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Last modified on February 28th, 2025
Let us imagine you are walking in a straight path in space. To describe your movement mathematically, we need a starting point and a direction. This is exactly what the vector equation of a line describes in mathematics.
In vector algebra, we use the vector equation of a line to describe lines in both 2-dimensional and 3-dimensional space. A vector is a quantity that has both magnitude (size) and direction. It is usually written in component form as:
v = (a, b), for 2D
Or,
v = (a, b, c), for 3D
Vectors represent direction and movement in space. When defining a line, we need:
A vector equation of a line is a representation of a straight line in 2D or 3D, which expresses all the points on the line using a position vector and a direction vector.
The vector equation of a line is given by:
r = r0 + td
Here,
This equation shows that as t increases or decreases, the vector r moves along the direction vector d and traces out all points on the line. A positive t moves forward, and a negative t moves backward along the line.
Note: The vector equation of a line can also be represented as:
r = (x0, y0, z0) + t(a, b, c)
⇒ (x, y, z) = (x0 + at, y0 + bt, z0 + ct)
Thus, x = x0 + at, y = y0 + bt, and z = z0 + ct
These equations are known as the parametric equations of a line that define the coordinates of any point on the line in terms of t.
Find the vector equation of a line passing through point A(1, 2) with a direction vector d = (3, -1).
As we know, the vector equation formula: r = r0 + td
Substituting r0 = (1, 2) and d = (3, -1)
Here,
r = (1, 2) + t(3, -1)
⇒ (x, y) = (1 + 3t, 2 – t), here t is a real number.
Thus, the vector equation is: r = (1 + 3t, 2 – t)
Find the vector equation of a line passing through point A(2, -1, 4) with a direction vector d = (3, 5, -2)
As we know, the vector equation formula: r = r0 + td
Substituting r0 = (2, -1, 4) and d = (3, 5, -2)
Here,
r = (2, -1, 4) + t(3, 5, -2)
⇒ (x, y, z) = (2 + 3t, -1 + 5t, 4 – 2t), here t is a real number
Thus, the vector equation is: r = (2 + 3t, -1 + 5t, 4 – 2t)
If two distinct points on the line are known, the vector equation of the line can be derived by first determining the direction vector and then applying the general equation of a line.
Let us consider two points A(x1, y1, z1) and B(x2, y2, z2).
The direction vector can be obtained by subtracting the coordinates of point A from those of point B.
d = B – A = (x2 – x1, y2 – y1, z2 – z1)
Thus, the general form of the vector equation of a line is:
r = A + t(B – A)
⇒ r = (x1, y1, z1) + t(x2 – x1, y2 – y1, z2 – z1)
Find the equation of a line passing through A(2, -1, 3) and B(5, 2, 6).
Here,
A = (2, -1, 3)
B = (5, 2, 6)
The direction vector is:
d = B – A
⇒ d = (5 – 2, 2 – (-1), 6 – 3)
⇒ d = (3, 3, 3)
As we know, the vector equation formula is:
r = A + t(B – A)
Here, substituting the values,
⇒ r = (2, -1, 3) + t(3, 3, 3)
⇒ r = (2 + 3t, -1 + 3t, 3 + 3t)
Thus, the vector equation is: r = (2 + 3t, -1 + 3t, 3 + 3t)
Vector equations can be converted into Cartesian equations, also known as the symmetric form. This is done by eliminating the parameter t from the vector equations.
Here are the key differences between the two forms:
Basis | Vector Equation | Cartesian Equation |
---|---|---|
General Form | r = r0 + td | ${\dfrac{x-x_{0}}{a}=\dfrac{y-y_{0}}{b}=\dfrac{z-z_{0}}{c}}$ |
For Two Points | r = A + t(B – A) | ${\dfrac{x-x_{0}}{x_{1}-x_{0}}=\dfrac{y-y_{0}}{y_{1}-y_{0}}=\dfrac{z-z_{0}}{z_{1}-z_{0}}}$ |
Representation | Expressed using position and direction vectors | Expressed using algebraic fractions by eliminating t |
Parameter | Includes the parameter t, which varies over all real numbers | Includes no parameter, only algebraic equality |
Last modified on February 28th, 2025