By considering \({\alpha _1}\)v\(_1 + {\alpha _2}\)v\(_2 + {\alpha _3}\)v\(_3 = 0\), show that v\(_1\), v\(_2\), v\(_3\) are linearly independent.

Explain briefly why v\(_1\), v\(_2\), v\(_3\) form a basis for vectors in \({\mathbb{R}^3}\).

Show that the vectors

\[\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right];{\text{ }}\left[ {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ 1 \end{array}} \right];{\text{ }}\left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right]\]

form an orthogonal basis.

Express the vector

\[\left[ {\begin{array}{*{20}{c}} 2 \\ 8 \\ 0 \end{array}} \right]\]

as a linear combination of these vectors.

let \({\alpha _1}\)v\(_1 + {\alpha _2}\)v\(_2 + {\alpha _3}\)v\(_3 = 0\)

take the dot product with v\(_1\)     M1

\({\alpha _1}\)v\(_1\) \( \bullet \) v\(_1 + {\alpha _2}\)v\(_2\) \( \bullet \) v\(_1 + {\alpha _3}\)v\(_3\) \( \bullet \) v\(_1 = 0\)     A1

because the vectors are orthogonal, v\(_2\) \( \bullet \) v\(_1\) = v\(_3\) \( \bullet \) v\(_1\) = 0      R1

and since v\(_1\) \( \bullet \) v\(_1\) > 0 it follows that \({\alpha _1} = 0\) and similarly, \({\alpha _2} = {\alpha _3} = 0\)     R1

so \({\alpha _1}\)v\(_1 + {\alpha _2}\)v\(_2 + {\alpha _3}\)v\(_3 = 0 \Rightarrow {\alpha _1} = {\alpha _2} = {\alpha _3} = 0\) therefore v\(_1\), v\(_2\), v\(_3\), are linearly independent     R1AG

[3 marks]

the three vectors form a basis for \({\mathbb{R}^3}\) because they are (linearly) independent     R1

[3 marks]

\(\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right] \bullet \left[ {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ 1 \end{array}} \right] = 0;{\text{ }}\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right] \bullet \left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right] = 0;{\text{ }}\left[ {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ 1 \end{array}} \right] \bullet \left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right] = 0\)     M1A1

therefore the vectors form an orthogonal basis     AG

[??? marks]

let \(\left[ {\begin{array}{*{20}{c}} 2 \\ 8 \\ 0 \end{array}} \right] = \lambda \left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right] + \mu \left[ {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ 1 \end{array}} \right] + v\left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right]\)     M1

\(\lambda  - \mu  + v = 2\)

\(\mu  + 2v = 8\)

\(\lambda  = \mu  - v = 0\)     A1

the solution is

\(\left[ {\begin{array}{*{20}{c}} \lambda  \\ \mu  \\ v \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \end{array}} \right]\,\,\,\,\,\left( {\left[ {\begin{array}{*{20}{c}} 2 \\ 8 \\ 0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right] + 2\left[ {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ 1 \end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right]} \right)\)     A1

[??? marks]