Find the shortest distance between the point D with coordinates (-2,-2,9) and the line passing through the points with coordinates B(2,4,6) and C(4,1,5)

Try reading this discussion. It is the same problem, different points.

http://math.stackexchange.com/questions/1300484/distance-between-line-and-a-point

To find the shortest distance between a point and a line, you can use the formula:

d = |(AD x AB) / |AB| |

where:
- d is the shortest distance,
- A is a point on the line,
- D is the given point,
- AB is the vector from A to B, and
- |AB| is the magnitude of AB.

Let's calculate the shortest distance using this formula.

1. Calculate AB, the vector from B to C:
AB = (4 - 2, 1 - 4, 5 - 6) = (2, -3, -1)

2. Find the magnitude of AB:
|AB| = sqrt(2^2 + (-3)^2 + (-1)^2) = sqrt(4 + 9 + 1) = sqrt(14)

3. Calculate AD, the vector from D to A:
AD = (-2 - 2, -2 - 4, 9 - 6) = (-4, -6, 3)

4. Calculate the cross product of AD and AB:
AD x AB = (6*(-1) - (-6)*(-1), 3*(-1) - (-4)*(-1), (-4)*(-3) - (-6)*2)
= (-6 + 6, -3 + 4, 12 - 12) = (0, 1, 0)

5. Calculate the magnitude of AD x AB:
|AD x AB| = sqrt(0^2 + 1^2 + 0^2) = sqrt(1) = 1

6. Calculate the shortest distance:
d = |(AD x AB) / |AB| |
= |(0, 1, 0) / sqrt(14)|
= 1 / sqrt(14)

Therefore, the shortest distance between point D and the line passing through points B and C is 1 / sqrt(14).

To find the shortest distance between a point and a line, we can use the formula:

d = | AP × AB | / | AB |

where d is the shortest distance, AP is the vector from point A on the line to the point P, AB is the vector representing the line, and | AP × AB | is the magnitude of the cross product of vectors AP and AB, and | AB | is the magnitude of the vector AB.

In this case, we need to find the shortest distance between the point D(-2,-2,9) and the line passing through the points B(2,4,6) and C(4,1,5).

1. Find the vector representing the line AB:
AB = B - A = (4,1,5) - (2,4,6) = (2,-3,-1)

2. Find the vector representing the line DC:
DC = C - D = (4,1,5) - (-2,-2,9) = (6,3,-4)

3. Find the cross product of vectors AB and DC:
AP × AB = (2,-3,-1) × (6,3,-4) = (3, 10, 12)

4. Find the magnitude of the cross product AP × AB:
| AP × AB | = √(3^2 + 10^2 + 12^2) = √(9 + 100 + 144) = √(253) = 15.91 (rounded to two decimal places)

5. Find the magnitude of the vector AB:
| AB | = √(2^2 + (-3)^2 + (-1)^2) = √(4 + 9 + 1) = √(14) = 3.74 (rounded to two decimal places)

6. Calculate the shortest distance, d:
d = | AP × AB | / | AB | = 15.91 / 3.74 = 4.25 (rounded to two decimal places)

Therefore, the shortest distance between point D and the line passing through points B and C is approximately 4.25 units.