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Vector Algebra

Practice MCQs

Vector Algebra is the branch of mathematics that deals with quantities having both magnitude and direction.

Elementary Mathematics Vector Algebra Dot Product and Cross Product Competitive Exams

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.

Quick idea: Scalar tells “how much”, while vector tells “how much and in which direction”.
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.”

Vector Algebra Tip
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.
vector magnitude unit vector dot product cross product work moment
Core Formula Bank
2D Vector
\[ \vec{a}=x\hat{i}+y\hat{j} \]
3D Vector
\[ \vec{a}=x\hat{i}+y\hat{j}+z\hat{k} \]
Magnitude
\[ |\vec{a}|=\sqrt{x^2+y^2+z^2} \]
Unit Vector
\[ \hat{a}=\frac{\vec{a}}{|\vec{a}|} \]
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
\[ \vec{M}=\vec{r}\times\vec{F} \]

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
Remember: Dot product gives a scalar result, while cross product gives a vector result.

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.
Important: In \(\vec{AB}\), subtract coordinates of \(A\) from coordinates of \(B\).

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 \(l=\frac{x}{|\vec{a}|},\ m=\frac{y}{|\vec{a}|},\ n=\frac{z}{|\vec{a}|}\) \(l^2+m^2+n^2=1\)
Exam tip: Magnitude is always non-negative. If \(|\vec{a}|=0\), the vector is a null vector.

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 \(-\vec{a}\)
Common trap: A null vector cannot be converted into a unit vector because its magnitude is zero.

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. \(\vec{R}=\vec{a}+\vec{b}\)
Remember: Vector addition follows triangle law and parallelogram law geometrically.

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.
Exam tip: If the dot product of two non-zero vectors is zero, the vectors are perpendicular.

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.
For \[ \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \] and \[ \vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \] the cross product is: \[ \vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix} \]
Common trap: Cross product is not commutative. In fact, \(\vec{a}\times\vec{b}=-(\vec{b}\times\vec{a})\).

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.
Important: Work done is a scalar quantity, while moment of force is a vector quantity.

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.
Important: Always check whether the final answer should be a scalar or a vector.

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 \(\frac{1}{2}|\vec{a}\times\vec{b}|\).
Exam approach: Identify whether the problem is based on magnitude, direction, unit vector, null vector, vector addition, scalar multiplication, dot product, cross product, work done, moment of force, or geometrical applications.

Practice

A) Multiple Choice Questions
  1. A vector has:
    Magnitude only Direction only Magnitude and direction Neither magnitude nor direction
  2. Magnitude of \(\vec{a}=6\hat{i}+8\hat{j}\) is:
    8 10 12 14
  3. Dot product of two vectors gives:
    A scalar A vector A matrix A triangle
  4. Cross product of two vectors gives:
    A scalar A vector Only a number Zero always
  5. Work done by a force is calculated using:
    Vector addition Dot product Cross product Scalar division
B) Solve the Higher-Order Problems
  1. Find the magnitude of \(\vec{a}=5\hat{i}+12\hat{j}\). (Hint: Use \(|\vec{a}|=\sqrt{x^2+y^2}\).)
  2. Find the unit vector in the direction of \(\vec{a}=8\hat{i}+6\hat{j}\). (Hint: Divide by magnitude.)
  3. 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.)
  4. If \(|\vec{a}|=7\), \(|\vec{b}|=2\), and angle between them is \(60^\circ\), find \(\vec{a}\cdot\vec{b}\). (Hint: Use \(\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta\).)
  5. 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
  1. Magnitude and direction
    A vector has both magnitude and direction.
  2. 10
    \[ |\vec{a}|=\sqrt{6^2+8^2}=\sqrt{36+64}=10 \]
  3. A scalar
    Dot product of two vectors gives a scalar.
  4. A vector
    Cross product of two vectors gives a vector.
  5. Dot product
    Work done by force is calculated using \(W=\vec{F}\cdot\vec{s}\).
Higher-Order Problems
  1. \[ |\vec{a}|=\sqrt{5^2+12^2}=\sqrt{25+144}=13 \] Answer = 13.
  2. \[ |\vec{a}|=\sqrt{8^2+6^2}=\sqrt{64+36}=10 \] \[ \hat{a}=\frac{\vec{a}}{|\vec{a}|} = \frac{8}{10}\hat{i}+\frac{6}{10}\hat{j} \] Answer = \(\frac{4}{5}\hat{i}+\frac{3}{5}\hat{j}\).
  3. \[ \vec{a}\cdot\vec{b}=3(1)+2(4)=3+8=11 \] Answer = 11.
  4. \[ \vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos 60^\circ \] \[ =7\times 2\times \frac{1}{2}=7 \] Answer = 7.
  5. \[ W=\vec{F}\cdot\vec{s} \] \[ W=2(4)+3(5)=8+15=23 \] Answer = 23 units.
Concept Matching
  1. Vector → Quantity having magnitude and direction
  2. Scalar → Quantity having magnitude only
  3. Unit Vector → Vector having magnitude 1
  4. Null Vector → Vector having magnitude 0
  5. Dot Product → \(\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta\)
  6. Cross Product → \(|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta\)
  7. Work Done → \(W=\vec{F}\cdot\vec{s}\)
  8. 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.
Practice MCQs