Answer to Question #210415 in Analytic Geometry for Ree

Question #210415

Determine projau the orthogonal projection of u and a and deduce ||projau|| for (2.1) u =<−1,3 >, a=<−1,−3>;

(2.2) u =<−2,1,−3 >, a =<−2,1,2 >.


1
Expert's answer
2021-06-25T07:15:01-0400

(2.1) u=1, 3,a=1, 3\mathbf{u} =\langle -1 ,\ 3\rangle ,\quad \mathbf{a} =\langle-1,\ -3\rangle


projau=(u  aa2) a\text{proj}_ \mathbf{a} \mathbf{u} =\big(\frac{ \mathbf{u}\ \cdot \ \mathbf{a}}{\| \mathbf{a}\|^2} \big)\ \mathbf{a} and projau=u  aa\| \text{proj}_ \mathbf{a} \mathbf{u}\|=\frac{|\mathbf{u}\ \cdot \ \mathbf{a}|}{\| \mathbf{a}\|}


u  a=19=8\mathbf{u}\ \cdot \ \mathbf{a}=1-9=-8

a2=1+9=10\| \mathbf{a}\|^2=1+9=10


projau=810a=0.8, 2.4\text{proj}_ \mathbf{a} \mathbf{u} =-\frac{8}{10} \mathbf{a}=\langle0.8,\ 2.4\rangle and projau=810=6.4\|\text{proj}_ \mathbf{a} \mathbf{u}\|=\frac{8}{\sqrt{10}}=\sqrt{6.4}


(2.2) u=2,1, 3,a=2,1, 2\mathbf{u} =\langle -2,1 ,\ - 3\rangle ,\quad \mathbf{a} =\langle-2,1,\ 2\rangle


projau=(u  aa2) a\text{proj}_ \mathbf{a} \mathbf{u} =\big(\frac{ \mathbf{u}\ \cdot \ \mathbf{a}}{\| \mathbf{a}\|^2} \big)\ \mathbf{a} and projau=u  aa\| \text{proj}_ \mathbf{a} \mathbf{u}\|=\frac{ |\mathbf{u}\ \cdot \ \mathbf{a}|}{\| \mathbf{a}\|}


u  a=4+16=1\mathbf{u}\ \cdot \ \mathbf{a}=4+1-6=-1

a2=4+1+4=9\| \mathbf{a}\|^2=4+1+4=9


projau=19a=29, 19, 29\text{proj}_ \mathbf{a} \mathbf{u} =-\frac{1}{9} \mathbf{a}=\langle \frac{2}{9},\ -\frac{1}{9},\ -\frac{2}{9} \rangle and projau=19=13\|\text{proj}_ \mathbf{a} \mathbf{u}\|=\frac{1}{\sqrt{9}}=\frac{1}{3}


Answer:

(2.1) projau=0.8, 2.4\text{proj}_ \mathbf{a} \mathbf{u}=\langle0.8,\ 2.4\rangle and projau=6.4\|\text{proj}_ \mathbf{a} \mathbf{u}\|=\sqrt{6.4} 

(2.2) projau=29, 19, 29\text{proj}_ \mathbf{a} \mathbf{u} =\langle \frac{2}{9},\ -\frac{1}{9},\ -\frac{2}{9} \rangle and projau=13\|\text{proj}_ \mathbf{a} \mathbf{u}\|=\frac{1}{3}



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment