Question

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

Short answer

The orthogonal projection is$pro{j}_v\overrightarrow{x}=\left[\begin{array}{ccc}19& 39& 64\end{array}\right]$ .

Step-by-step solution

Determine the orthonormal basis

Consider a set $span\left\{{\overrightarrow{v}}_1=\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}3\\ -6\\ 2\end{array}\right]\right\}$spanned ${\mathrm{ℝ}}^3{\mathrm{ℝ}}^3$and a vector$\overrightarrow{x}=\left[\begin{array}{c}49\\ 49\\ 49\end{array}\right]$.

If the vector $\overrightarrow{\mathbf{x}}$is perpendicular$\overrightarrow{\mathbf{y}}$then$\overrightarrow{\mathbf{x}}\mathbf{.}\overrightarrow{\mathbf{y}}\mathbf{=}\mathbf{0}$ .

By the theorem of orthogonal basis, the orthogonal${\overrightarrow{u}}_1$is defined as follows.

${\overrightarrow{u}}_1=\frac{{\overrightarrow{v}}_1}{||{\overrightarrow{v}}_1||}$

Substitute the value$\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right]$for${\overrightarrow{v}}_1$in the equation${\overrightarrow{u}}_1=\frac{{\overrightarrow{v}}_1}{||{\overrightarrow{v}}_1||}$as follows.

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{{\overrightarrow{v}}_1}{||{\overrightarrow{v}}_1||} \\ {\overrightarrow{u}}_1=\frac{\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right]}{||\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right]||} \\ {\overrightarrow{u}}_1=\frac{1}{\sqrt{2^2+3^2+6^2}}\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right] \\ {\overrightarrow{u}}_1=\frac{1}{7}\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{1}{7}\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right] \\ {\overrightarrow{u}}_1=\left[\begin{array}{c}2/7\\ 3/7\\ 6/7\end{array}\right] \end{gathered}$$

Similarly, the orthogonal is defined as follows.

${\overrightarrow{u}}_2=\frac{{\overrightarrow{v}}_2}{||{\overrightarrow{v}}_2||}$

Substitute the value$\left[\begin{array}{c}2\\ 3\\ 6\end{array}\right]$for${\overrightarrow{v}}_2$in the equation${\overrightarrow{u}}_2=\frac{{\overrightarrow{v}}_2}{||{\overrightarrow{v}}_2||}$as follows.

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{{\overrightarrow{v}}_2}{||{\overrightarrow{v}}_2||} \\ {\overrightarrow{u}}_2=\frac{\left[\begin{array}{c}3\\ -6\\ 2\end{array}\right]}{||\left[\begin{array}{c}3\\ -6\\ 2\end{array}\right]||} \\ {\overrightarrow{u}}_2=\frac{1}{\sqrt{{\left(3\right)}^2+{\left(-6\right)}^2+{\left(2\right)}^2}}\left[\begin{array}{c}3\\ -6\\ 2\end{array}\right] \\ {\overrightarrow{u}}_2=\frac{1}{7}\left[\begin{array}{c}3\\ -6\\ 2\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{1}{7}\left[\begin{array}{c}3\\ -6\\ 2\end{array}\right] \\ {\overrightarrow{u}}_2=\left[\begin{array}{c}3/7\\ -6/7\\ 2/7\end{array}\right] \end{gathered}$$

Therefore, the orthogonal basis is$span\left\{{\overrightarrow{u}}_1=\left[\begin{array}{c}2/7\\ 3/7\\ 6/7\end{array}\right],{\overrightarrow{u}}_2=\left[\begin{array}{c}3/7\\ -6/7\\ 2/7\end{array}\right]\right\}$.

Determine the orthogonal projection

${\overrightarrow{u}}_2$By the theorem of orthogonal projection, the orthogonal projection of vector$\overrightarrow{x}$ is defined as follows.

$pro{j}_v\overrightarrow{x}=\left({\overrightarrow{u}}_1.\overrightarrow{x}\right){\overrightarrow{u}}_1+\left({\overrightarrow{u}}_2.\overrightarrow{x}\right){\overrightarrow{u}}_2$

Substitute the value $\left[\begin{array}{ccc}2/7& 3/7& 6/7\end{array}\right]$for ${\overrightarrow{u}}_1,\left[\begin{array}{ccc}3/7& -6/7& 2/7\end{array}\right]$for${\overrightarrow{u}}_2$ and$\left[\begin{array}{c}49\\ 49\\ 49\end{array}\right]$ for$\overrightarrow{x}$ in the equation$pro{j}_v\overrightarrow{x}=\left({\overrightarrow{u}}_1.\overrightarrow{x}\right){\overrightarrow{u}}_1+\left({\overrightarrow{u}}_2.\overrightarrow{x}\right){\overrightarrow{u}}_2$ as follows.

$$\begin{gathered} pro{j}_v\overrightarrow{x}=\left({\overrightarrow{u}}_1.\overrightarrow{x}\right){\overrightarrow{u}}_1+\left({\overrightarrow{u}}_2.\overrightarrow{x}\right){\overrightarrow{u}}_2 \\ pro{j}_v\overrightarrow{x}=\left(\left[\begin{array}{ccc}2/7& 3/7& 6/7\end{array}\right].\left[\begin{array}{c}49\\ 49\\ 49\end{array}\right]\right)\left[\begin{array}{ccc}2/7& 3/7& 6/7\end{array}\right]+\left(\left[\begin{array}{ccc}3/7& -6/7& 2/7\end{array}\right].\left[\begin{array}{c}49\\ 49\\ 49\end{array}\right]\right)\left[\begin{array}{ccc}3/7& -6/7& 2/7\end{array}\right] \\ pro{j}_v\overrightarrow{x}=\left(\frac{2}{7}.49+\frac{3}{7}.49+\frac{6}{7}.49\right)\left[\begin{array}{ccc}2/7& 3/7& 6/7\end{array}\right]+\left(\frac{3}{7}.49-\frac{6}{7}.49+\frac{2}{7}.49\right)\left[\begin{array}{ccc}3/7& -6/7& 2/7\end{array}\right] \\ pro{j}_v\overrightarrow{x}=77\left[\begin{array}{ccc}2/7& 3/7& 6/7\end{array}\right]-7\left[\begin{array}{ccc}3/7& -6/7& 2/7\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} pro{j}_v\overrightarrow{x}=77\left[\begin{array}{ccc}2/7& 3/7& 6/7\end{array}\right]-7\left[\begin{array}{ccc}3/7& -6/7& 2/7\end{array}\right] \\ pro{j}_v\overrightarrow{x}=\left[\begin{array}{ccc}22& 33& 66\end{array}\right]-\left[\begin{array}{ccc}3& -6& 2\end{array}\right] \\ pro{j}_v\overrightarrow{x}=\left[\begin{array}{ccc}19& 39& 64\end{array}\right] \end{gathered}$$

Hence, the orthogonal projection of the vector$\overrightarrow{x}$ is$pro{j}_v\overrightarrow{x}=\left[\begin{array}{ccc}19& 39& 64\end{array}\right]$ .