Answer:
(a) 6 units
(b) 4 units
(c) 7 units
(d) 9.22 units
(e) 7.21 units
(f) 8.06 units
Step-by-step explanation:
The distance between two points, (x₁, y₁, z₁) and (x₂, y₂, z₂), can be calculated using;
d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]
According to the problem;
(a) The distance from (4, -7, 6) to the xy-plane
The xy-plane corresponds to where z equals 0, so
xy-plane = (4, -7, 0).
Thus, the distance d is calculated from (4, -7, 6) to (4, -7, 0)
d = √[(4 - 4)² + (-7 - (-7))² + (0 - 6)²]
d = √[(0)² + (0)² + (-6)²]
d = √(-6)²
d = √36
d = 6
Thus, the distance to the xy-plane is 6 units
(b) The distance from (4, -7, 6) to the yz-plane
The yz-plane is located where x is 0, hence
yz-plane = (0, -7, 6).
So, the distance d is from (4, -7, 6) to (0, -7, 6)
d = √[(4 - 0)² + (-7 - (-7))² + (6 - 6)²]
d = √[(4)² + (0)² + (0)²]
d = √(4)²
d = √16
d = 4
Thus, the distance to the yz-plane is 4 units
(c) The distance from (4, -7, 6) to the xz-plane
The xz-plane exists where y is 0, meaning
xz-plane = (4, 0, 6).
The distance d from (4, -7, 6) to (4, 0, 6)
d = √[(4 - 4)² + (-7 - 0)² + (6 - 6)²]
d = √[(0)² + (-7)² + (0)²]
d = √[(-7)²]
d = √49
d = 7
Thus, the distance to the xz-plane is 7 units
(d) The distance from (4, -7, 6) to the x-axis
The x-axis is defined by y and z being 0, which implies
x-axis = (4, 0, 0).
Thus, the distance d is from (4, -7, 6) to (4, 0, 0)
d = √[(4 - 4)² + (-7 - 0)² + (6 - 0)²]
d = √[(0)² + (-7)² + (6)²]
d = √[(-7)² + (6)²]
d = √[(49 + 36)]
d = √(85)
d = 9.22
Hence, the distance to the x-axis is 9.22 units
(e) The distance from (4, -7, 6) to the y-axis
The y-axis is defined where x and z are both 0, thus
y-axis = (0, -7, 0).
Thus, the distance d is from (4, -7, 6) to (0, -7, 0)
d = √[(4 - 0)² + (-7 - (-7))² + (6 - 0)²]
d = √[(4)² + (0)² + (6)²]
d = √[(4)² + (6)²]
d = √[(16 + 36)]
d = √(52)
d = 7.22
Thus, the distance to the y-axis is 7.21 units
(f) The distance from (4, -7, 6) to the z-axis
The z-axis is defined by x and y being 0, which gives
z-axis = (0, 0, 6).
Thus, the distance d is calculated from (4, -7, 6) to (0, 0, 6)
d = √[(4 - 0)² + (-7 - 0)² + (6 - 6)²]
d = √[(4)² + (-7)² + (0)²]
d = √[(4)² + (-7)²]
d = √[(16 + 49)]
d = √(65)
d = 8.06
Thus, the distance to the z-axis is 8.06 units
