Question

Suppose that two asteroids in the asteroid belt are moving in a straight line and at the same speed toward each other. If a 3D coordinate system is placed with the origin at the sun, then the position of one asteroid can be described as 2.8 astronomical units (AU) in the direction of positive x, 0.3 AU in the direction of positive p, and 0.15 AU in the direction of positive z. At the same time, the position of the other asteroid can be described as 2.7 AU in the direction of positive x, 0.23 AU in the direction of positiven, and 0.12 XU in the direction of positive z. At what point in this 3D coordinate system will they collide? Use the midpoint formula given below.

Answer 1

Answer:

• 2.75 astronomical units (AU) in the direction of positive x

,

• 0.265 AU in the direction of positive y

,

• 0.135 AU in the direction of positive z.

Explanation:

First Asteroid

• 2.8 astronomical units (AU) in the direction of positive x

,

• 0.3 AU in the direction of positive y

,

• 0.15 AU in the direction of positive z.

$\implies\text{Asteriod 1: )}(x_1,y_1,z_1)=(2.8,0.3,0.15)$

Second Asteroid

The position of the asteroids is given below:

• 2.7 AU in the direction of positive x

,

• 0.23 AU in the direction of positive y

,

• 0.12 AU in the direction of positive z.

$\implies\text{Asteriod 2: }(x_2,y_2,z_2)=(2.7,0.23,0.12)$

To find the point where they collide, we use the given midpoint formula.

$\mleft(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2},\frac{z_{1}+z_{2}}{2}\mright)$

Substitute the values:

$\begin{gathered} \text{Midpoint}=\mleft(\frac{2.8+2.7}{2},\frac{0.3+0.23}{2},\frac{0.15+0.12}{2}\mright) \\ =\mleft(\frac{5.5}{2},\frac{0.53}{2},\frac{0.27}{2}\mright) \\ =\mleft(2.75,0.265,0.135\mright) \end{gathered}$

Thus, in this coordinate system, the asteroid will collide at:

• 2.75 astronomical units (AU) in the direction of positive x

,

• 0.265 AU in the direction of positive y

,

• 0.135 AU in the direction of positive z.