how to find a vector from two points in 3d

We all know the very pop equation of the straight line Y = chiliad . 10 + C which a direct line in a aeroplane. Just here we are going to discuss the Equation of a Direct Line in 3-dimensional space. A Straight Line is uniquely characterized if it passes through the 2 unique points or it passes through a unique signal in a definite direction. In 3 Dimensional Geometry lines (straight lines) are usually represented in the ii forms Cartesian Grade and Vector form. Here we are going to hash out the two-point form of a straight line in 3-dimensions using both cartesian too as vector course.

Equation of a Straight Line in Cartesian Course

For writing the equation of a straight line in the cartesian grade nosotros crave the coordinates of a minimum of ii points through which the direct line passes. Let's say (x₁, y_ane, z₁) and (x₂, y₂, z₂) are the position coordinates of the two fixed points in the 3-dimensional space through which the line passes.

At present to obtain the equation we have to follow these three steps:

Pace ane: Find the DR's (Direction Ratios) past taking the divergence of the corresponding position coordinates of the two given points. l = (x₂ – 10₁), chiliad = (y₂ – y₁), n = (z₂ – z_i); Here l, one thousand, northward are the DR'due south.
Step 2: Choose either of the two given points say, we choose (ten₁, y_i, z₁).
Step 3: Write the required equation of the directly line passing through the points (x_ane, y_i, z_i) and (x₂, y₂, z_two). L : (x – 10₁)/l = (y – y₁)/m = (z – z₁)/n

Where (ten, y, z) are the position coordinates of whatsoever variable signal lying on the straight line.

Example ane: If a direct line is passing through the two fixed points in the three-dimensional whose position coordinates are P (two, iii, 5) and Q (4, 6, 12) so its cartesian equation using the two-bespeak form is given past

Solution:

fifty = (four – 2), m = (6 – 3), n = (12 – five)

50 = two, m = 3, north = vii

Choosing the point P (2, 3, v)

The required equation of the line

L : (ten – 2) / 2 = (y – iii) / 3 = (z – 5) / seven

Example 2: If a direct line is passing through the two fixed points in the three-dimensional whose position coordinates are A (two, -i, iii) and B (4, two, 1) then its cartesian equation using the two-point form is given by

Solution:

l = (4 – 2), thou = (two – (-one)), n = (ane – 3)

l = two, m = 3, due north = -2

Choosing the point A (2, -1, 3)

The required equation of the line

50 : (10 – 2) / 2 = (y + i) / 3 = (z – 3) / -2 or

Fifty : (10 – 2) / 2 = (y + 1) / three = (3 – z) / 2

Example three: If a direct line is passing through the two fixed points in the 3-dimensional whose position coordinates are X (2, 3, 4) and Y (5, 3, 10) then its cartesian equation using the 2-point form is given past

Solution:

fifty = (5 – 2), m = (3 – 3), north = (10 – 4)

l = 3, m = 0, n = 6

Choosing the betoken X (ii, 3, four)

The required equation of the line

L : (ten – two) / iii = (y – three) / 0 = (z – iv) / half dozen or

L : (ten – 2) / 1 = (y – iii) / 0 = (z – 4) / 2

Equation of a Directly Line in Vector Form

For writing the equation of a directly line in the vector grade nosotros require the position vectors of a minimum of two points through which the straight line passes. Allow'south say ${\vec {a}}$ and ${\vec {n}}$ are the position vectors of the two fixed points in the three-dimensional space through which the line passes.

At present to obtain the equation we accept to follow these three steps:

Where ${\vec {r}}$ is the position vector of whatsoever variable point lying on the directly line and t is the parameter whose value is used to locate any bespeak on the line uniquely.

Case 1: If a straight line is passing through the two fixed points in the 3-dimensional whose position vectors are (ii i + 3 j + v k) and (4 i + half-dozen j + 12 m) then its Vector equation using the two-betoken form is given by

Solution:

${\vec {p}}$ = (4 i + six j + 12 chiliad) – (2 i + 3 j + 5 k)

${\vec {p}}$ = (two i + iii j + 7 k) ; Here ${\vec {p}}$ is a vector parallel to the straight line

Choosing the position vector (2 i + 3 j + 5 thousand)

The required equation of the straight line

L : ${\vec {r}}$ = (ii i + 3 j + 5 k) + t . (2 i + 3 j + 7 k)

Example 2: If a straight line is passing through the two fixed points in the 3-dimensional space whose position coordinates are (iii, 4, -7) and (1, -1, 6) then its vector equation using the two-betoken class is given by

Solution:

Position vectors of the given points volition be (three i + four j – 7 m) and (i – j + half dozen m)

${\vec {p}}$ = (3 i + 4 j – vii k) – (i – j + six k)

${\vec {p}}$ = (2 i + five j – 13 thou) ; Here ${\vec {p}}$ is a vector parallel to the directly line

Choosing the position vector (i – j + six chiliad)

The required equation of the straight line

L : ${\vec {r}}$ = (i – j + six k) + t . (two i + 5 j – xiii one thousand)

Case 3: If a direct line is passing through the ii stock-still points in the iii-dimensional whose position vectors are (5 i + iii j + 7 k) and (2 i + j – 3 one thousand) and then its Vector equation using the two-point grade is given by

Solution:

${\vec {p}}$ = (5 i + 3 j + 7 k) – (2 i + j – 3 k)

${\vec {p}}$ = (iii i + 2 j + 10 k) ; Here ${\vec {p}}$ is a vector parallel to the straight line

Choosing the position vector (2 i + j – 3 k)

The required equation of the straight line

L : ${\vec {r}}$ = (ii i + j – 3 k) + t . (2 i + 3 j + 7 k)

Source: https://www.geeksforgeeks.org/equation-of-a-line-in-3d/

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