Answer
$||{\bf{e}} - 4{\bf{f}}|| = \sqrt {13} $
Work Step by Step
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} $
