Explain why the set of position vectors of points whose coordinates satisfy $x - y - z = 1$ does not form a vector subspace of ${\mathbb{R}^3}$.

(i)     Show that the set of position vectors of points whose coordinates satisfy $x - y - z = 0$ forms a vector subspace, $V$, of ${\mathbb{R}^3}$.

(ii)     Determine an orthogonal basis for $V$ of which one member is $\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right)$.

(iii)     Augment this basis with an orthogonal vector to form a basis for ${\mathbb{R}^3}$.

(iv)     Express the position vector of the point with coordinates $(4,{\text{ }}0,{\text{ }} - 2)$ as a linear combination of these basis vectors.

Accept any valid reasoning:

Example 1:

$(1,{\text{ }}0,{\text{ }}0)$ lies on the plane, however linear combinations of this do not (for example $(2,{\text{ }}0,{\text{ }}0)$)     R1

hence the position vectors of the points on the plane do not form a vector space     AG

Example 2:

the given plane does not pass through the origin (or the zero vector is not the position vector of any point on the plane)     R1

hence the position vectors of the points on the plane do not form a vector space     AG

[1 mark]

(i)     (the set of position vectors is non-empty)

let $x$$_1 = \left( {\begin{array}{*{20}{c}} {{x_1}} \\ {{y_1}} \\ {{z_1}} \end{array}} \right)$ be the position vector of a point on the plane and $a \in \mathbb{R}$

then the coordinates of the position vector of $ax$ satisfy the equation for the plane because $a{x_1} - a{y_1} - a{z_1} = a({x_1} - {y_1} - {z_1}) = 0$     M1A1

let $x$$_2 = \left( {\begin{array}{*{20}{c}} {{x_2}} \\ {{y_2}} \\ {{x_2}} \end{array}} \right)$ be the position vector of another point on the plane

consider $x$$_3 =$ $x$$_1 +$ $x$$_2$

then the coordinates of $x$$_3 = \left( {\begin{array}{*{20}{l}} {{x_3} = {x_1} + {x_2}} \\ {{y_1} = {y_1} + {y_2}} \\ {{z_3} = {z_1} + {z_2}} \end{array}} \right)$ satisfy     M1

${x_3} - {y_3} - {z_3} = ({x_1} + {x_2}) - ({y_1} + {y_2}) - ({z_1} + {z_2})$

$= 0$    A1

subspace conditions established     AG

Note:     The above conditions may be combined in one calculation.

(ii)     if $\left( {\begin{array}{*{20}{c}} a \\ b \\ c \end{array}} \right)$ is the position vector of a second point on the plane orthogonal to the given vector, then     (M1)

$a - b - c = 0$ and $a + 2b - c = 0$     (A1)(A1)

for example $\left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right)$ completes the basis     A1

(iii)     the basis for $({\mathbb{R}^3})$ can be augmented to an orthogonal basis for ${\mathbb{R}^3}$ by adjoining $\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right)$     (M1)

$= \left( {\begin{array}{*{20}{c}} 2 \\ { - 2} \\ { - 2} \end{array}} \right)$    A1

(iv)     attempt to solve $\alpha = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right) + \beta \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right) + \gamma \left( {\begin{array}{*{20}{c}} 2 \\ { - 2} \\ { - 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 4 \\ 0 \\ { - 2} \end{array}} \right)$     M1

obtain $\alpha  = \beta  = \gamma  = 1$     A2

[13 marks]

Again this was found difficult by many candidates and resulted in no attempt being made. For those who were able to start, parts (a) and (b)(i) showed a reasonable degree of understanding. After that it was only a significant minority of candidates who were able to proceed successfully with many ignoring or not realising the significance of the word “orthogonal”.

Again this was found difficult by many candidates and resulted in no attempt being made. For those who were able to start, parts (a) and (b)(i) showed a reasonable degree of understanding. After that it was only a significant minority of candidates who were able to proceed successfully with many ignoring or not realising the significance of the word “orthogonal”.