Find the projection of u onto v.
$\begin{array} { l }
\mathbf { u } = \langle 3,6 \rangle \
\mathbf { v } = \langle 4,6 \rangle
\end{array}$

A) $13 \langle 72,48 \rangle$
B) $\frac { 1 } { 13 } \langle 3,4 \rangle$
C) $\frac { 13 } { 12 } \langle 48,72 \rangle$
D) $\frac { 1 } { 13 } \langle 48,72 \rangle$
E) $\frac { 12 } { 13 } \langle 48,72 \rangle$

Question

Find the projection of u onto v. 
 $\begin{array} { l } 
\mathbf { u } = \langle 3,6 \rangle \
\mathbf { v } = \langle 4,6 \rangle
\end{array}$

A) $13 \langle 72,48 \rangle$
B) $\frac { 1 } { 13 } \langle 3,4 \rangle$
C) $\frac { 13 } { 12 } \langle 48,72 \rangle$
D) $\frac { 1 } { 13 } \langle 48,72 \rangle$
E) $\frac { 12 } { 13 } \langle 48,72 \rangle$

Quizplus · Accepted Answer

The projection of u onto v is given by $$	ext{proj}_\mathbf{v}\mathbf{u}=\left(\frac{\mathbf{u}\cdot\mathbf{v}}{\Vert\mathbf{v}\Vert^2}ight)\mathbf{v}$$So, $$\begin{split} 	ext{proj}_\mathbf{v}\mathbf{u} &=\left(\frac{\langle 3,6 angle\cdot\langle 4,6 angle}{\Vert\langle 4,6 angle\Vert^2}ight)\langle 4,6 angle \\ &=\left(\frac{3\cdot4+6\cdot6}{4^2+6^2}ight)\langle 4,6 angle \\ &=\left(\frac{12+36}{52}ight)\langle 4,6 angle \\ &=\frac{1}{13}\langle 4,6 angle \end{split}$$

Find the Projection of U onto V u=⟨3,6⟩v=⟨4,6⟩\begin{array} { l } \mathbf { u } = \langle 3,6 \rangle \\\mathbf { v } = \langle 4,6 \rangle\end{array}u=⟨3,6⟩v=⟨4,6⟩​

Find the Projection of U onto V $\begin{array} { l } \mathbf { u } = \langle 3,6 \rangle \\\mathbf { v } = \langle 4,6 \rangle\end{array}$