Calculus I, II and III

To solve this problem, we first need to determine the vector \( \mathbf{u} = \overrightarrow{PQ} \). Given the points \( P(0, 0) \) and \( Q(9, 12) \), the vector \( \mathbf{u} \) can be calculated as follows: \[ \mathbf{u} = Q - P = (9 - 0, 12 - 0) = (9, 12) \] Next, we consider the scalar multiplication of \( \mathbf{u} \) by \( c \), where \( c < 0 \). The vector resulting from this operation is \( c\mathbf{u} \). The components of \( c\mathbf{u} \) are given by: \[ c\mathbf{u} = (c \times 9, c \times 12) \] Since \( c \) is negative, multiplying \( c \) with the positive components of \( \mathbf{u} \) will result in negative components. Therefore, the vector \( c\mathbf{u} \) will have the form: \[ c\mathbf{u} = (-|c| \times 9, -|c| \times 12) \] This vector points in the direction opposite to \( \mathbf{u} \). Since \( \mathbf{u} \) points from the origin (0,0) to the point (9,12) in Quadrant I, the vector \( c\mathbf{u} \), being the negative of \( \mathbf{u} \), will point from the origin to a point in the direction opposite to Quadrant I, which is Quadrant III. Thus, the terminal point of \( c\mathbf{u} \) lies in Quadrant III. This is because both components of \( c\mathbf{u} \) are negative, placing the terminal point in the quadrant where both \( x \) and \( y \) coordinates are negative. Therefore, the correct statement is: - The terminal point of \( c\mathbf{u} \) lies in Quadrant III.

Math

Let \( \mathbf{u} = \overrightarrow{PQ} \) be the directed line segment from \( P(0, 0) \) to \( Q(9, 12) \), and let \( c \) be a scalar such that \( c < 0 \). Which statement best describes \( c\mathbf{u} \)? - The terminal point of \( c\mathbf{u} \) lies in Quadrant III. - The terminal point of \( c\mathbf{u} \) lies in Quadrant II. - The terminal point of \( c\mathbf{u} \) lies in Quadrant I. - The terminal point of \( c\mathbf{u} \) lies in Quadrant IV.

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