For non-zero vectors \({\boldsymbol{a}}\) and \({\boldsymbol{b}}\), show that

(i)     if \(\left| {{\boldsymbol{a}} - {\boldsymbol{b}}} \right| = \left| {{\boldsymbol{a}} + {\boldsymbol{b}}} \right|\), then \({\boldsymbol{a}}\) and \({\boldsymbol{b}}\) are perpendicular;

(ii)     \({\left| {{\boldsymbol{a}} \times {\boldsymbol{b}}} \right|^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} - {({\boldsymbol{a}} \cdot {\boldsymbol{b}})^2}\).

The points A, B and C have position vectors \({\boldsymbol{a}}\), \({\boldsymbol{b}}\) and \({\boldsymbol{c}}\).

(i)     Show that the area of triangle ABC is \(\frac{1}{2}\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|\).

(ii)     Hence, show that the shortest distance from B to AC is

\[\frac{{\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|}}{{\left| {{\boldsymbol{c}} - {\boldsymbol{a}}} \right|}}{\text{.}}\]

(i)     \(\left| {{\boldsymbol{a}} - {\boldsymbol{b}}} \right| = \left| {{\boldsymbol{a}} + {\boldsymbol{b}}} \right|\)

\( \Rightarrow \left( {{\boldsymbol{a}} - {\boldsymbol{b}}} \right) \cdot \left( {{\boldsymbol{a}} - {\boldsymbol{b}}} \right) = \left( {{\boldsymbol{a}} + {\boldsymbol{b}}} \right) \cdot \left( {{\boldsymbol{a}} + {\boldsymbol{b}}} \right)\)     (M1)

\( \Rightarrow {\left| {\boldsymbol{a}} \right|^2} - 2{\boldsymbol{a}} \cdot {\boldsymbol{b}} + {\left| {\boldsymbol{b}} \right|^2} = {\left| {\boldsymbol{a}} \right|^2} + 2{\boldsymbol{a}} \cdot {\boldsymbol{b}} + {\left| {\boldsymbol{b}} \right|^2}\)     A1

\( \Rightarrow 4{\boldsymbol{a}} \cdot {\boldsymbol{b}} = 0 \Rightarrow {\boldsymbol{a}} \cdot {\boldsymbol{b}} = 0\)     A1

therefore \({\boldsymbol{a}}\) and \({\boldsymbol{b}}\) are perpendicular     R1

Note: Allow use of 2-d components.

Note: Do not condone sloppy vector notation, so we must see something to the effect that \({\left| {\boldsymbol{c}} \right|^2} = {\boldsymbol{c}} \cdot {\boldsymbol{c}}\) is clearly being used for the M1.

Note: Allow a correct geometric argument, for example that the diagonals of a parallelogram have the same length only if it is a rectangle.

(ii)     \({\left| {{\boldsymbol{a}} \times {\boldsymbol{b}}} \right|^2} = {\left( {\left| {\boldsymbol{a}} \right|\left| {\boldsymbol{b}} \right|\sin \theta } \right)^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}{\sin ^2}\theta \)     M1A1

\({\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} - {\left( {{\boldsymbol{a}} \cdot {\boldsymbol{b}}} \right)^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} - {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}{\cos ^2}\theta \)     M1

\( = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}\left( {1 - {{\cos }^2}\theta } \right)\)     A1
\( = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}\left( {{{\sin }^2}\theta } \right)\)

\( \Rightarrow {\left| {{\boldsymbol{a}} \times {\boldsymbol{b}}} \right|^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} - {\left( {{\boldsymbol{a}} \cdot {\boldsymbol{b}}} \right)^2}\)     AG

[8 marks]

(i)     area of triangle \( = \frac{1}{2}\left| {\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} } \right|\)     (M1)

\( = \frac{1}{2}\left| {\left( {{\boldsymbol{b}} - {\boldsymbol{a}}} \right) \times \left( {{\boldsymbol{c}} - {\boldsymbol{a}}} \right)} \right|\)     A1

\( = \frac{1}{2}\left| {{\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{b}} \times {\boldsymbol{ - a}} + {\boldsymbol{ - a}} \times {\boldsymbol{c}} + {\boldsymbol{ - a}} \times {\boldsymbol{ - a}}} \right|\)     A1

\({\boldsymbol{b}} \times {\boldsymbol{ - a}} = {\boldsymbol{a}} \times {\boldsymbol{b}}\); \({\boldsymbol{c}} \times {\boldsymbol{a}} = {\boldsymbol{ - a}} \times {\boldsymbol{c}}\); \({\boldsymbol{ - a}} \times {\boldsymbol{ - a}} = 0\)     M1

hence, area of triangle is \(\frac{1}{2}\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|\)     AG

(ii)     D is the foot of the perpendicular from B to AC

area of triangle \({\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BD}}} } \right|\)     A1

therefore

\(\frac{1}{2}\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BD}}} } \right| = \frac{1}{2}\left| {\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} } \right|\)     M1

hence, \(\left| {\overrightarrow {{\text{BD}}} } \right| = \frac{{\left| {\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} } \right|}}{{\left| {\overrightarrow {{\text{AC}}} } \right|}}\)     A1

\( = \frac{{\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|}}{{\left| {{\boldsymbol{c}} - {\boldsymbol{a}}} \right|}}\)     AG

[7 marks]

(i) The majority of candidates were very sloppy in their use of vector notation. Some candidates used Cartesian coordinates, which was acceptable. Part (ii) was well done.

Part (i) was usually well started, but not completed satisfactorily. Many candidates understood the geometry involved in this part.