Finding Orthogonal Projection of \overrightarrow{x}

[VinnyCee]

Question (5.1, #26 -> Bretscher, O.):

Find the orthogonal projection of $$\left[\begin{array}{c} 49 \\ 49 \\ 49 \end{array}\right]$$ onto the subspace of $$\mathbb{R}^3$$ spanned by $$\left[\begin{array}{c} 2 \\ 3 \\ 6 \end{array}\right]$$ and $$\left[\begin{array}{c} 3 \\ -6 \\ 2 \end{array}\right]$$.

$$\overrightarrow{x} = \left[\begin{array}{c} 49 \\ 49 \\ 49 \end{array}\right]$$

My Answer:

Magnitude -> $$\sqrt{(2)^2 + (3)^2 + (6)^2} = 7\,\,=\,\,\sqrt{(3)^2 + (-6)^2 + (2)^2}$$


$$\overrrightarrow{u_1} = \left[\begin{array}{c} \frac{2}{7} \\ \frac{3}{7} \\ \frac{6}{7}\end{array}\right]$$

$$\overrrightarrow{u_2} = \left[\begin{array}{c} \frac{3}{7} \\ \frac{-6}{7} \\ \frac{2}{7}\end{array}\right]$$

$$proj_v \overrightarrow{x} = \left( \overrightarrow{u_1} \cdot \overrightarrow{x} \right) \overrightarrow{u_1} + \left( \overrightarrow{u_2} \cdot \overrightarrow{x} \right) \overrightarrow{u_2}$$

$$proj_v \overrightarrow{x} = \frac{539}{7} \overrightarrow{u_1} + \frac{7}{7} \overrightarrow{u_2} = 77 \overrightarrow{u_1} + \overrightarrow{u_2}$$

$$proj_v \overrightarrow{x} = \left[\begin{array}{c} \frac{157}{7} \\ \frac{225}{7} \\ \frac{464}{7}\end{array}\right]$$

Does this look correct?

EDIT: IT is incorrect, I now figured it out. Thanks

 

[lippyka]

proj_v \overrightarrow{x} = \left( \overrightarrow{u_1} \cdot \overrightarrow{x} \right) \overrightarrow{u_1} + \left( \overrightarrow{u_2} \cdot \overrightarrow{x} \right) \overrightarrow{u_2}proj_v \overrightarrow{x} = \frac{196}{7} \overrightarrow{u_1} + \frac{343}{7} \overrightarrow{u_2} = 28 \overrightarrow{u_1} + 49 \overrightarrow{u_2}proj_v \overrightarrow{x} = \left[\begin{array}{c} 28 \\ -36 \\ 28 \end{array}\right]