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]