Let's analyze the given problem step-by-step:
1. Define the Vector [tex]\( \mathbf{u} \)[/tex]:
The vector [tex]\( \mathbf{u} \)[/tex] is given as the directed line segment from [tex]\( P(0,0) \)[/tex] to [tex]\( Q(9,12) \)[/tex]. This means the vector [tex]\( \mathbf{u} \)[/tex] has components:
[tex]\[
\mathbf{u} = \langle 9, 12 \rangle
\][/tex]
2. Scalar Multiplication:
Since [tex]\( c \)[/tex] is a scalar such that [tex]\( c < 0 \)[/tex], multiplying [tex]\( \mathbf{u} \)[/tex] by [tex]\( c \)[/tex] will reverse the direction of the vector.
If we denote the scalar multiplication as [tex]\( c \mathbf{u} = \langle c \cdot 9, c \cdot 12 \rangle \)[/tex].
3. Assume a Specific [tex]\( c \)[/tex]:
To simplify our understanding, let's assume [tex]\( c = -1 \)[/tex], which clearly satisfies [tex]\( c < 0 \)[/tex].
4. Calculate [tex]\( c \mathbf{u} \)[/tex]:
By multiplying the vector components by [tex]\( c \)[/tex]:
[tex]\[
c \mathbf{u} = \langle -1 \cdot 9, -1 \cdot 12 \rangle = \langle -9, -12 \rangle
\][/tex]
5. Analyze the Resulting Vector:
The vector [tex]\( \langle -9, -12 \rangle \)[/tex] has both components negative.
6. Determine the Quadrant:
A point with negative [tex]\( x \)[/tex]-coordinate and negative [tex]\( y \)[/tex]-coordinate lies in Quadrant III of the Cartesian coordinate system.
Thus, the terminal point of [tex]\( c \mathbf{u} \)[/tex] lies in Quadrant III.
Conclusion:
Given the vector [tex]\( \mathbf{u} = \overrightarrow{PQ} \)[/tex] from [tex]\( P(0,0) \)[/tex] to [tex]\( Q(9,12) \)[/tex] and a scalar [tex]\( c < 0 \)[/tex], the best statement describing [tex]\( c \mathbf{u} \)[/tex] is:
[tex]\[
\boxed{\text{The terminal point of \( c \mathbf{u} \) lies in Quadrant III.}}
\][/tex]
