How to Calculate Distance Point to Point, Line, and Field

How to Calculate Distance Titikke Points, Lines and Fields - Have you ever played a rubik? Rubik is a puzzle game that has a 3-dimensional shape. Shape Rubik in general is a cube, as you can see below:

Do you know how long diagonal and diagonal field in a rubik? to answer that you must understand concepts and formulas are looking diagonal field and diagonal space. long diagonal field and diagonal space is a long distance from point to point in a cube. in this post Mathematical formulas Basic will discuss it. Listen carefully following explanation:

How to Calculate Distance Point to Point, Line, and Field

There are three possibilities occur to notch a point to point, line, or areas, namely:

DISTANCE POINT TO POINT THE OTHERS

Will you observe the image the following:

In this figure, there are two points, namely point A and point B. the distance of these two points can be ten tukan by connecting point A and point B with a line. The length of the line that determines the distance the two points. Thus, the distance from point A to point B is the length of a line segment that connects the two.

Consider the example problems below:

Example Problem 1:

Note cube image ABCD.EFGH following:

If the length of the ribs on the cube is 6 cm above and X is a mid point between ribs AB. Then calculate the distance:

a. the point H to point A

b. the point H to point X

c. H point to point B

d. Point E to point X

Solved:

a.) The point H to the point A is poanjang AH line. Line AH is the diagonal length of the sides of the cube so we can use the theorem of Pythagoras following:

AH = √ (EH2 + AE ² )

AH = √ (6 ² + 6 ² )

AH = √ (36 + 36)

AH = √72 [1945921million]

AH = 6√2 [1945921million]

[1945921million]

b.) within a point H to point X is the length of the line HX. Long AX equal to half the length of the ribs AB, then:

AX = 1/2 AB = 1/2 x 6 x m = 3 cm

using theorem Pythagoras:

HX = √ (AH ² + AX ​​ ² )

HX = √ ((6√2) 2 + 3 ² )

HX = √ (72 + 9)

HX = √81

HX = 9 cm

c.) distance point H to point B is the length of the line BH. Line BH is the diagonal length of the space in the cube, and thus we can use the theorem of Pythagoras:

BH = √ (AH ² + AB ² ) [1945921million]

BH = √ ((6√ ² ) 2 + 6 ² ) [1945921million]

BH = √ (72 + 36 )

BH = √108

BH = 6√3 cm

d.) Distance to point E to point X aalah length of the line EAX. AX length equal to half the length of the ribs AB, then:

AX = 1/2 AB = 1/2 x 6 x m = 3 cm

By using the theorem of Pythagoras:

EX = √ (AE ² + AX ​​ ² )

EX = √ (6 ² + 3 ² )

EX = √ (36 + 9)

EX = √45 [1945921million]

EX = 3√5 cm [1945921million]

[1945921million]

POINT DISTANCE TO OUTLINE

Observe the following picture:

[1945921million]

In this figure, there is a point A and line g. The distance between point A at line g obtained by drawing haris from point A to line g, the line stops at the point P so that creates the AP line perpendicular to line g. the distance from point A to line g is the length of the line AP. Thus, the distance between the point of the segment length of the line is a line drawn from that point is perpendicular to the line.

Consider the example to the following:

Example Problem 2:

Note ABCD.EFGH cube image below:

if the length of ribs on the cube above is 6 cm and point X is halfway between the ribs AB, then calculate: [1945921million]

[1945921million]

a. distance of the point X to the line DE [1945921million]

b. distance point X to the line CE

Solved: [1945921million]

Because of this problem exactly the same as a matter of first instance, it will be used on the calculation of the sample questions 1. [1945921million]

We create a first picture like this:

a.) Distance [1945923million] point X to the line DE is the length of the line from point X to point M whose position perpendicular to the line DE, as shown below:

DE = AH and ME = ½ AH = DE = ½ ½ 6√2 = 3√2 [1945921million]

By using the Pythagorean theorem: [1945921million]

MX = √ (EX ² - ME ² )

MX = √ ((3√5) 2 - (3√2) ² )

MX = √ (45-18)

MX = √27 [1945921million]

MX = 3√3 cm [1945921million]

[1945921million]

b) Distance to point X to the CE line is [1945923million] length of the line from point X to point N whose position perpendicular to the CE, as shown below:

CE = BH and NE = CE = ½ ½ ½ BH = 6√3 = 3√3 [1945921million]

By using the Pythagorean theorem: [1945921million]

NX = √ (EX ² - NE ² ) [1945921million]

NX = √ ((3√5) ² - (3√3) ² )

NX = √ (45-27)

NX = √18

NX = 3√2 cm

pOINT DISTANCE tO fIELD

Consider the following picture:

in the picture there a tiktik a and α field. Distance from point A to field α can be determined by connecting point A is perpendicular to the plane α. Thus, the distance from a point to a plane is the distance from that point to the projection on the field.

Consider the example to the following:

Example Problem 3:

Note ABCD.EFGH cube image below:

If long-cube on top is 6 cm and X is the mid-point between ribs AB. Then calculate the distance from point X to the field CDEF!

Solved:

Draw a picture like the following:

[1945921million]

the point X to the field CDEF is the length of the line from point X to point Z that is perpendicular to the plane CDEF.

XZ = AH = ½ ½ 6√2 = 3√2 cm

So presumably the explanation long enough about How to Calculate Distance point to point, Line, and field. I hope you can understand it well.

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