## Calculus (3rd Edition)

$||{\bf{e}} - 4{\bf{f}}|| = \sqrt {13}$
We have $||{\bf{e}} + {\bf{f}}|| = \sqrt 3$. Since $||{\bf{e}} + {\bf{f}}|{|^2} = \left( {{\bf{e}} + {\bf{f}}} \right)\cdot\left( {{\bf{e}} + {\bf{f}}} \right)$, so $||{\bf{e}} + {\bf{f}}|{|^2} = {\bf{e}}\cdot{\bf{e}} + {\bf{f}}\cdot{\bf{e}} + {\bf{e}}\cdot{\bf{f}} + {\bf{f}}\cdot{\bf{f}}$ $||{\bf{e}} + {\bf{f}}|{|^2} = ||{\bf{e}}|{|^2} + 2{\bf{e}}\cdot{\bf{f}} + ||{\bf{f}}|{|^2}$ Since ${\bf{e}}$ and ${\bf{f}}$ are unit vectors, it follows that $3 = 1 + 2{\bf{e}}\cdot{\bf{f}} + 1$ ${\bf{e}}\cdot{\bf{f}} = \frac{1}{2}$ Evaluate $||{\bf{e}} - 4{\bf{f}}|{|^2}$. $||{\bf{e}} - 4{\bf{f}}|{|^2} = \left( {{\bf{e}} - 4{\bf{f}}} \right)\cdot\left( {{\bf{e}} - 4{\bf{f}}} \right)$ $||{\bf{e}} - 4{\bf{f}}|{|^2} = {\bf{e}}\cdot{\bf{e}} - 4{\bf{f}}\cdot{\bf{e}} - 4{\bf{e}}\cdot{\bf{f}} + 16{\bf{f}}\cdot{\bf{f}}$ $||{\bf{e}} - 4{\bf{f}}|{|^2} = ||{\bf{e}}|{|^2} - 8{\bf{e}}\cdot{\bf{f}} + 16||{\bf{f}}|{|^2}$ Since ${\bf{e}}$ and ${\bf{f}}$ are unit vectors and from previous result ${\bf{e}}\cdot{\bf{f}} = \frac{1}{2}$, so $||{\bf{e}} - 4{\bf{f}}|{|^2} = 1 - 8\cdot\frac{1}{2} + 16 = 13$ $||{\bf{e}} - 4{\bf{f}}|| = \sqrt {13}$