Question

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 >.

Answer 1

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

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

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

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

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

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

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

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

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

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

Answer:

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

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