To solve the given problem, let's break it down into the following steps:
1. Identify the original vector [tex]\( u = \overrightarrow{PQ} \)[/tex] and its coordinates from points [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]:
- Point [tex]\( P \)[/tex] has coordinates [tex]\( (0, 0) \)[/tex].
- Point [tex]\( Q \)[/tex] has coordinates [tex]\( (9, 12) \)[/tex].
- Therefore, the vector [tex]\( u \)[/tex] can be represented as [tex]\( u = (9, 12) \)[/tex].
2. Introduce a scalar [tex]\( c \)[/tex] such that [tex]\( c < 0 \)[/tex]:
3. Calculate the new coordinates of the vector [tex]\( cu \)[/tex] when multiplied by the scalar [tex]\( c \)[/tex]:
- If [tex]\( u = (9, 12) \)[/tex] and [tex]\( c \)[/tex] is a negative scalar, then:
[tex]\[
cu = c \cdot (9, 12) = (c \cdot 9, c \cdot 12)
\][/tex]
- Since [tex]\( c \)[/tex] is negative ([tex]\( c < 0 \)[/tex]), the coordinates [tex]\( (c \cdot 9) \)[/tex] and [tex]\( (c \cdot 12) \)[/tex] will both be negative numbers because multiplying a positive number by a negative scalar results in a negative number.
4. Determine the quadrant where the terminal point of [tex]\( cu \)[/tex] lies:
- The standard coordinate system is divided into four quadrants:
1. Quadrant I: Both [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are positive.
2. Quadrant II: [tex]\( x \)[/tex] is negative and [tex]\( y \)[/tex] is positive.
3. Quadrant III: Both [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are negative.
4. Quadrant IV: [tex]\( x \)[/tex] is positive and [tex]\( y \)[/tex] is negative.
- Since both coordinates of [tex]\( cu \)[/tex], [tex]\( (c \cdot 9) \)[/tex] and [tex]\( (c \cdot 12) \)[/tex], are negative, the point [tex]\( (c \cdot 9, c \cdot 12) \)[/tex] will lie in Quadrant III.
Therefore, the statement that best describes the point [tex]\( c u \)[/tex] is:
The terminal point of [tex]\( c u \)[/tex] lies in Quadrant III.
