Given that a \( \times \) b \( = \) b \( \times \) c \( \ne \) 0 prove that a \( + \) c \( = \) sb where s is a scalar.

METHOD 1

a \( \times \) b = b \( \times \) c

(a \( \times \) b) \( - \) (b \( \times \) c) = 0

(a \( \times \) b) + (c \( \times \) b) = 0     M1A1

(a + c) \( \times \) b = 0     A1

(a + c) is parallel to b \( \Rightarrow \) a + c = sb     R1AG

Note:     Condone absence of arrows, underlining, or other otherwise “correct” vector notation throughout this question.

Note:     Allow “is in the same direction to”, for the final R mark.

METHOD 2

a \( \times \) b = b \( \times \) c \( \Rightarrow \left( {\begin{array}{*{20}{c}} {{a_2}{b_3} - {a_3}{b_2}} \\ {{a_3}{b_1} - {a_1}{b_3}} \\ {{a_1}{b_2} - {a_2}{b_1}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{b_2}{c_3} - {b_3}{c_2}} \\ {{b_3}{c_1} - {b_1}{c_3}} \\ {{b_1}{c_2} - {b_2}{c_1}} \end{array}} \right)\)     M1A1

\({a_2}{b_3} - {a_3}{b_2} = {b_2}{c_3} - {b_3}{c_2} \Rightarrow {b_3}({a_2} + {c_2}) = {b_2}({a_3} + {c_3})\)

\({a_3}{b_1} - {a_1}{b_3} = {b_3}{c_1} - {b_1}{c_3} \Rightarrow {b_1}({a_3} + {c_3}) = {b_3}({a_1} + {c_1})\)

\({a_1}{b_2} - {a_2}{b_1} = {b_1}{c_2} - {b_2}{c_1} \Rightarrow {b_2}({a_1} + {c_1}) = {b_1}({a_2} + {c_2})\)

\(\frac{{({a_1} + {c_1})}}{{{b_1}}} = \frac{{({a_2} + {c_2})}}{{{b_2}}} = \frac{{({a_3} + {c_3})}}{{{b_3}}} = s\)     A1

\( \Rightarrow {a_1} + {c_1} = s{b_1}\)

\( \Rightarrow {a_2} + {c_2} = s{b_2}\)

\( \Rightarrow {a_3} + {c_3} = s{b_3}\)

\( \Rightarrow \left( {\begin{array}{*{20}{c}} {{a_1}} \\ {{a_2}} \\ {{a_3}} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} {{c_1}} \\ {{c_2}} \\ {{c_3}} \end{array}} \right) = s\left( {\begin{array}{*{20}{c}} {{b_1}} \\ {{b_2}} \\ {{b_3}} \end{array}} \right)\)     A1

\( \Rightarrow \) a + c = sb     AG

[4 marks]