Answer:
The component of [tex]\vec u[/tex] orthogonal to [tex]\vec a[/tex] is [tex]\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)[/tex].
Step-by-step explanation:
Let [tex]\vec u[/tex] and [tex]\vec a[/tex], from Linear Algebra we get that component of [tex]\vec u[/tex] parallel to [tex]\vec a[/tex] by using this formula:
[tex]\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a[/tex] (Eq. 1)
Where [tex]\|\vec a\|[/tex] is the norm of [tex]\vec a[/tex], which is equal to [tex]\|\vec a\| = \sqrt{\vec a\bullet \vec a}[/tex]. (Eq. 2)
If we know that [tex]\vec u =(2,1,1,2)[/tex] and [tex]\vec a=(4,-4,2,-2)[/tex], then we get that vector component of [tex]\vec u[/tex] parallel to [tex]\vec a[/tex] is:
[tex]\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)[/tex]
[tex]\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)[/tex]
[tex]\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)[/tex]
Lastly, we find the vector component of [tex]\vec u[/tex] orthogonal to [tex]\vec a[/tex] by applying this vector sum identity:
[tex]\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}[/tex] (Eq. 3)
If we get that [tex]\vec u =(2,1,1,2)[/tex] and [tex]\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)[/tex], the vector component of [tex]\vec u[/tex] is:
[tex]\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)[/tex]
[tex]\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)[/tex]
The component of [tex]\vec u[/tex] orthogonal to [tex]\vec a[/tex] is [tex]\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)[/tex].
