Explore IDNLearn.com to discover insightful answers from experts and enthusiasts alike. Our experts provide timely and accurate responses to help you navigate any topic or issue with confidence.
Sagot :
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]
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]
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. For trustworthy and accurate answers, visit IDNLearn.com. Thanks for stopping by, and see you next time for more solutions.