Let's start by analyzing the given problem step-by-step:
1. Define Points [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]:
- [tex]\( P \)[/tex] is at the origin: [tex]\( P(0, 0) \)[/tex]
- [tex]\( Q \)[/tex] is at: [tex]\( Q(9, 12) \)[/tex]
2. Determine the Direction Vector [tex]\( u \)[/tex]:
- The direction vector [tex]\( u \)[/tex] from [tex]\( P \)[/tex] to [tex]\( Q \)[/tex] is obtained by subtracting the coordinates of [tex]\( P \)[/tex] from those of [tex]\( Q \)[/tex]:
[tex]\[
u = Q - P = (9 - 0, 12 - 0) = (9, 12)
\][/tex]
3. Scaling the Vector [tex]\( u \)[/tex] by Scalar [tex]\( c \)[/tex]:
- Given that [tex]\( c \)[/tex] is a scalar such that [tex]\( c < 0 \)[/tex], we need to find the vector [tex]\( cu \)[/tex].
- Since [tex]\( c \)[/tex] is negative, multiplying the vector [tex]\( u \)[/tex] by [tex]\( c \)[/tex] will reverse its direction.
- If [tex]\( c = -1 \)[/tex] (a common choice for simplicity when [tex]\( c \)[/tex] is negative), the scaled vector [tex]\( cu \)[/tex] is:
[tex]\[
cu = c \cdot u = -1 \cdot (9, 12) = (-9, -12)
\][/tex]
4. Determining the Terminal Point and Its Quadrant:
- The terminal point of the vector [tex]\( (-9, -12) \)[/tex] indicates where the vector points in coordinate space.
- The coordinates [tex]\( (-9, -12) \)[/tex] represent a point that is 9 units to the left and 12 units down from the origin.
- The negative x-coordinate and negative y-coordinate tell us that the terminal point lies in Quadrant III.
Therefore, the best statement that describes [tex]\( cu \)[/tex] is:
The terminal point of [tex]\( cu \)[/tex] lies in Quadrant III.
