(a)     Show that the following system of equations has an infinite number of solutions.

     $x + y + 2z =  - 2$

     $3x - y + 14z = 6$

     $x + 2y =  - 5$

The system of equations represents three planes in space.

(b)     Find the parametric equations of the line of intersection of the three planes.

(a)     EITHER

$\left( {\begin{array}{*{20}{c}}1&1&2\\3&{ - 1}&{14}\\1&2&0\end{array}\left| \begin{array}{c} - 2\\6\\ - 5\end{array} \right.} \right) \to \left( {\begin{array}{*{20}{c}}1&1&2\\0&1&{ - 2}\\0&0&0\end{array}\left| \begin{array}{c} - 2\\ - 3\\0\end{array} \right.} \right)$     M1

 

row of zeroes implies infinite solutions, (or equivalent).     R1

 

Note:     Award M1 for any attempt at row reduction.

 

OR

$\left| {\begin{array}{*{20}{c}}1&1&2\\3&{ - 1}&{14}\\1&2&0\end{array}} \right| = 0$     M1

$\left| {\begin{array}{*{20}{c}}1&1&2\\3&{ - 1}&{14}\\1&2&0\end{array}} \right| = 0$ with one valid point     R1

OR

$x + y + 2z =  - 2$

$3x - y + 14z = 6$

$x + 2y =  - 5$   $\Rightarrow x =  - 5 - 2y$

substitute $x =  - 5 - 2y$ into the first two equations:

$- 5 - 2y + y + 2z =  - 2$

$3( - 5 - 2y) - y + 14z = 6$     M1

$- y + 2z = 3$

$- 7y + 14z = 21$

the latter two equations are equivalent (by multiplying by 7) therefore an infinite number of solutions.     R1

OR

for example, $7 \times {{\text{R}}_1} - {{\text{R}}_2}$ gives $4x + 8y =  - 20$     M1

this equation is a multiple of the third equation, therefore an infinite

number of solutions.     R1

[2 marks]

 

(b)     let $y = t$     M1

then $x =  - 5 - 2t$     A1

$z = \frac{{t + 3}}{2}$     A1

OR

let $x = t$     M1

then $y = \frac{{ - 5 - t}}{2}$     A1

$z = \frac{{1 - t}}{4}$     A1

OR

let $z = t$     M1

then $x = 1 - 4t$     A1

$y =  - 3 + 2t$     A1

OR

attempt to find cross product of two normal vectors:

eg: $\left| {\begin{array}{*{20}{c}}i&j&k\\1&1&2\\1&2&0\end{array}} \right| =  - 4i + 2j + k$     M1A1

$x = 1 - 4t$

$y =  - 3 + 2t$

$z = t$     A1

(or equivalent)

[3 marks]

 

Total [5 marks]

[N/A]