Question

Let \({\mathbb{R}^{\bf{2}}}\) have the inner product of Example 1, and let \({\bf{x}} = \left( {{\bf{1}},{\bf{1}}} \right)\) and \({\bf{y}} = \left( {{\bf{5}}, - {\bf{1}}} \right)\).

a. Find\(\left\| {\bf{x}} \right\|\),\(\left\| {\bf{y}} \right\|\), and\({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^{\bf{2}}}\).

b. Describe all vectors\(\left( {{z_{\bf{1}}},{z_{\bf{2}}}} \right)\), that are orthogonal to y.

Short answer

(a) \(\left\| {\bf{x}} \right\| = 3\),\(\left\| {\bf{y}} \right\| = \sqrt {105} \)and \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^2} = 225\)

(b) The set of vectors \(\left( {{z_1},{z_2}} \right)\) are in the form of \(\frac{{{z_1}}}{1} = \frac{{{z_2}}}{4}\).

Step-by-step solution

Find answer for part (a)

Find the value of \(\left\| {\bf{x}} \right\|\).

\(\begin{aligned}\left\| {\bf{x}} \right\| &= \sqrt {\left\langle {{\bf{x}},{\bf{x}}} \right\rangle } \\ &= \sqrt {\left\langle {\left( {1,1} \right),\left( {1,1} \right)} \right\rangle } \\ &= \sqrt {4\left( 1 \right)\left( 1 \right) + 5\left( 1 \right)\left( 1 \right)} \\ &= \sqrt 9 \\ &= 3\end{aligned}\)

Thus, the value of \(\left\| {\bf{x}} \right\|\) is 3.

Find the value of \(\left\| {\bf{y}} \right\|\).

\(\begin{aligned}\left\| {\bf{y}} \right\| &= \sqrt {\left\langle {{\bf{y}},{\bf{y}}} \right\rangle } \\ &= \sqrt {\left\langle {\left( {5, - 1} \right),\left( {5, - 1} \right)} \right\rangle } \\ &= \sqrt {4\left( 5 \right)\left( 5 \right) + 5\left( { - 1} \right)\left( { - 1} \right)} \\ &= \sqrt {105} \end{aligned}\)

Thus, the value of \(\left\| {\bf{y}} \right\|\) is \(\sqrt {105} \).

Find the value of \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^2}\).

\(\begin{aligned}{\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^2} &= {\left| {\left\langle {\left( {1,1} \right),\left( {5, - 1} \right)} \right\rangle } \right|^2}\\ &= {\left| {4\left( 1 \right)\left( 5 \right) + 5\left( 1 \right)\left( { - 1} \right)} \right|^2}\\ &= {\left| {15} \right|^2}\\ &= 225\end{aligned}\)

Thus, the value of \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^2}\) is 225.

Find answer for part (b)

The vectors \(\left( {{z_1},{z_2}} \right)\) are orthogonal to \(y\), if and only if \(\left\langle {z,{\bf{y}}} \right\rangle = 0\). Therefore,

\(\begin{aligned}\left\langle {z,{\bf{y}}} \right\rangle &= 0\\\left\langle {\left( {{z_1},{z_2}} \right),\left( {5, - 1} \right)} \right\rangle &= 0\\4\left( {{z_1}} \right)\left( 5 \right) + 5\left( { - 1} \right)\left( {{z_2}} \right) &= 0\\20{z_1} &= 5{z_2}\\\frac{{{z_1}}}{1} &= \frac{{{z_2}}}{4}\end{aligned}\)

Thus, the set of vectors \(\left( {{z_1},{z_2}} \right)\) are in the form of \(\frac{{{z_1}}}{1} = \frac{{{z_2}}}{4}\).

So, all the multiples of the vector \(\left( {\begin{aligned}1\\4\end{aligned}} \right)\) are orthogonal to the vector y.