In exercises 7-10, show that u₁,u₂ or u₁,u₂,u₃ is an orthogonal basis for \({\mathbb{R}^2}\) or \({\mathbb{R}^3}\), respectively. Then express x as a linear combination of the u’s.

8. \[{u_1} = \left[ {\begin{align}3\\1\end{align}} \right]\], \[{u_2} = \left[ {\begin{align}{ - 2}\\6\end{align}} \right]\], and \[x = \left[ {\begin{align}{ - 6}\\3\end{align}} \right]\]

Let the set of vectors \({u_1},.....,{u_p}\) be an orthogonal basis for a subspace \(W\) of \({\mathbb{R}^n}\) and the linear combination is given by \(y = {c_1}{u_1} + ..... + {c_p}{u_p}\) , then the weights in the linear combination are given as \({c_j} = \frac{{y \cdot {u_j}}}{{{u_j} \cdot {u_j}}}\), for each \(y\) in \(W\).

Find \({u_1} \cdot {u_2}\) as follows:

\(\begin{align}{c}{u_1} \cdot {u_2} = \left( 3 \right)\left( { - 2} \right) + \left( 1 \right)\left( 6 \right)\\ = - 6 + 6\\ = 0\end{align}\)

Hence, the vectors are orthogonal to each other as the vectors are non-zero and linearly independent. Therefore, the given set form a basis for \({\mathbb{R}^2}\).

The vector x can be expressed as a linear combination as follows:

\[\begin{align}{c}x = \left( {\frac{{x \cdot {u_1}}}{{{u_1} \cdot {u_1}}}} \right){u_1} + \left( {\frac{{x \cdot {u_2}}}{{{u_2} \cdot {u_2}}}} \right){u_2}\\ = \left( {\frac{{\left( { - 6} \right)\left( 3 \right) + \left( 3 \right)\left( 1 \right)}}{{\left( 3 \right)\left( 3 \right) + \left( 1 \right)\left( 1 \right)}}} \right){u_1} + \left( {\frac{{\left( { - 6} \right)\left( { - 2} \right) + \left( 3 \right)\left( 6 \right)}}{{\left( { - 2} \right)\left( { - 2} \right) + \left( 6 \right)\left( 6 \right)}}} \right){u_2}\\ = \left( {\frac{{ - 18 + 3}}{{9 + 1}}} \right){u_1} + \left( {\frac{{12 + 18}}{{4 + 36}}} \right){u_2}\\ = \left( {\frac{{ - 15}}{{10}}} \right){u_1} + \left( {\frac{{30}}{{40}}} \right){u_2}\,\\ = \frac{{ - 3}}{2}{u_1} + \frac{3}{4}{u_2}\end{align}\]

Hence, \[x = - \frac{3}{2}{u_1} + \frac{3}{4}{u_2}\].