Question

Find the distance from (3, −4, 8) to each of the following. (a) the xy-plane √7 (b) the yz-plane (0,−4,8) (c) the xz-plane (3,0,8) (d) the x-axis (3,0,0) (e) the y-axis (0,−4,0) (f) the z-axis (0,0,8)

Answer 1

Answer:

a) 8 units

b) 3 units

c) 4 units

d) $4\sqrt{5}\text{ units}$

e) $\sqrt{73}\text{ units}$

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:

$(x_1,y_1,z_1),(x_2,y_2,z_2)\\\\d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

(a) the xy-plane

We have to find the distance from (3, −4, 8) to (3, −4, 0)

$d = \sqrt{(3-3)^2 + (-4+4)^2 + (0-8)^2} = 8\text{ units}$

(b) the yz-plane

We have to find the distance from (3, −4, 8) to (0, −4, 8)

$d = \sqrt{(0-3)^2 + (-4+4)^2 + (8-8)^2} = 3\text{ units}$

(c) the xz-plane

We have to find the distance from (3, −4, 8) to (3, 0, 8)

$d = \sqrt{(3-3)^2 + (0+4)^2 + (8-8)^2} = 4\text{ units}$

(d) the x-axis

We have to find the distance from (3, −4, 8) to (3, 0, 0)

$d = \sqrt{(3-3)^2 + (0+4)^2 + (0-8)^2} = \sqrt{80} = 4\sqrt{5}\text{ units}$

(e) the y-axis (0,−4,0)

We have to find the distance from (3, −4, 8) to (0, -4, 0)

$d = \sqrt{(0-3)^2 + (-4+4)^2 + (0-8)^2} = \sqrt{73}\text{ units}$

(f) the z-axis (0,0,8)

We have to find the distance from (3, −4, 8) to (0, 0, 8)

$d = \sqrt{(0-3)^2 + (0+4)^2 + (8-8)^2} = \sqrt{25} = 5\text{ units}$