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)

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

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

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

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

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