To find the projection of the vector [tex]\( u = \langle -6, -7 \rangle \)[/tex] onto the vector [tex]\( v = \langle 1, 1 \rangle \)[/tex], we follow these steps:
1. Find the dot product of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[
u \cdot v = (-6) \cdot 1 + (-7) \cdot 1 = -6 - 7 = -13
\][/tex]
2. Calculate the magnitude squared of [tex]\( v \)[/tex] (also called the norm of [tex]\( v \)[/tex] squared):
[tex]\[
\|v\|^2 = v \cdot v = 1 \cdot 1 + 1 \cdot 1 = 1 + 1 = 2
\][/tex]
3. Use the dot product and the norm squared to find the projection of [tex]\( u \)[/tex] onto [tex]\( v \)[/tex]:
[tex]\[
\text{Proj}_v u = \left( \frac{u \cdot v}{\|v\|^2} \right) v
\][/tex]
Substitute the values:
[tex]\[
\text{Proj}_v u = \left( \frac{-13}{2} \right) \langle 1, 1 \rangle
\][/tex]
4. Distribute the scalar to the vector [tex]\( v \)[/tex]:
[tex]\[
\text{Proj}_v u = \left\langle \frac{-13}{2} \cdot 1, \frac{-13}{2} \cdot 1 \right\rangle = \left\langle -6.5, -6.5 \right\rangle
\][/tex]
Thus, the projection of [tex]\( u \)[/tex] onto [tex]\( v \)[/tex] is:
[tex]\[
\boxed{\left\langle -6.5, -6.5 \right\rangle}
\][/tex]
