Question

select the correct answer
vector u has its initial point at (15, 22) and its terminal point at (5, -4). vector v points in a direction opposite that of u, and its magnitude is twice the magnitude of u. what is the component form of v?
a. ( mathbf{v} = langle -20, 36
angle )
b. ( mathbf{v} = langle -20, 52
angle )
c. ( mathbf{v} = langle 20, 36
angle )
d. ( mathbf{v} = langle 20, 52
angle )

Explanation:

Step1: Find component form of u

To find the component form of vector \( \mathbf{u} \) with initial point \( (x_1, y_1) = (15, 22) \) and terminal point \( (x_2, y_2) = (5, -4) \), we use the formula \( \mathbf{u} = \langle x_2 - x_1, y_2 - y_1
angle \).

So, \( x \)-component: \( 5 - 15 = -10 \)

\( y \)-component: \( -4 - 22 = -26 \)

Thus, \( \mathbf{u} = \langle -10, -26
angle \)

Step2: Determine vector v

Vector \( \mathbf{v} \) points in the opposite direction of \( \mathbf{u} \) and has twice the magnitude.

A vector in the opposite direction of \( \mathbf{u} \) is \( -\mathbf{u} \), and twice the magnitude means we multiply by 2. So, \( \mathbf{v} = -2\mathbf{u} \)

Substitute \( \mathbf{u} = \langle -10, -26
angle \) into the equation:

\( \mathbf{v} = -2\langle -10, -26
angle = \langle (-2)(-10), (-2)(-26)
angle = \langle 20, 52
angle \) Wait, no, wait. Wait, opposite direction of \( \mathbf{u} \) is \( -\mathbf{u} \), and magnitude twice of \( \mathbf{u} \). Wait, magnitude of \( \mathbf{u} \) is \( \sqrt{(-10)^2 + (-26)^2} \), magnitude of \( \mathbf{v} \) is twice that, and direction opposite. So, the component form of \( \mathbf{v} \) should be \( -2\times \mathbf{u} \)? Wait, no. Wait, if \( \mathbf{u} = \langle a, b
angle \), then a vector opposite in direction and twice the magnitude is \( -2\langle a, b
angle \)? Wait, no. Wait, direction opposite: so the vector is \( k\mathbf{u} \) where \( k < 0 \). Magnitude twice: \( |k| \times ||\mathbf{u}|| = 2 ||\mathbf{u}|| \), so \( |k| = 2 \), so \( k = -2 \). Therefore, \( \mathbf{v} = -2\mathbf{u} \)

So, \( \mathbf{u} = \langle 5 - 15, -4 - 22
angle = \langle -10, -26
angle \)

Then \( \mathbf{v} = -2 \times \langle -10, -26
angle = \langle (-2)(-10), (-2)(-26)
angle = \langle 20, 52
angle \)? Wait, but let's check the options. Option D is \( \langle 20, 52
angle \). Wait, but let's recheck the calculation of \( \mathbf{u} \).

Initial point (15,22), terminal point (5,-4). So, the vector \( \mathbf{u} \) is terminal - initial: (5 - 15, -4 - 22) = (-10, -26). Correct. Then, opposite direction: so the vector would be (10, 26) (since opposite direction is multiplying by -1). Then twice the magnitude: multiply by 2, so (20, 52). Yes, that's correct. So \( \mathbf{v} = \langle 20, 52
angle \), which is option D.

Answer:

D. \( \mathbf{v} = \langle 20, 52
angle \)