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SIZIF
7 days ago
11

Find the distance from (4, −7, 6) to each of the following.

Mathematics
1 answer:
Zina [3.9K]7 days ago
5 0

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

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