To solve this problem, let's start by finding the components of vector [tex]\( u \)[/tex].
1. Identify the initial and terminal points of vector [tex]\( u \)[/tex]:
[tex]\[
\text{Initial Point of } u: (15, 22)
\][/tex]
[tex]\[
\text{Terminal Point of } u: (5, -4)
\][/tex]
2. Calculate the components of vector [tex]\( u \)[/tex] by subtracting the coordinates of the initial point from the coordinates of the terminal point:
[tex]\[
u_x = 5 - 15 = -10
\][/tex]
[tex]\[
u_y = -4 - 22 = -26
\][/tex]
So, the component form of vector [tex]\( u \)[/tex] is:
[tex]\[
u = \langle -10, -26 \rangle
\][/tex]
3. Calculate the magnitude of vector [tex]\( u \)[/tex]:
[tex]\[
\text{Magnitude of } u = \sqrt{(-10)^2 + (-26)^2} = \sqrt{100 + 676} = \sqrt{776} \approx 27.86
\][/tex]
4. Since vector [tex]\( v \)[/tex] points in the opposite direction of [tex]\( u \)[/tex] and has twice the magnitude, first, we determine the vector components for a vector that would have twice the magnitude of [tex]\( u \)[/tex]:
[tex]\[
\text{Magnitude of } v = 2 \times 27.86 = 55.72
\][/tex]
5. The components of vector [tex]\( v \)[/tex] are:
[tex]\[
v_x = -2 \times (-10) = 20
\][/tex]
[tex]\[
v_y = -2 \times (-26) = 52
\][/tex]
Therefore, the component form of [tex]\( v \)[/tex] is [tex]\( \langle 20, 52 \rangle \)[/tex].
The correct answer is:
[tex]\[
\boxed{D \quad v = \langle 20, 52 \rangle}
\][/tex]
