If each of these equations defines a plane, show that, for any value of a , the planes do not intersect at a unique point.

Find the value of b for which the intersection of the planes is a straight line.

METHOD 1

$\det \left( {\begin{array}{*{20}{c}}   1&3&{a - 1} \\   2&2&{a - 2} \\   3&1&{a - 3} \end{array}} \right)$     M1

$= 1\left( {2(a - 3) - (a - 2)} \right) - 3\left( {2(a - 3) - 3(a - 2)} \right) + (a - 1)(2 - 6)$

(or equivalent)     A1

= 0 (therefore there is no unique solution)     A1

[3 marks] 

METHOD 2 

$\left( {\begin{array}{*{20}{c}} 1&3&{a - 1}\\ 2&2&{a - 2}\\ 3&1&{a - 3} \end{array}} \right|\left. {\begin{array}{*{20}{c}} 1\\ 1\\ b \end{array}} \right):\left( {\begin{array}{*{20}{c}} 1&3&{a - 1}\\ 0&{ - 4}&{ - a}\\ 0&{ - 8}&{ - 2a} \end{array}} \right|\left. {\begin{array}{*{20}{c}} 1\\ { - 1}\\ {b - 3} \end{array}} \right)$     M1A1

$:\left( {\begin{array}{*{20}{c}} 1&3&{a - 1}\\ 0&{ - 4}&{ - a}\\ 0&0&0 \end{array}} \right|\left. {\begin{array}{*{20}{c}} 1\\ { - 1}\\ {b - 1} \end{array}} \right)$ (and 3 zeros imply no unique solution)     A1

[3 marks]

METHOD 1

$\left( {\begin{array}{*{20}{c}} 1&3&{a - 1}\\ 2&2&{a - 2}\\ 3&1&{a - 3} \end{array}} \right|\left. {\begin{array}{*{20}{c}} 1\\ 1\\ b \end{array}} \right):\left( {\begin{array}{*{20}{c}} 1&3&{a - 1}\\ 0&{ - 4}&{ - a}\\ 0&{ - 8}&{ - 2a} \end{array}} \right|\left. {\begin{array}{*{20}{c}} 1\\ { - 1}\\ {b - 3} \end{array}} \right)$     M1A1

$:\left( {\begin{array}{*{20}{c}} 1&3&{a - 1}\\ 0&{ - 4}&{ - a}\\ 0&0&0 \end{array}} \right|\left. {\begin{array}{*{20}{c}} 1\\ { - 1}\\ {b - 1} \end{array}} \right)$     A1

b = 1     A1     N2

Note: Award M1 for an attempt to use row operations.

[4 marks]

METHOD 2

b = 1     A4

Note: Award A4 only if “ b −1 ” seen in (a).

[4 marks]

The best candidates used row reduction correctly in part a) and were hence able to deduce b = 1 in part b) for an easy final 4 marks. The determinant method was often usefully employed in part a).

The best candidates used row reduction correctly in part a) and were hence able to deduce b = 1 in part b) for an easy final 4 marks. The determinant method was often usefully employed in part a).