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 d from one point (x₁, y₁, z₁) to another point (x₂, y₂, z₂) is given by;
d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]
Now from the question;
<em>(a) The distance from (4, -7, 6) to the xy-plane</em>
The xy-plane is the point where z is 0. i.e
xy-plane = (4, -7, 0).
Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, -7, 0)</em>
d = √[(4 - 4)² + (-7 - (-7))² + (0 - 6)²]
d = √[(0)² + (0)² + (-6)²]
d = √(-6)²
d = √36
d = 6
Hence, the distance to the xy plane is 6 units
<em>(b) The distance from (4, -7, 6) to the yz-plane</em>
The yz-plane is the point where x is 0. i.e
yz-plane = (0, -7, 6).
Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 6)</em>
d = √[(4 - 0)² + (-7 - (-7))² + (6 - 6)²]
d = √[(4)² + (0)² + (0)²]
d = √(4)²
d = √16
d = 4
Hence, the distance to the yz plane is 4 units
<em>(c) The distance from (4, -7, 6) to the xz-plane</em>
The xz-plane is the point where y is 0. i.e
xz-plane = (4, 0, 6).
Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 6)</em>
d = √[(4 - 4)² + (-7 - 0)² + (6 - 6)²]
d = √[(0)² + (-7)² + (0)²]
d = √[(-7)²]
d = √49
d = 7
Hence, the distance to the xz plane is 7 units
<em>(d) The distance from (4, -7, 6) to the x axis</em>
The x axis is the point where y and z are 0. i.e
x-axis = (4, 0, 0).
Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 0)</em>
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
<em>(e) The distance from (4, -7, 6) to the y axis</em>
The x axis is the point where x and z are 0. i.e
y-axis = (0, -7, 0).
Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 0)</em>
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
Hence, the distance to the y axis is 7.21 units
<em>(f) The distance from (4, -7, 6) to the z axis</em>
The z axis is the point where x and y are 0. i.e
z-axis = (0, 0, 6).
Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, 0 6)</em>
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
Hence, the distance to the z axis is 8.06 units
