Respuesta :
Answer:
a) 8 units
b) 3 units
c) 4 units
d) [tex]4\sqrt{5}\text{ units}[/tex]
e) [tex]\sqrt{73}\text{ units}[/tex]
f) 5 units
Step-by-step explanation:
We are given the following:
Point (3, −4, 8)
We have to find the distance of the point from the following:
Distance formula:
[tex](x_1,y_1,z_1),(x_2,y_2,z_2)\\\\d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}[/tex]
(a) the xy-plane
We have to find the distance from (3, −4, 8) to (3, −4, 0)
[tex]d = \sqrt{(3-3)^2 + (-4+4)^2 + (0-8)^2} = 8\text{ units}[/tex]
(b) the yz-plane
We have to find the distance from (3, −4, 8) to (0, −4, 8)
[tex]d = \sqrt{(0-3)^2 + (-4+4)^2 + (8-8)^2} = 3\text{ units}[/tex]
(c) the xz-plane
We have to find the distance from (3, −4, 8) to (3, 0, 8)
[tex]d = \sqrt{(3-3)^2 + (0+4)^2 + (8-8)^2} = 4\text{ units}[/tex]
(d) the x-axis
We have to find the distance from (3, −4, 8) to (3, 0, 0)
[tex]d = \sqrt{(3-3)^2 + (0+4)^2 + (0-8)^2} = \sqrt{80} = 4\sqrt{5}\text{ units}[/tex]
(e) the y-axis (0,−4,0)
We have to find the distance from (3, −4, 8) to (0, -4, 0)
[tex]d = \sqrt{(0-3)^2 + (-4+4)^2 + (0-8)^2} = \sqrt{73}\text{ units}[/tex]
(f) the z-axis (0,0,8)
We have to find the distance from (3, −4, 8) to (0, 0, 8)
[tex]d = \sqrt{(0-3)^2 + (0+4)^2 + (8-8)^2} = \sqrt{25} = 5\text{ units}[/tex]
