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112 LearnersLast updated on October 21, 2025

A vector is a quantity that has both magnitude and direction. It is represented using an arrow. The length of the arrow reflects the magnitude, and the orientation indicates the direction.
Vector equations use direction vectors and variables to describe planes or lines in three-dimensional space. A vector with a magnitude of 1 is a unit vector. In a three-dimensional space, all positions are described using 3 axes, the x-axis, y-axis, and z-axis, and each axis has a unit vector. The unit vector for the x-axis is \(\hat {i}\), the unit vector for the y-axis is \(\hat {j}\), and the unit vector for the z-axis is \(\hat {k}\).
In a 3D space, a vector is written as:
\(\vec{r} = {x {\hat {i}}} + {y {\hat {j}}} + {z {\hat {k}}}\)Where x, y, and z represent the scalar components of the vector.
The vector equation of a line in three-dimensional space is: \(\vec{r} = \vec{a} + \lambda{{\vec{b}} }\) where \(λ∈R\)
The vector equation of a plane in three-dimensional space is: \(\vec{r} \cdot {\vec n} = d\).
A Cartesian equation is an algebraic expression that shows the relationship between variables (typically x, y, and z) using the Cartesian coordinate system. It is used to describe geometric figures such as lines, curves, and surfaces in terms of their positions on a coordinate plane or in space. Here are some key differences between vector and Cartesian equations.
|
Feature |
Vector Equation |
Cartesian Equation |
|
Form |
Uses vectors and vector units. |
Uses x, y, and z coordinates directly. |
|
Representation |
Line : \({\vec r = \vec a + \lambda \vec b}\) Plane: \(\vec r \cdot \hat n = d\) |
Line: \({{x - x_1} \over a} = {{y - y_1} \over b} = {{z-z_1} \over c} \) Plane: \(ax + by + cz = d\) |
|
Nature |
Vector-based, compact, and geometric |
coordinate-based, algebraic |
|
Variables used |
Vectors and scalars |
Coordinates |
|
Visualization |
Easier to visualize in space using direction and position vectors |
Useful for calculations and plotting on a coordinate grid |
|
Used in |
Physics and mechanics |
Algebraic calculations and graphing |
In a three-dimensional space, the vector equation of a line uses vectors to describe all the points that lie on a straight line.
The vector equation of a line that passes through a single point:
\(\vec r = \vec a + \lambda \vec b \)
Here,
\(\vec r\) represents the position vector of any point on the line
While \(\vec a\) represents the position vector of a fixed point through which the line passes
\(\vec b\) = direction vector of the line
and \(\lambda\) = scalar parameter that varies.
Vector equation for a line passing through two points:
\(\vec r = \vec a + \lambda(\vec b - \vec a)\)
Here,
\(\vec a\) (position vector of point A)
\(\vec a = x_1 \hat i + y_1 \hat j + z_1 \hat k \)
\(\vec b\) (position vector of point B)
\(\vec b = x_2 \hat i + y_2 \hat j + z_2 \hat k \)
The direction vector is \(\vec b - \vec a\), pointing from point A to point B.
\(λ∈R\) is a scalar parameter
The vector equation represents motion from a fixed point in a plane along two independent directions.
\(\vec r = \vec a + s \vec u + t \vec v\)
Here, \(\vec r\) is the position vector of any point P(x, y, z) lying on the plane
\(\vec a\) is vector of a specific point on the plane
\(\vec u, \vec v\) are two non-parallel direction vectors that lie on the plane.
s and t are scalar parameters; they can be any real number.
Vector equations help us represent lines, planes, and directions in space using vectors. In this section, we will discuss some tips and tricks to master vector equations. This help students to solve problems in physics and mathematics much more easily.
Always separate the components, that is, break the vectors into x, y, and z components before solving.
use unit vectors, express the vectors as combinations of i, j, and k. It simplifies addition, subtraction, and scalar multiplication.
Memorize the basic formulas like:
Angle between vectors: \(cos{\theta} = {{{\vec a \cdot \vec b} \over |\vec a| |\vec b|}}\).
Perpendicular vectors: Dot product = 0 \({(\vec a \cdot \vec b = 0 )}\)Magnitude: \({{|\vec v| = \sqrt {x^2 + y^2 + z^2}}}\)
If a vector is a multiple of another vector, then the vectors are parallel. For example, (2, 4) is parallel to (1, 2).
Students can draw a simple diagram to visualize how the vectors look, where they point, and how they relate to each other. This help students to understand if they are parallel, perpendicular, or intersecting.
Vector equations are crucial in solving 3D geometry problems and various subjects like physics and engineering. To ensure a better understanding and precise results, avoid the following frequent errors that get overlooked:
Vector equations represent quantities having both magnitude and direction. They play a crucial role in various real-life, direction-based calculations:
\(\vec p_{new} = \vec p _{original} + \vec v \cdot t \)
Find the vector equation of a line passing through a point A(1,2) and having a direction vector d = 34
\(\vec r (\lambda) = \begin{bmatrix} 1 + 3\lambda \\ 2 + 4\lambda \end{bmatrix} \)
The vector equation of a line is:
\(\vec r (\lambda) = \vec a + \lambda \vec d\)
Where:
\(\vec a\) = \(\begin{bmatrix} 1 \\ 2 \end{bmatrix} \) position vector of A
\(\vec d = \begin{bmatrix} 3 \\ 4 \end{bmatrix} \)
\(\vec r (\lambda) = \begin{bmatrix} 1 \\ 2 \end{bmatrix} + \lambda\begin{bmatrix} 4 \\ 3 \end{bmatrix} \)
\(= \begin{bmatrix} 1 + 3\lambda \\ 2 + 4\lambda \end{bmatrix} \)
Let u = [2, -1, 4] and v = [-3, 0, 1]. Find u + v
\(\vec u + \vec v = {{\begin{bmatrix} -1 \\ -1 \\ 5 \\ \end{bmatrix} }}\)
\(\vec u + \vec v = \begin{bmatrix} 2 \\ -1 \\ 4 \\ \end{bmatrix} + \begin{bmatrix} -3 \\ 0\\ 1\\ \end{bmatrix} \)
\(\begin{bmatrix} 2 - 3 \\ -1 + 0\\ 4 + 1 \\ \end{bmatrix} \)
\(\begin{bmatrix} -1 \\ -1\\ 5 \\ \end{bmatrix} \)
Let a = [3 4] and b = [4 3]. Find the angle Θ between them
\(\theta = 16.26 ^\circ\)
Finding \(\vec a \cdot \vec b\) to find the angle \({{{\theta }}}\),
\({{\vec a \cdot \vec b }} = {3 \times 4} + {4 \times 3}\\ = 12 + 12 \\ = 24\)
Finding the magnitudes of \(\vec a {\text { and }} \vec b\)
\(|\vec a| = \sqrt {3^2 + 4^2} = 5,\\ |\vec b| = \sqrt {4^2 + 3^2} = 5\\ \)
To find the value of \({{{\theta }}}\), we use the formula:
\({{\vec a \cdot \vec b }} = {{|\vec a| |\vec b| cos {\theta}}}\)
\({{\vec a \cdot \vec b }} = {{|\vec a| |\vec b| cos {\theta}}} \implies 24= 5 \times 5 cos {\theta} \\\implies cos {\theta } = {24 \over 25}\\ \implies {\theta } = cos^{-1} {{({24\over 25})}} \approx 16.26 ^\circ\)
Project vector a = [2 4] onto vector b = [1 2]
\({{{\text { Proj} _ {\vec b} {\vec a}}}} = \begin{bmatrix} 2 \\ 4 \\ \end {bmatrix} \)
\( \text{Proj}_{\vec{b}} \vec{a} = {({\vec a \cdot \vec b \over \vec b \cdot \vec b})} \vec b \)
Finding \({\vec a \cdot \vec b} = {2 \times 1 + 4 \times 2} = 10\\ {\vec b \cdot \vec b} = {1^2 + 2^2} = 5\\ \)
\( \text{Proj}_{\vec{b}} \vec{a} = {{({10 \over 5})}} \begin{bmatrix}1 \\ 2\\ \end {bmatrix}\\ = 2 \begin{bmatrix}1 \\ 2\\ \end {bmatrix} \\ = \begin{bmatrix}2 \\ 4\\ \end {bmatrix}\)
Find the vector equation of a plane passing through point A(1,0,2) with direction vectors
\(\vec r (s, t) = \begin {bmatrix} 1 + s \\ 2s + t\\ 2 + 3t\\ \end {bmatrix}\)
The vector equation for the palne is:
\(\vec r (s, t) = \vec a + s \vec u + t \vec v\)
Here, a = (1, 0, 2)
\(\vec u = \begin {bmatrix} 1\\ 2\\ 0\\ \end {bmatrix}\)
\(\vec v = \begin {bmatrix} 0\\ 1\\ 3\\ \end {bmatrix}\)
So, \(\vec r (s, t) = \begin {bmatrix} 1\\ 0\\ 2\\ \end {bmatrix} + s \begin {bmatrix} 1\\ 2\\ 0\\ \end {bmatrix} + t \begin {bmatrix} 0\\ 1\\ 3\\ \end {bmatrix}\)
\(= \begin {bmatrix} 1 + s\\ 2s + t \\ 2s + 3t\\ \end {bmatrix}\)
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.






