Question

Question: Let\[y = \left[ {\begin{aligned}5\\{ - 9}\\5\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}{ - 3}\\{ - 5}\\1\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}{ - 3}\\2\\1\end{aligned}} \right]\]. Find the distance from\[y\]to plane in\[{\mathbb{R}^3}\]spanned by\[{{\bf{u}}_1}\], and\[{{\bf{u}}_2}\].

Short answer

The distance is \[2\sqrt {10} \].

Step-by-step solution

Write the definition

Orthogonal set: If \[{{\bf{u}}_i} \cdot {{\bf{u}}_j} = 0\] for \[i \ne j\], then the set of vectors \[\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\} \in {\mathbb{R}^n}\] is said to be orthogonal.

Check the set of given vectors is orthogonal or not

Given vectors are,\[{{\bf{u}}_1} = \left[ {\begin{aligned}{ - 3}\\{ - 5}\\1\end{aligned}} \right]\], and\[{{\bf{u}}_2} = \left[ {\begin{aligned}{ - 3}\\2\\1\end{aligned}} \right]\].

Find\[{{\bf{u}}_1} \cdot {{\bf{u}}_2}\].

\[\begin{aligned}{{\bf{u}}_1} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}{ - 3}\\{ - 5}\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 3}\\2\\1\end{aligned}} \right]\\ &= - 3\left( { - 3} \right) + \left( { - 5} \right)\left( 2 \right) + \left( 1 \right)\left( 1 \right)\\ &= 0\end{aligned}\]

As \[{{\bf{u}}_1} \cdot {{\bf{u}}_2} = 0\], so the set of vectors \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\] is orthogonal.

The Best Approximation Theorem

Here, \[{\bf{\hat y}}\] is said to be the closest point in \[w\] to \[{\bf{y}}\] for which \[\left\| {{\bf{y}} - {\bf{\hat y}}} \right\| < \left\| {{\bf{y}} - {\bf{v}}} \right\|\] for all \[{\bf{v}}\] in \[w\] distinct from \[{\bf{\hat y}}\], if \[{\bf{\hat y}}\] be the orthogonal projection of \[{\bf{y}}\] onto \[w\], where \[w\] is a subspace of \[{\mathbb{R}^n}\], and \[{\bf{y}} \in w\]. \[{\bf{\hat y}}\] can be found as:

\[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\]

Find the distance

The distance from\[{\bf{y}}\]to the closest point is also the distance between the point\[{\bf{y}}\]in\[{\mathbb{R}^3}\]to subspace\[w\], so the distance can be found by\[\left\| {{\bf{y}} - {\bf{\hat y}}} \right\|\].

Find\[{\bf{y}} \cdot {{\bf{u}}_1}\],\[{\bf{y}} \cdot {{\bf{u}}_2}\],\[{{\bf{u}}_1} \cdot {{\bf{u}}_1}\], and\[{{\bf{u}}_2} \cdot {{\bf{u}}_2}\]first, where\[{\bf{y}} = \left[ {\begin{aligned}5\\{ - 9}\\5\end{aligned}} \right]\].

\[\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_1} &= \left[ {\begin{aligned}5\\{ - 9}\\5\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 3}\\{ - 5}\\1\end{aligned}} \right]\\ &= 5\left( { - 3} \right) + \left( { - 9} \right)\left( { - 5} \right) + 5\left( 1 \right)\\ &= 35\end{aligned}\]

\[\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}5\\{ - 9}\\5\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 3}\\2\\1\end{aligned}} \right]\\ &= 5\left( { - 3} \right) + \left( { - 9} \right)\left( 2 \right) + 5\left( 1 \right)\\ &= - 28\end{aligned}\]

\[\begin{aligned}{{\bf{u}}_1} \cdot {{\bf{u}}_1} &= \left[ {\begin{aligned}{ - 3}\\{ - 5}\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 3}\\{ - 5}\\1\end{aligned}} \right]\\ &= - 3\left( { - 3} \right) + \left( { - 5} \right)\left( { - 5} \right) + \left( 1 \right)\left( 1 \right)\\ &= 35\end{aligned}\]

\[\begin{aligned}{{\bf{u}}_2} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}{ - 3}\\2\\1\end{aligned}} \right] \cdot {{\bf{u}}_2}\left[ {\begin{aligned}{ - 3}\\2\\1\end{aligned}} \right]\\ &= - 3\left( { - 3} \right) + \left( 2 \right)\left( 2 \right) + 1\left( 1 \right)\\ &= 14\end{aligned}\]

Now, find the closest point to\[{\bf{y}}\], that is\[{\bf{\hat y}}\]by using the formula\[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\].

\[\begin{aligned}{c}{\bf{\hat y}} = \frac{{35}}{{35}}{{\bf{u}}_1} - \frac{{28}}{{14}}{{\bf{u}}_2}\\ = {{\bf{u}}_1} - 2{{\bf{u}}_2}\\ = \left[ {\begin{aligned}{ - 3}\\{ - 5}\\1\end{aligned}} \right] - 2\left[ {\begin{aligned}{ - 3}\\2\\1\end{aligned}} \right]\\ = \left[ {\begin{aligned}{ - 3}\\{ - 5}\\1\end{aligned}} \right] - \left[ {\begin{aligned}{ - 6}\\4\\2\end{aligned}} \right]\\ = \left[ {\begin{aligned}3\\{ - 9}\\{ - 1}\end{aligned}} \right]\end{aligned}\]

Hence,theclosest point is\[\left[ {\begin{aligned}3\\{ - 9}\\{ - 1}\end{aligned}} \right]\].

Now, find the distance by using\[\left\| {{\bf{y}} - {\bf{\hat y}}} \right\|\].

\[\begin{aligned}\left\| {{\bf{y}} - {\bf{\hat y}}} \right\| &= \left\| {\left[ {\begin{aligned}5\\{ - 9}\\5\end{aligned}} \right] - \left[ {\begin{aligned}3\\{ - 9}\\{ - 1}\end{aligned}} \right]} \right\|\\ &= \left\| {\left[ {\begin{aligned}2\\0\\6\end{aligned}} \right]} \right\|\\ &= \sqrt {{2^2} + 0 + {6^2}} \\ &= \sqrt {40} \\ &= 2\sqrt {10} \end{aligned}\]

So, the distance is\[2\sqrt {10} \].