Answer
The projection of ${\bf{u}}$ along the vector $2{\bf{v}}$.
The answer is (b) ${{\bf{u}}_{||{\bf{v}}}}$.
The projection of $2{\bf{u}}$ along ${\bf{v}}$.
The answer is (c) $2{{\bf{u}}_{||{\bf{v}}}}$.
Work Step by Step
By Eq. (4) of Theorem 3, the projection of ${\bf{u}}$ along ${\bf{v}}$ is given by
${{\bf{u}}_{||{\bf{v}}}} = \left( {\frac{{{\bf{u}}\cdot{\bf{v}}}}{{{\bf{v}}\cdot{\bf{v}}}}} \right){\bf{v}}$.
1. So, the projection of ${\bf{u}}$ along the vector $2{\bf{v}}$ is
${{\bf{u}}_{||2{\bf{v}}}} = \left( {\frac{{{\bf{u}}\cdot\left( {2{\bf{v}}} \right)}}{{\left( {2{\bf{v}}} \right)\cdot\left( {2{\bf{v}}} \right)}}} \right)2{\bf{v}} = \left( {\frac{{{\bf{u}}\cdot{\bf{v}}}}{{{\bf{v}}\cdot{\bf{v}}}}} \right){\bf{v}} = {{\bf{u}}_{||{\bf{v}}}}$
The answer is (b) ${{\bf{u}}_{||{\bf{v}}}}$.
2. the projection of $2{\bf{u}}$ along ${\bf{v}}$ is
${\left( {2{\bf{u}}} \right)_{||{\bf{v}}}} = \left( {\frac{{\left( {2{\bf{u}}} \right)\cdot{\bf{v}}}}{{{\bf{v}}\cdot{\bf{v}}}}} \right){\bf{v}} = 2\left( {\frac{{{\bf{u}}\cdot{\bf{v}}}}{{{\bf{v}}\cdot{\bf{v}}}}} \right){\bf{v}} = 2{{\bf{u}}_{||{\bf{v}}}}$.
The answer is (c) $2{{\bf{u}}_{||{\bf{v}}}}$.
