The three-distance theorem states that there are at most three distinct gaps between consecutive elements in the set of fractional parts of the first n multiples of any real number. In a real world of three - dimensions, we must extend our knowledge of geometry ti three dimensional space.
To find the distance between two points whose co-ordinates are given. Let P( x1 , y1 , z1 ) and Q ( a2, b2, c2 ) be the two given points. Through P and Q draw planes parallel to the co-ordinate planes to form a rectangular box whose one diagonal is PQ. Geometrically to find the distance PQ is the computing of the length of the diagonal PQ of the box ( as shown in the figure) by means of Pythagoras theorem.
Since MQ is perpendicular to the plane PAMB and PM lies in the plane so MQ is perpendicular to PM `=>` `angle` PMQ = 900 .
in triangle PMQ, `angle` PMQ = 900 , therefore by pythagoras theorem we get,
PQ2 = PM2 + MQ2 ........(1)
since, AM is perpendicular to AP, so `angle` MAP = 900
In triangle AMP, `angle` MAP = 900 , therefore, by pythagoras theorem
PM2 = AP2 + AM2 .....(2)
From (1) and (2), we get
PQ2 = AP2+ AM2+ MQ2 ......(3)
Clearly from the figure
AP = x2 - x1 , AM = y2 -y1 , MQ = z2-z1
Substituting these values in (3) we get
PQ2 = (x2 - x1 ) + ( y2 - y1 ) + (z 2- z1)
PQ = `sqrt(( x_2-x_1)+ ( y_2-y_1) + (z_2-z_1))`
Distance from origin: The distance of P(x,1y,1z1) from the origin o(0,0,0) is `sqrt((x_1-0)^2 +( y_2 - 0)^2 + (z_2 - 0)^2)` = `sqrt((x_1)^2 + (y_1)^2 + (z_1)^2)`
Example Problems on three Distance Theorem:
Problem1: Find ' a' so that the distance between ( -2,1,-3) and ( 1, 3, -6) be 7 units?
Solution:
Let A, B be ( -2,1,3) ,( 1, 3, -6) respectively then
`=>` `sqrt((a-2)^2 + (3-1)^2 + (-6 + 3 )^2)` = 7
`=>` a2 + a + 4 + 4 + 9 = 49 `=>` a2 + 4a - 32 = 0
`=>` (a-4) (a+ 8) = 0 `=>` a= 4 , -8
Problem2: Find the distance between the points A(1,2,3) from he point B(3, 6, 8)?
Solution: The distance
`sqrt((3-1)^2 +(6-3)^2 + ( 8-3)^2)`
AB = `sqrt(2^2 + 3 ^2 + 5^2)`
AB = `sqrt(4 + 9 + 25)`
AB = `sqrt(38)` <br>
Practice Problems on three Distance Theorem
Problem2: Find the distance of the point P( 4, 6, 8) from the point Q ( 9,5,2)? ( Answer: 3`sqrt(7)` )
problem3: By using the distance formula prove that the points A( 3, -5, 1), B ( -1, 0 , 8) and C( 7, -10, -6) are collinear ("Answer : The sum of two side is equal to the third side so the points are collinear)
To find the distance between two points whose co-ordinates are given. Let P( x1 , y1 , z1 ) and Q ( a2, b2, c2 ) be the two given points. Through P and Q draw planes parallel to the co-ordinate planes to form a rectangular box whose one diagonal is PQ. Geometrically to find the distance PQ is the computing of the length of the diagonal PQ of the box ( as shown in the figure) by means of Pythagoras theorem.
Since MQ is perpendicular to the plane PAMB and PM lies in the plane so MQ is perpendicular to PM `=>` `angle` PMQ = 900 .
in triangle PMQ, `angle` PMQ = 900 , therefore by pythagoras theorem we get,
PQ2 = PM2 + MQ2 ........(1)
since, AM is perpendicular to AP, so `angle` MAP = 900
In triangle AMP, `angle` MAP = 900 , therefore, by pythagoras theorem
PM2 = AP2 + AM2 .....(2)
From (1) and (2), we get
PQ2 = AP2+ AM2+ MQ2 ......(3)
Clearly from the figure
AP = x2 - x1 , AM = y2 -y1 , MQ = z2-z1
Substituting these values in (3) we get
PQ2 = (x2 - x1 ) + ( y2 - y1 ) + (z 2- z1)
PQ = `sqrt(( x_2-x_1)+ ( y_2-y_1) + (z_2-z_1))`
Distance from origin: The distance of P(x,1y,1z1) from the origin o(0,0,0) is `sqrt((x_1-0)^2 +( y_2 - 0)^2 + (z_2 - 0)^2)` = `sqrt((x_1)^2 + (y_1)^2 + (z_1)^2)`
Example Problems on three Distance Theorem:
Problem1: Find ' a' so that the distance between ( -2,1,-3) and ( 1, 3, -6) be 7 units?
Solution:
Let A, B be ( -2,1,3) ,( 1, 3, -6) respectively then
`=>` `sqrt((a-2)^2 + (3-1)^2 + (-6 + 3 )^2)` = 7
`=>` a2 + a + 4 + 4 + 9 = 49 `=>` a2 + 4a - 32 = 0
`=>` (a-4) (a+ 8) = 0 `=>` a= 4 , -8
Problem2: Find the distance between the points A(1,2,3) from he point B(3, 6, 8)?
Solution: The distance
`sqrt((3-1)^2 +(6-3)^2 + ( 8-3)^2)`
AB = `sqrt(2^2 + 3 ^2 + 5^2)`
AB = `sqrt(4 + 9 + 25)`
AB = `sqrt(38)` <br>
Practice Problems on three Distance Theorem
Problem2: Find the distance of the point P( 4, 6, 8) from the point Q ( 9,5,2)? ( Answer: 3`sqrt(7)` )
problem3: By using the distance formula prove that the points A( 3, -5, 1), B ( -1, 0 , 8) and C( 7, -10, -6) are collinear ("Answer : The sum of two side is equal to the third side so the points are collinear)
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