To find the projection of [tex]\( \mathbf{u} \)[/tex] onto [tex]\( \mathbf{v} \)[/tex], denoted as [tex]\( \operatorname{proj}_{\mathbf{v}} \mathbf{u} \)[/tex], we need to follow some specific steps.
1. Find the dot product [tex]\( \mathbf{u} \cdot \mathbf{v} \)[/tex]:
Given [tex]\( \mathbf{u} = \langle -12, -5 \rangle \)[/tex] and [tex]\( \mathbf{v} = \langle 3, 9 \rangle \)[/tex],
[tex]\[
\mathbf{u} \cdot \mathbf{v} = (-12) \cdot 3 + (-5) \cdot 9 = -36 - 45 = -81
\][/tex]
2. Find the dot product [tex]\( \mathbf{v} \cdot \mathbf{v} \)[/tex]:
[tex]\[
\mathbf{v} \cdot \mathbf{v} = 3 \cdot 3 + 9 \cdot 9 = 9 + 81 = 90
\][/tex]
3. Calculate the scalar projection of [tex]\( \mathbf{u} \)[/tex] onto [tex]\( \mathbf{v} \)[/tex]:
The scalar projection is given by
[tex]\[
\frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} = \frac{-81}{90} = -\frac{9}{10}
\][/tex]
4. Calculate the vector projection [tex]\( \operatorname{proj}_{\mathbf{v}} \mathbf{u} \)[/tex]:
[tex]\[
\operatorname{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \right) \mathbf{v} = \left( -\frac{9}{10} \right) \langle 3, 9 \rangle = \langle -\frac{27}{10}, -\frac{81}{10} \rangle
\][/tex]
So, the projection of [tex]\( \mathbf{u} \)[/tex] onto [tex]\( \mathbf{v} \)[/tex] is:
[tex]\[
\operatorname{proj}_{\mathbf{v}} \mathbf{u} = \left\langle -\frac{27}{10}, -\frac{81}{10} \right\rangle
\][/tex]
The correct choice is:
[tex]\[
\left\langle -\frac{27}{10}, -\frac{81}{10} \right\rangle
\][/tex]
