A point \(P\), relative to an origin \(O\), has position vector \(\overrightarrow {{\text{OP}}} = \left( {\begin{array}{*{20}{c}} {1 + s} \\ {3 + 2s} \\ {1 - s} \end{array}} \right),{\text{ }}s \in \mathbb{R}\).

Find the minimum length of \(\overrightarrow {{\text{OP}}} \).

METHOD 1

\(\left| {\overrightarrow {{\text{OP}}} } \right| = \sqrt {{{(1 + s)}^2} + {{(3 + 2s)}^2} + {{(1 - s)}^2}} \;\;\;\left( { = \sqrt {6{s^2} + 12s + 11} } \right)\)     A1

Note:     Award A1 if the square of the distance is found.

EITHER

attempt to differentiate: \(\frac{{\text{d}}}{{{\text{d}}s}}{\left| {\overrightarrow {{\text{OP}}} } \right|^2}\;\;\;( = 12s + 12)\)     M1

attempting to solve \(\frac{{\text{d}}}{{{\text{d}}s}}{\left| {\overrightarrow {{\text{OP}}} } \right|^2} = 0\;\;\;{\text{for }}s\)     (M1)

\(s =  - 1\)     (A1)

OR

attempt to differentiate: \(\frac{{\text{d}}}{{{\text{d}}s}}\left| {\overrightarrow {{\text{OP}}} } \right|\;\;\;\left( { = \frac{{6s + 6}}{{\sqrt {6{s^2} + 12s + 11} }}} \right)\)     M1

attempting to solve \(\frac{{\text{d}}}{{{\text{d}}s}}\left| {\overrightarrow {{\text{OP}}} } \right| = 0\;\;\;{\text{for }}s\)     (M1)

\(s =  - 1\)     (A1)

OR

attempt at completing the square: \(\left( {{{\left| {\overrightarrow {{\text{OP}}} } \right|}^2} = 6{{(s + 1)}^2} + 5} \right)\)     M1

minimum value     (M1)

occurs at \(s =  - 1\)     (A1)

THEN

the minimum length of \(\overrightarrow {{\text{OP}}} \) is \(\sqrt 5 \)     A1

METHOD 2

the length of \(\overrightarrow {{\text{OP}}} \) is a minimum when \(\overrightarrow {{\text{OP}}} \) is perpendicular to \(\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right)\)     (R1)

\(\left( {\begin{array}{*{20}{c}} {1 + s} \\ {3 + 2s} \\ {1 - s} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right) = 0\)     A1

attempting to solve \(1 + s + 6 + 4s - 1 + s = 0\;\;\;(6s + 6 = 0)\;\;\;{\text{for }}s\)     (M1)

\(s =  - 1\)     (A1)

\(\left| {\overrightarrow {{\text{OP}}} } \right| = \sqrt 5 \)     (A1)

[5 marks]

Generally well done. But there was a significant minority who didn’t realise that they had to use calculus or completion of squares to minimise the length. Trying random values of \(s\) gained no marks. A number of candidates wasted time showing that their answer gave a minimum rather than a maximum value of the length.