Vector Algebra
Practice MCQsVector Algebra is the branch of mathematics that deals with quantities having both magnitude and direction.
Vector Algebra is the branch of mathematics that deals with quantities having both magnitude and direction. Vectors are used in geometry, physics, engineering and mechanics to represent displacement, velocity, acceleration, force, work and moment. This chapter covers vectors in two and three dimensions, magnitude and direction of a vector, unit and null vectors, addition of vectors, scalar multiplication, scalar product, vector product and applications such as work done by a force, moment of a force and geometrical problems.
What is a Vector?
A vector is a quantity that has both magnitude and direction. For example, displacement, velocity, acceleration and force are vector quantities because direction is important along with size.
A quantity that has only magnitude and no direction is called a scalar. For example, mass, time, temperature, distance and speed are scalar quantities.
| Concept | Meaning | Example |
|---|---|---|
| Scalar Quantity | Quantity having magnitude only. | Mass, time, speed, temperature. |
| Vector Quantity | Quantity having magnitude and direction. | Force, displacement, velocity. |
| Two-Dimensional Vector | Vector with two components. | \(\vec{a}=x\hat{i}+y\hat{j}\) |
| Three-Dimensional Vector | Vector with three components. | \(\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}\) |
| Position Vector | Vector representing position of a point from origin. | For \(P(x,y,z)\), \(\vec{OP}=x\hat{i}+y\hat{j}+z\hat{k}\) |
“Vectors describe both size and direction, making them powerful tools for geometry and mechanics.”
Key points
- Vectors have magnitude and direction.
- Scalars have magnitude only.
- Unit vector has magnitude 1.
- Null vector has magnitude 0.
- Vectors can be added component-wise.
- Scalar multiplication changes magnitude and possibly direction.
- Dot product gives a scalar.
- Cross product gives a vector.
- Work done uses dot product.
- Moment of force uses cross product.
Core Formula Bank
Tip: Dot product is linked with projection and work. Cross product is linked with area, direction and moment.
Vector Formula and Application Guide
Use this guide to quickly identify which vector operation is required in a problem.
| Question Type | What to Use | Typical Clue |
|---|---|---|
| Length of vector | Magnitude formula | Find size or modulus of vector |
| Direction only | Unit vector | Find vector of magnitude 1 |
| Resultant vector | Vector addition | Add two or more vectors |
| Angle between vectors | Dot product | Contains \(\cos\theta\) |
| Perpendicular vectors | Dot product equals zero | \(\vec{a}\cdot\vec{b}=0\) |
| Area of parallelogram | Cross product magnitude | \(|\vec{a}\times\vec{b}|\) |
| Work done by force | Dot product | Force and displacement |
| Moment of force | Cross product | Position vector and force |
Vectors in Two and Three Dimensions
Vectors can be represented using unit vectors along coordinate axes. In two dimensions, the standard unit vectors are \(\hat{i}\) and \(\hat{j}\). In three dimensions, the standard unit vectors are \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\).
| Vector Type | Representation | Example |
|---|---|---|
| Two-Dimensional Vector | \(\vec{a}=x\hat{i}+y\hat{j}\) | \(\vec{a}=3\hat{i}+4\hat{j}\) |
| Three-Dimensional Vector | \(\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}\) | \(\vec{a}=2\hat{i}-3\hat{j}+5\hat{k}\) |
| Position Vector in 2D | For point \(P(x,y)\), \(\vec{OP}=x\hat{i}+y\hat{j}\) | For \(P(2,5)\), \(\vec{OP}=2\hat{i}+5\hat{j}\) |
| Position Vector in 3D | For point \(P(x,y,z)\), \(\vec{OP}=x\hat{i}+y\hat{j}+z\hat{k}\) | For \(P(1,2,3)\), \(\vec{OP}=\hat{i}+2\hat{j}+3\hat{k}\) |
| Vector Joining Two Points | If \(A(x_1,y_1,z_1)\), \(B(x_2,y_2,z_2)\), then \(\vec{AB}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}\) | Subtract initial point from final point. |
Magnitude and Direction of a Vector
The magnitude of a vector is its length. The direction tells where the vector points. Direction in two dimensions is often represented using the angle made with the positive x-axis.
| Vector | Magnitude | Direction Information |
|---|---|---|
| \(\vec{a}=x\hat{i}+y\hat{j}\) | \(|\vec{a}|=\sqrt{x^2+y^2}\) | \(\tan\theta=\frac{y}{x}\), where \(\theta\) is direction angle. |
| \(\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}\) | \(|\vec{a}|=\sqrt{x^2+y^2+z^2}\) | Direction can be described using direction cosines. |
| Direction Cosines | \(l=\frac{x}{|\vec{a}|},\ m=\frac{y}{|\vec{a}|},\ n=\frac{z}{|\vec{a}|}\) | \(l^2+m^2+n^2=1\) |
Unit Vector and Null Vector
A unit vector has magnitude 1 and gives direction only. A null vector has magnitude 0 and no definite direction.
| Vector Type | Definition | Formula / Example |
|---|---|---|
| Unit Vector | A vector whose magnitude is 1. | \(\hat{a}=\frac{\vec{a}}{|\vec{a}|}\), where \(\vec{a}\neq \vec{0}\) |
| Null Vector | A vector whose magnitude is 0. | \(\vec{0}=0\hat{i}+0\hat{j}+0\hat{k}\) |
| Equal Vectors | Vectors having same magnitude and same direction. | Position may differ, but magnitude and direction are same. |
| Negative Vector | Vector having same magnitude but opposite direction. | Negative of \(\vec{a}\) is \(-\vec{a}\) |
Addition of Vectors and Scalar Multiplication
Vectors are added component-wise. Multiplication of a vector by a scalar changes its magnitude. If the scalar is negative, the direction of the vector is reversed.
| Operation | Rule | Example |
|---|---|---|
| Addition | If \(\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}\) and \(\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\), then \(\vec{a}+\vec{b}=(a_1+b_1)\hat{i}+(a_2+b_2)\hat{j}+(a_3+b_3)\hat{k}\) | Add corresponding components. |
| Subtraction | \(\vec{a}-\vec{b}=(a_1-b_1)\hat{i}+(a_2-b_2)\hat{j}+(a_3-b_3)\hat{k}\) | Subtract corresponding components. |
| Scalar Multiplication | If \(k\) is a scalar, then \(k\vec{a}=ka_1\hat{i}+ka_2\hat{j}+ka_3\hat{k}\) | Multiply every component by \(k\). |
| Resultant Vector | Single vector equivalent to two or more vectors. | \(\vec{R}=\vec{a}+\vec{b}\) |
Scalar Product or Dot Product
The scalar product or dot product of two vectors gives a scalar quantity. It is used to find angle between vectors, projection, perpendicularity and work done.
| Form | Formula | Use |
|---|---|---|
| Geometrical Form | \(\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta\) | Used when magnitudes and angle are known. |
| Component Form | If \(\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}\), \(\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\), then \(\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2+a_3b_3\) | Used when components are given. |
| Angle Between Vectors | \(\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\) | Used to find angle. |
| Perpendicular Vectors | \(\vec{a}\cdot\vec{b}=0\) | Condition for non-zero perpendicular vectors. |
| Parallel Same Direction | \(\theta=0^\circ\), so \(\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\) | Maximum positive dot product. |
| Parallel Opposite Direction | \(\theta=180^\circ\), so \(\vec{a}\cdot\vec{b}=-|\vec{a}||\vec{b}|\) | Maximum negative dot product. |
Vector Product or Cross Product
The vector product or cross product of two vectors gives another vector. The resulting vector is perpendicular to both the given vectors.
| Form | Formula | Use |
|---|---|---|
| Magnitude Form | \(|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta\) | Used when magnitudes and angle are known. |
| Direction | Direction is perpendicular to both \(\vec{a}\) and \(\vec{b}\). | Found using right-hand rule. |
| Area of Parallelogram | Area \(=|\vec{a}\times\vec{b}|\) | Geometrical application. |
| Area of Triangle | Area \(=\frac{1}{2}|\vec{a}\times\vec{b}|\) | Triangle formed by two vectors. |
| Parallel Vectors | \(\vec{a}\times\vec{b}=\vec{0}\) | Condition for non-zero parallel vectors. |
| Unit Vector Cross Products | \(\hat{i}\times\hat{j}=\hat{k}\), \(\hat{j}\times\hat{k}=\hat{i}\), \(\hat{k}\times\hat{i}=\hat{j}\) | Cyclic order gives positive direction. |
Applications of Vector Algebra
Vector algebra has important applications in mechanics and geometry. Dot product is commonly used for work done by a force, while cross product is used for moment of a force and area calculations.
| Application | Formula | Meaning |
|---|---|---|
| Work Done by a Force | \(W=\vec{F}\cdot\vec{s}=|\vec{F}||\vec{s}|\cos\theta\) | Work is scalar. It depends on force, displacement and angle between them. |
| Moment of a Force | \(\vec{M}=\vec{r}\times\vec{F}\) | Moment measures turning effect of force. |
| Area of Parallelogram | \(|\vec{a}\times\vec{b}|\) | Area formed by two adjacent side vectors. |
| Area of Triangle | \(\frac{1}{2}|\vec{a}\times\vec{b}|\) | Area formed by two side vectors from same vertex. |
| Projection of Vector | Projection of \(\vec{a}\) on \(\vec{b}\) is \(\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\) | Component of one vector along another. |
| Angle Between Vectors | \(\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\) | Geometrical relationship between vectors. |
Step-by-Step Solving Method
| Step | Action | Example Focus |
|---|---|---|
| Step 1 | Identify what is asked. | Magnitude, unit vector, sum, dot product, cross product, work, moment. |
| Step 2 | Write vectors in component form. | \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) form. |
| Step 3 | Select the correct operation. | Dot product for scalar result, cross product for vector result. |
| Step 4 | Apply the formula carefully. | Use component-wise calculation or magnitude-angle formula. |
| Step 5 | Interpret the result. | Scalar, vector, area, work, moment, angle or direction. |
Solved Examples
| Question | Method | Answer |
|---|---|---|
| Find the magnitude of \(\vec{a}=3\hat{i}+4\hat{j}\). | \[ |\vec{a}|=\sqrt{3^2+4^2}=\sqrt{9+16}=5 \] | 5 |
| Find the magnitude of \(\vec{a}=2\hat{i}-3\hat{j}+6\hat{k}\). | \[ |\vec{a}|=\sqrt{2^2+(-3)^2+6^2} = \sqrt{4+9+36} = 7 \] | 7 |
| Find unit vector in the direction of \(\vec{a}=3\hat{i}+4\hat{j}\). | \[ |\vec{a}|=5 \] \[ \hat{a}=\frac{\vec{a}}{|\vec{a}|} = \frac{3}{5}\hat{i}+\frac{4}{5}\hat{j} \] | \(\frac{3}{5}\hat{i}+\frac{4}{5}\hat{j}\) |
| If \(\vec{a}=2\hat{i}+3\hat{j}\) and \(\vec{b}=4\hat{i}-\hat{j}\), find \(\vec{a}+\vec{b}\). | \[ \vec{a}+\vec{b}=(2+4)\hat{i}+(3-1)\hat{j} \] | \(6\hat{i}+2\hat{j}\) |
| If \(\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}\) and \(\vec{b}=\hat{i}-2\hat{j}+3\hat{k}\), find \(\vec{a}\cdot\vec{b}\). | \[ \vec{a}\cdot\vec{b}=2(1)+3(-2)+4(3) \] \[ =2-6+12=8 \] | 8 |
| Find angle between \(\vec{a}=\hat{i}\) and \(\vec{b}=\hat{j}\). | \[ \vec{a}\cdot\vec{b}=0 \] Since both are non-zero and dot product is zero, vectors are perpendicular. | \(90^\circ\) |
| Find \(\hat{i}\times\hat{j}\). | By unit vector cross product rule: \[ \hat{i}\times\hat{j}=\hat{k} \] | \(\hat{k}\) |
| If \(|\vec{a}|=5\), \(|\vec{b}|=6\), and angle between them is \(30^\circ\), find \(\vec{a}\cdot\vec{b}\). | \[ \vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos 30^\circ \] \[ =5\times 6\times \frac{\sqrt{3}}{2}=15\sqrt{3} \] | \(15\sqrt{3}\) |
| If \(|\vec{a}|=4\), \(|\vec{b}|=5\), and angle between them is \(90^\circ\), find \(|\vec{a}\times\vec{b}|\). | \[ |\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin 90^\circ \] \[ =4\times 5\times 1=20 \] | 20 |
| A force \(\vec{F}=3\hat{i}+4\hat{j}\) moves a body through displacement \(\vec{s}=5\hat{i}+2\hat{j}\). Find work done. | \[ W=\vec{F}\cdot\vec{s} \] \[ W=3(5)+4(2)=15+8=23 \] | 23 units |
Note: Use dot product when the answer is scalar. Use cross product when the answer has direction.
Common Traps and Shortcuts
Common Traps
- Confusing scalar and vector quantities.
- Forgetting that magnitude is always non-negative.
- Trying to find unit vector of a null vector.
- Adding vectors without matching components.
- Confusing dot product with cross product.
- Writing dot product answer as a vector.
- Writing cross product answer as a scalar.
- Forgetting that \(\vec{a}\times\vec{b}=-(\vec{b}\times\vec{a})\).
- Using cross product for work done instead of dot product.
- Using dot product for moment of force instead of cross product.
Useful Shortcuts
- For magnitude, square components and add.
- For unit vector, divide vector by its magnitude.
- For vector addition, add like components.
- For perpendicular vectors, dot product is zero.
- For parallel vectors, cross product is zero.
- For work done, use \(W=\vec{F}\cdot\vec{s}\).
- For moment of force, use \(\vec{M}=\vec{r}\times\vec{F}\).
- For area of parallelogram, use \(|\vec{a}\times\vec{b}|\).
- For area of triangle, use \(\frac{1}{2}|\vec{a}\times\vec{b}|\).
Practice
A) Multiple Choice Questions
-
A vector has:
Magnitude only Direction only Magnitude and direction Neither magnitude nor direction
-
Magnitude of \(\vec{a}=6\hat{i}+8\hat{j}\) is:
8 10 12 14
-
Dot product of two vectors gives:
A scalar A vector A matrix A triangle
-
Cross product of two vectors gives:
A scalar A vector Only a number Zero always
-
Work done by a force is calculated using:
Vector addition Dot product Cross product Scalar division
B) Solve the Higher-Order Problems
- Find the magnitude of \(\vec{a}=5\hat{i}+12\hat{j}\). (Hint: Use \(|\vec{a}|=\sqrt{x^2+y^2}\).)
- Find the unit vector in the direction of \(\vec{a}=8\hat{i}+6\hat{j}\). (Hint: Divide by magnitude.)
- If \(\vec{a}=3\hat{i}+2\hat{j}\) and \(\vec{b}=\hat{i}+4\hat{j}\), find \(\vec{a}\cdot\vec{b}\). (Hint: Multiply corresponding components and add.)
- If \(|\vec{a}|=7\), \(|\vec{b}|=2\), and angle between them is \(60^\circ\), find \(\vec{a}\cdot\vec{b}\). (Hint: Use \(\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta\).)
- A force \(\vec{F}=2\hat{i}+3\hat{j}\) produces displacement \(\vec{s}=4\hat{i}+5\hat{j}\). Find work done. (Hint: Work done \(=\vec{F}\cdot\vec{s}\).)
C) Match the Concept with the Correct Rule
| Concept | Correct Rule / Meaning |
|---|---|
| Vector | Quantity having magnitude and direction |
| Scalar | Quantity having magnitude only |
| Unit Vector | Vector having magnitude 1 |
| Null Vector | Vector having magnitude 0 |
| Dot Product | \(\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta\) |
| Cross Product | \(|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta\) |
| Work Done | \(W=\vec{F}\cdot\vec{s}\) |
| Moment of Force | \(\vec{M}=\vec{r}\times\vec{F}\) |
Vector Algebra Reminder
Vector Algebra deals with quantities having both magnitude and direction. A vector may be represented in two or three dimensions using unit vectors. Magnitude gives the length of a vector, and a unit vector gives direction. Vector addition and scalar multiplication are performed component-wise. Dot product gives a scalar and is used in work done and angle problems. Cross product gives a vector and is used in area and moment of force problems.
Task: Create five Vector Algebra questions using one question each from magnitude, unit vector, vector addition, dot product, cross product, work done and moment of force.
Show Suggested Answers
Multiple Choice
-
Magnitude and direction
A vector has both magnitude and direction. -
10
\[ |\vec{a}|=\sqrt{6^2+8^2}=\sqrt{36+64}=10 \] -
A scalar
Dot product of two vectors gives a scalar. -
A vector
Cross product of two vectors gives a vector. -
Dot product
Work done by force is calculated using \(W=\vec{F}\cdot\vec{s}\).
Higher-Order Problems
- \[ |\vec{a}|=\sqrt{5^2+12^2}=\sqrt{25+144}=13 \] Answer = 13.
- \[ |\vec{a}|=\sqrt{8^2+6^2}=\sqrt{64+36}=10 \] \[ \hat{a}=\frac{\vec{a}}{|\vec{a}|} = \frac{8}{10}\hat{i}+\frac{6}{10}\hat{j} \] Answer = \(\frac{4}{5}\hat{i}+\frac{3}{5}\hat{j}\).
- \[ \vec{a}\cdot\vec{b}=3(1)+2(4)=3+8=11 \] Answer = 11.
- \[ \vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos 60^\circ \] \[ =7\times 2\times \frac{1}{2}=7 \] Answer = 7.
- \[ W=\vec{F}\cdot\vec{s} \] \[ W=2(4)+3(5)=8+15=23 \] Answer = 23 units.
Concept Matching
- Vector → Quantity having magnitude and direction
- Scalar → Quantity having magnitude only
- Unit Vector → Vector having magnitude 1
- Null Vector → Vector having magnitude 0
- Dot Product → \(\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta\)
- Cross Product → \(|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta\)
- Work Done → \(W=\vec{F}\cdot\vec{s}\)
- Moment of Force → \(\vec{M}=\vec{r}\times\vec{F}\)
Clue Explanation
Vector Algebra questions are solved by first identifying whether the required result is a scalar or a vector. Use magnitude formula for length, unit vector formula for direction, dot product for angle and work, and cross product for area, direction and moment of force.
Exam tips
- Check whether the answer should be scalar or vector.
- Magnitude is never negative.
- Unit vector has magnitude 1.
- Null vector has no definite direction.
- Add vectors component-wise.
- Use dot product for angle and work.
- Use cross product for area and moment.
- If dot product is zero, vectors are perpendicular.
- If cross product is zero, vectors are parallel.