Chapter 13 - Vector Geometry - 13.3 Dot Product and the Angle Between Two Vectors - Exercises - Page 668: 85

Since $||{\bf{v}} + {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} + 2{\bf{v}}\cdot{\bf{w}} + ||{\bf{w}}|{|^2}$ and $||{\bf{v}} - {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} - 2{\bf{v}}\cdot{\bf{w}} + ||{\bf{w}}|{|^2}$ so $||{\bf{v}} + {\bf{w}}|{|^2} - ||{\bf{v}} - {\bf{w}}|{|^2} = 4{\bf{v}}\cdot{\bf{w}}$.

Write $||{\bf{v}} + {\bf{w}}|{|^2} = \left( {{\bf{v}} + {\bf{w}}} \right)\cdot\left( {{\bf{v}} + {\bf{w}}} \right) = {\bf{v}}\cdot{\bf{v}} + {\bf{w}}\cdot{\bf{v}} + {\bf{v}}\cdot{\bf{w}} + {\bf{w}}\cdot{\bf{w}}$ $||{\bf{v}} + {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} + 2{\bf{v}}\cdot{\bf{w}} + ||{\bf{w}}|{|^2}$ Write $||{\bf{v}} - {\bf{w}}|{|^2} = \left( {{\bf{v}} - {\bf{w}}} \right)\cdot\left( {{\bf{v}} - {\bf{w}}} \right) = {\bf{v}}\cdot{\bf{v}} - {\bf{w}}\cdot{\bf{v}} - {\bf{v}}\cdot{\bf{w}} + {\bf{w}}\cdot{\bf{w}}$ $||{\bf{v}} - {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} - 2{\bf{v}}\cdot{\bf{w}} + ||{\bf{w}}|{|^2}$ So, $||{\bf{v}} + {\bf{w}}|{|^2} - ||{\bf{v}} - {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} + 2{\bf{v}}\cdot{\bf{w}} + ||{\bf{w}}|{|^2} - \left( {||{\bf{v}}|{|^2} - 2{\bf{v}}\cdot{\bf{w}} + ||{\bf{w}}|{|^2}} \right)$ Hence, $||{\bf{v}} + {\bf{w}}|{|^2} - ||{\bf{v}} - {\bf{w}}|{|^2} = 4{\bf{v}}\cdot{\bf{w}}$.