}{|\vec a|^2}\vec a=\dfrac{-1}{9}\langle-2,1,2\rangle$

$=\langle\dfrac{2}{9},-\dfrac{1}{9},-\dfrac{2}{9}\rangle$ $|proj_{\vec a}\vec u|=\sqrt{(\dfrac{2}{9})^2+(-\dfrac{1}{9})^2+(-\dfrac{2}{9})^2}=\dfrac{1}{3}$