Find the distance from (3,7,-5) to each of the following.
a. the xy-plane
b. the yz-plane
c. the xz-plane
d. the x-axis
e. the y-axis
f. the z-axis
**I just wanna check my answers because I didn't really understand it when the teacher was explaining it and it's an even numbered problem so the answers aren't in the back of the book. thank you!
What are your answers?
To find the distance from a point to a plane or an axis, you can use the distance formula in three-dimensional space. The distance formula is derived from the Pythagorean theorem.
The distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by the formula:
d = √((x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²)
Now let's calculate the distance from the point (3, 7, -5) to each plane or axis:
a. The xy-plane: To calculate the distance between a point and a plane, you need to find the perpendicular distance from the point to the plane. Since the xy-plane has a fixed z-coordinate of 0, you can simply subtract the z-coordinate of the point from 0 to get the distance:
Distance from (3, 7, -5) to the xy-plane = |-5 - 0| = 5 units.
b. The yz-plane: Similar to the previous case, the yz-plane has a fixed x-coordinate of 0. So you can subtract the x-coordinate of the point from 0 to get the distance:
Distance from (3, 7, -5) to the yz-plane = |3 - 0| = 3 units.
c. The xz-plane: The xz-plane has a fixed y-coordinate of 0. So you can subtract the y-coordinate of the point from 0 to get the distance:
Distance from (3, 7, -5) to the xz-plane = |7 - 0| = 7 units.
d. The x-axis: To find the distance between a point and the x-axis, you need to calculate the perpendicular distance from the point to the x-axis. Since the x-axis has a constant y-coordinate of 0 and z-coordinate of 0, you can find the distance between the point (3, 7, -5) and (3, 0, 0) on the x-axis. Using the distance formula:
Distance from (3, 7, -5) to the x-axis = √((3 - 3)² + (7 - 0)² + (-5 - 0)²) = √(0 + 49 + 25) = √74 units.
e. The y-axis: The distance between the point (3, 7, -5) and the y-axis can be found by calculating the perpendicular distance from the point to the y-axis. Since the y-axis has a fixed x-coordinate of 0 and z-coordinate of 0, you can find the distance between the point (3, 7, -5) and (0, 7, 0) on the y-axis. Using the distance formula:
Distance from (3, 7, -5) to the y-axis = √((3 - 0)² + (7 - 7)² + (-5 - 0)²) = √(9 + 0 + 25) = √34 units.
f. The z-axis: To calculate the distance between a point and the z-axis, you need to find the perpendicular distance from the point to the z-axis. Since the z-axis has a fixed x-coordinate of 0 and y-coordinate of 0, you can find the distance between the point (3, 7, -5) and (0, 0, -5) on the z-axis. Using the distance formula:
Distance from (3, 7, -5) to the z-axis = √((3 - 0)² + (7 - 0)² + (-5 + 5)²) = √(9 + 49 + 0) = √58 units.
So, the distances are:
a. Distance to xy-plane = 5 units
b. Distance to yz-plane = 3 units
c. Distance to xz-plane = 7 units
d. Distance to x-axis = √74 units
e. Distance to y-axis = √34 units
f. Distance to z-axis = √58 units