3 Forms Of A Line

We all know the very popular equation of the straight line Y = 1000 . X + C which a direct line in a plane. But here we are going to discuss the Equation of a Straight Line in iii-dimensional space. A Straight Line is uniquely characterized if it passes through the two unique points or it passes through a unique point in a definite direction. In Iii Dimensional Geometry lines (straight lines) are commonly represented in the 2 forms Cartesian Form and Vector form. Here we are going to hash out the 2-signal class of a direct line in 3-dimensions using both cartesian too every bit 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 two points through which the direct line passes. Permit'southward say (x₁, y₁, z₁) and (x₂, y₂, z₂) are the position coordinates of the ii fixed points in the 3-dimensional space through which the line passes.

Now to obtain the equation we take to follow these three steps:

Footstep i: Notice the DR's (Direction Ratios) by taking the difference of the corresponding position coordinates of the two given points. fifty = (x_two – x_i), thou = (y₂ – y₁), northward = (z₂ – z_one); Here fifty, 1000, due north are the DR's.
Step ii: Choose either of the 2 given points say, we cull (10₁, y₁, z₁).
Step 3: Write the required equation of the direct line passing through the points (x_one, y₁, z₁) and (x₂, y₂, z₂). L : (10 – 10_ane)/l = (y – y₁)/m = (z – z_i)/n

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

Example one: If a directly line is passing through the two fixed points in the iii-dimensional whose position coordinates are P (2, iii, 5) and Q (4, 6, 12) then its cartesian equation using the two-indicate form is given by

Solution:

l = (4 – 2), m = (6 – iii), n = (12 – 5)

l = 2, m = 3, n = 7

Choosing the point P (2, 3, 5)

The required equation of the line

50 : (ten – 2) / two = (y – 3) / 3 = (z – five) / seven

Instance 2: If a straight line is passing through the two fixed points in the 3-dimensional whose position coordinates are A (two, -ane, 3) and B (4, 2, 1) and then its cartesian equation using the 2-point grade is given by

Solution:

l = (4 – two), m = (ii – (-1)), n = (1 – 3)

l = 2, yard = 3, northward = -2

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

The required equation of the line

L : (x – 2) / 2 = (y + one) / three = (z – three) / -ii or

L : (10 – 2) / ii = (y + 1) / iii = (iii – z) / 2

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

Solution:

l = (5 – 2), thou = (3 – three), n = (10 – 4)

l = 3, m = 0, n = six

Choosing the signal 10 (2, three, 4)

The required equation of the line

L : (x – two) / three = (y – 3) / 0 = (z – 4) / 6 or

L : (ten – 2) / 1 = (y – 3) / 0 = (z – 4) / 2

Equation of a Directly Line in Vector Course

For writing the equation of a direct line in the vector grade we require the position vectors of a minimum of ii points through which the direct line passes. Let'south say ${\vec {a}}$ and ${\vec {n}}$ are the position vectors of the two fixed points in the three-dimensional infinite through which the line passes.

Now to obtain the equation we have 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 whatsoever point 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 (2 i + 3 j + v k) and (4 i + 6 j + 12 k) then its Vector equation using the two-betoken course is given by

Solution:

${\vec {p}}$ = (4 i + half dozen j + 12 k) – (ii i + 3 j + 5 k)

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

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

The required equation of the straight line

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

Case 2: If a directly line is passing through the two stock-still points in the three-dimensional space whose position coordinates are (iii, 4, -7) and (one, -1, 6) and then its vector equation using the 2-point form is given by

Solution:

Position vectors of the given points volition be (iii i + 4 j – 7 k) and (i – j + 6 k)

${\vec {p}}$ = (three i + iv j – seven k) – (i – j + half-dozen k)

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

Choosing the position vector (i – j + 6 k)

The required equation of the straight line

L : ${\vec {r}}$ = (i – j + 6 k) + t . (2 i + v j – xiii k)

Example 3: If a straight line is passing through the 2 stock-still points in the 3-dimensional whose position vectors are (5 i + three j + seven thou) and (2 i + j – 3 one thousand) then its Vector equation using the ii-signal form is given past

Solution:

${\vec {p}}$ = (5 i + 3 j + seven yard) – (2 i + j – 3 chiliad)

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

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

The required equation of the straight line

50 : ${\vec {r}}$ = (2 i + j – 3 m) + t . (2 i + 3 j + seven thousand)