# Last vector-plane problem for the night

• prace
In summary, to show that two planes are parallel to each other, their normal vectors must be the same or a multiple of each other. To find the distance between them, one can pick a point in one plane and use the perpendicular distance formula to find the distance between that point and the closest point on the other plane.
prace
I would like to show that two planes are || and then find the distance between the two planes. I think I have the distance between the two planes covered pretty good, but I am not sure about the || part.

For example:

$$P_1$$ = 4x-2y+6z=3 and $$P_2$$ = -6x+3y-9z=4

then $$\vec{n}P_1$$ = <4,-2,6> and $$\vec{n}P_2$$ = <-6,3,-9>. So, my thinking is that if these two planes were || then they should have the same normal vector or at least a multiple of that normal vector. Is this correct, should I be looking at something else as well?

P2 is: -3/2* (4x-2y+6z) = 4

Or 4x-2y+6z = -8/3

Does that help?

prace said:
I would like to show that two planes are || and then find the distance between the two planes. I think I have the distance between the two planes covered pretty good, but I am not sure about the || part.

For example:

$$P_1$$ = 4x-2y+6z=3 and $$P_2$$ = -6x+3y-9z=4

then $$\vec{n}P_1$$ = <4,-2,6> and $$\vec{n}P_2$$ = <-6,3,-9>. So, my thinking is that if these two planes were || then they should have the same normal vector or at least a multiple of that normal vector. Is this correct, should I be looking at something else as well?
Yes, that's correct. And, just as obviously, as Office_Shredder showed, one is a multiple of the other. Now, to find the distance between them, pick any point in one plane, find the line through that point perpendicular to the plane (in the direction of the normal vector), find the point where that line intersects the other plane, and find the distance between the two points.

## 1. What is the "last vector-plane problem for the night"?

The "last vector-plane problem for the night" is a phrase commonly used in aviation to describe the final navigational challenge that a pilot must complete before the end of their flight. It typically involves calculating the vector (direction and magnitude) of the last leg of the flight to reach the destination airport.

## 2. Why is the "last vector-plane problem for the night" important?

The "last vector-plane problem for the night" is important because it ensures that the pilot arrives at the correct destination airport and can safely land the aircraft. It also helps the pilot to manage their fuel consumption and make any necessary adjustments to their flight plan.

## 3. How do pilots solve the "last vector-plane problem for the night"?

Pilots use a combination of navigational tools, such as compasses, GPS systems, and flight computers, to solve the "last vector-plane problem for the night". They also rely on their training and experience to make accurate calculations and adjustments.

## 4. What factors can affect the accuracy of the "last vector-plane problem for the night"?

Several factors can affect the accuracy of the "last vector-plane problem for the night", including changes in weather conditions, air traffic control instructions, and mechanical issues with the aircraft. It is important for pilots to constantly monitor these factors and make adjustments as needed.

## 5. Is the "last vector-plane problem for the night" different from other navigational challenges in aviation?

Yes, the "last vector-plane problem for the night" is different from other navigational challenges in aviation because it is usually the final calculation that a pilot must make before reaching their destination. It is also important because it is typically done at night, when visibility is reduced and pilots must rely on their instruments and calculations to navigate.

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