Distance formula for between a point and a plane

In summary, the problem is to find the distance from a point P to a plane passing through points Q, R, and S. Using the definitions of dot and cross products, it can be shown that the distance d is equal to the magnitude of the cross product of vectors b and c, divided by the magnitude of vector a. However, it is unclear how to incorporate the dot product into this solution.
  • #1
ProPatto16
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Homework Statement



Let P be a point not on a plane that passes through points Q, R, S. show that the distance, d, from P to the plane is:

d = |a.(b x c)| over |a x b| where a=QR, b=QS and c=QP

Homework Equations



definition of dot product is a.b=|a||b|cos(theta)
definition of cross product is a x b= |a||b|sin(theta)
triple scalar product is |a.(b x c)|=|b x c||a|cos(theta)

The Attempt at a Solution



putting point P above point S in plane, gives d = |PS| =|PQ|sin(theta) = |c|sin(theta)
theta is angle between |QP| and |QS| which is c and b so by definition of the cross product sin(theta)=|a x b| over |a||b| gives d = csin(theta) = |c||b x c| over |c||b| = |b x c| over |b|

to incorporate dot product i need a cos theta and the only relavant one i can figure is angle between QR and QS which is a and b. which wouldn't be in the plane. I am not sure where to go from here??

thanks
 
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  • #2
i can do it with 2 points on a line if that's any help i will show it.

let P be a point not on the line that passes through Q and R. show that the distance d from the point P to the line L is

d = |a x b| over |a|

have line QR with point P drawn above and an arbitrary point S on the line where the perpendicular of P meets. let QR be vector a and QP be vector b.

the distance d is PS. which is also QPsin(theta) = |b|sin(theta). theta is angle between QP=b and QR=a therefore by definition of cross product, sin(theta) = |a x b| over |a||b|
and so d = |b|sin(theta) = |b||a x b| over |a||b| = |a x b| over |a|


but i don't know how to incorporate that into a plane with a dot product in the solution?
 
  • #3
help?
 

1. What is the distance formula for finding the distance between a point and a plane?

The distance formula for finding the distance between a point (x1, y1, z1) and a plane ax + by + cz + d = 0 is:

d = |ax1 + by1 + cz1 + d| / √(a² + b² + c²)

2. How is the distance formula for a point and a plane derived?

The distance formula for a point and a plane is derived from the Pythagorean theorem. The shortest distance between a point and a plane is a perpendicular line, which forms a right angle with the plane. This perpendicular line then becomes the hypotenuse of a right triangle, with the distance "d" being the length of the hypotenuse. By using the Pythagorean theorem, the distance formula can be derived.

3. Can the distance formula be applied to any point and plane in 3-dimensional space?

Yes, the distance formula for a point and a plane can be applied to any point and plane in 3-dimensional space. The formula takes into account the x, y, and z coordinates of the point and the coefficients of the plane's equation, making it applicable to any point and plane in 3-dimensional space.

4. What does the "d" in the distance formula represent?

The "d" in the distance formula represents the shortest distance between the given point and the given plane. It is the length of the perpendicular line that forms a right angle with the plane and intersects with the point.

5. Are there any other methods for finding the distance between a point and a plane?

Yes, there are other methods for finding the distance between a point and a plane, such as using the vector projection formula or the dot product formula. However, the distance formula is the most commonly used and efficient method for finding the distance between a point and a plane.

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