Quick answer: The dot product a ⃗ ⋅ b ⃗ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ = a 1 b 1 + a 2 b 2 + a 3 b 3 \vec{a} \cdot \vec{b} = \lvert\vec{a}\rvert\lvert\vec{b}\rvert\cos\theta = a_1b_1 + a_2b_2 + a_3b_3 a ⋅ b = ∣ a ∣ ∣ b ∣ cos θ = a 1 b 1 + a 2 b 2 + a 3 b 3 cross product a ⃗ × b ⃗ \vec{a} \times \vec{b} a × b ∣ a ⃗ ∣ ∣ b ⃗ ∣ sin θ \lvert\vec{a}\rvert\lvert\vec{b}\rvert\sin\theta ∣ a ∣ ∣ b ∣ sin θ 0 ⃗ \vec{0} 0 ∣ a ⃗ × b ⃗ ∣ \lvert\vec{a} \times \vec{b}\rvert ∣ a × b ∣ a ⃗ \vec{a} a b ⃗ \vec{b} b
Syllabus note: Several portals report that Vectors was removed in the revised NIMCET syllabus. Confirm with the official information brochure, and see our revised syllabus update .
This builds on position vectors and direction cosines .
Dot (scalar) product
Property Result Definition a ⃗ ⋅ b ⃗ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ \vec{a} \cdot \vec{b} = \lvert\vec{a}\rvert\lvert\vec{b}\rvert\cos\theta a ⋅ b = ∣ a ∣ ∣ b ∣ cos θ Components a 1 b 1 + a 2 b 2 + a 3 b 3 a_1b_1 + a_2b_2 + a_3b_3 a 1 b 1 + a 2 b 2 + a 3 b 3 Commutative a ⃗ ⋅ b ⃗ = b ⃗ ⋅ a ⃗ \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} a ⋅ b = b ⋅ a Perpendicular a ⃗ ⋅ b ⃗ = 0 \vec{a} \cdot \vec{b} = 0 a ⋅ b = 0 Self a ⃗ ⋅ a ⃗ = ∣ a ⃗ ∣ 2 \vec{a} \cdot \vec{a} = \lvert\vec{a}\rvert^2 a ⋅ a = ∣ a ∣ 2 Unit vectors i ^ ⋅ i ^ = 1 \hat{i} \cdot \hat{i} = 1 i ^ ⋅ i ^ = 1 i ^ ⋅ j ^ = 0 \hat{i} \cdot \hat{j} = 0 i ^ ⋅ j ^ = 0
Angle between vectors:
cos θ = a ⃗ ⋅ b ⃗ ∣ a ⃗ ∣ ∣ b ⃗ ∣ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{\lvert\vec{a}\rvert\lvert\vec{b}\rvert} cos θ = ∣ a ∣ ∣ b ∣ a ⋅ b
Projection of a ⃗ \vec{a} a b ⃗ \vec{b} b a ⃗ ⋅ b ⃗ ∣ b ⃗ ∣ \frac{\vec{a} \cdot \vec{b}}{\lvert\vec{b}\rvert} ∣ b ∣ a ⋅ b
Work done: W = F ⃗ ⋅ d ⃗ W = \vec{F} \cdot \vec{d} W = F ⋅ d
Useful identity: ∣ a ⃗ + b ⃗ ∣ 2 = ∣ a ⃗ ∣ 2 + ∣ b ⃗ ∣ 2 + 2 a ⃗ ⋅ b ⃗ \lvert\vec{a} + \vec{b}\rvert^2 = \lvert\vec{a}\rvert^2 + \lvert\vec{b}\rvert^2 + 2\vec{a} \cdot \vec{b} ∣ a + b ∣ 2 = ∣ a ∣ 2 + ∣ b ∣ 2 + 2 a ⋅ b
Cross (vector) product
a ⃗ × b ⃗ = ∣ i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3 ∣ \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} a × b = i ^ a 1 b 1 j ^ a 2 b 2 k ^ a 3 b 3
Property Result Magnitude ∣ a ⃗ ∣ ∣ b ⃗ ∣ sin θ \lvert\vec{a}\rvert\lvert\vec{b}\rvert\sin\theta ∣ a ∣ ∣ b ∣ sin θ Direction Perpendicular to both, by the right-hand rule Anti-commutative a ⃗ × b ⃗ = − b ⃗ × a ⃗ \vec{a} \times \vec{b} = -\vec{b} \times \vec{a} a × b = − b × a Parallel a ⃗ × b ⃗ = 0 ⃗ \vec{a} \times \vec{b} = \vec{0} a × b = 0 Unit vectors i ^ × j ^ = k ^ \hat{i} \times \hat{j} = \hat{k} i ^ × j ^ = k ^ j ^ × k ^ = i ^ \hat{j} \times \hat{k} = \hat{i} j ^ × k ^ = i ^ k ^ × i ^ = j ^ \hat{k} \times \hat{i} = \hat{j} k ^ × i ^ = j ^ Self a ⃗ × a ⃗ = 0 ⃗ \vec{a} \times \vec{a} = \vec{0} a × a = 0
Unit vector perpendicular to both: ± a ⃗ × b ⃗ ∣ a ⃗ × b ⃗ ∣ \pm\frac{\vec{a} \times \vec{b}}{\lvert\vec{a} \times \vec{b}\rvert} ± ∣ a × b ∣ a × b
Lagrange's identity: ∣ a ⃗ × b ⃗ ∣ 2 + ( a ⃗ ⋅ b ⃗ ) 2 = ∣ a ⃗ ∣ 2 ∣ b ⃗ ∣ 2 \lvert\vec{a} \times \vec{b}\rvert^2 + (\vec{a} \cdot \vec{b})^2 = \lvert\vec{a}\rvert^2\lvert\vec{b}\rvert^2 ∣ a × b ∣ 2 + ( a ⋅ b ) 2 = ∣ a ∣ 2 ∣ b ∣ 2
Areas
Figure Area Parallelogram with sides a ⃗ , b ⃗ \vec{a}, \vec{b} a , b ∣ a ⃗ × b ⃗ ∣ \lvert\vec{a} \times \vec{b}\rvert ∣ a × b ∣ Parallelogram with diagonals d 1 ⃗ , d 2 ⃗ \vec{d_1}, \vec{d_2} d 1 , d 2 1 2 ∣ d 1 ⃗ × d 2 ⃗ ∣ \frac{1}{2}\lvert\vec{d_1} \times \vec{d_2}\rvert 2 1 ∣ d 1 × d 2 ∣ Triangle ABC 1 2 ∣ A B → × A C → ∣ \frac{1}{2}\lvert\overrightarrow{AB} \times \overrightarrow{AC}\rvert 2 1 ∣ A B × A C ∣
Worked examples
Example 1: angle
Find the angle between a ⃗ = i ^ + j ^ \vec{a} = \hat{i} + \hat{j} a = i ^ + j ^ b ⃗ = j ^ + k ^ \vec{b} = \hat{j} + \hat{k} b = j ^ + k ^
cos θ = 0 + 1 + 0 2 2 = 1 2 \cos\theta = \frac{0 + 1 + 0}{\sqrt{2}\sqrt{2}} = \frac{1}{2} cos θ = 2 2 0 + 1 + 0 = 2 1
So θ = 60 ∘ \theta = 60^\circ θ = 6 0 ∘
Example 2: perpendicular condition
Find λ if 2 i ^ + λ j ^ + k ^ 2\hat{i} + \lambda\hat{j} + \hat{k} 2 i ^ + λ j ^ + k ^ i ^ − 2 j ^ + 3 k ^ \hat{i} - 2\hat{j} + 3\hat{k} i ^ − 2 j ^ + 3 k ^
2 − 2 λ + 3 = 0 2 - 2\lambda + 3 = 0 2 − 2 λ + 3 = 0 λ = 5 2 \lambda = \frac{5}{2} λ = 2 5
Example 3: cross product and area
Find the area of the triangle with vertices A ( 1 , 0 , 0 ) A(1, 0, 0) A ( 1 , 0 , 0 ) B ( 0 , 1 , 0 ) B(0, 1, 0) B ( 0 , 1 , 0 ) C ( 0 , 0 , 1 ) C(0, 0, 1) C ( 0 , 0 , 1 )
Step 1: A B → = − i ^ + j ^ \overrightarrow{AB} = -\hat{i} + \hat{j} A B = − i ^ + j ^ A C → = − i ^ + k ^ \overrightarrow{AC} = -\hat{i} + \hat{k} A C = − i ^ + k ^
Step 2: A B → × A C → = i ^ + j ^ + k ^ \overrightarrow{AB} \times \overrightarrow{AC} = \hat{i} + \hat{j} + \hat{k} A B × A C = i ^ + j ^ + k ^
Step 3: Area = 1 2 3 = 3 2 = \frac{1}{2}\sqrt{3} = \frac{\sqrt{3}}{2} = 2 1 3 = 2 3
Check: ABC is equilateral with side 2 \sqrt{2} 2 3 4 ⋅ 2 = 3 2 \frac{\sqrt{3}}{4} \cdot 2 = \frac{\sqrt{3}}{2} 4 3 ⋅ 2 = 2 3
Example 4: magnitude from data
If ∣ a ⃗ ∣ = 3 \lvert\vec{a}\rvert = 3 ∣ a ∣ = 3 ∣ b ⃗ ∣ = 4 \lvert\vec{b}\rvert = 4 ∣ b ∣ = 4 a ⃗ ⋅ b ⃗ = 6 \vec{a} \cdot \vec{b} = 6 a ⋅ b = 6 ∣ a ⃗ × b ⃗ ∣ \lvert\vec{a} \times \vec{b}\rvert ∣ a × b ∣
By Lagrange's identity, ∣ a ⃗ × b ⃗ ∣ 2 = 144 − 36 = 108 \lvert\vec{a} \times \vec{b}\rvert^2 = 144 - 36 = 108 ∣ a × b ∣ 2 = 144 − 36 = 108 ∣ a ⃗ × b ⃗ ∣ = 6 3 \lvert\vec{a} \times \vec{b}\rvert = 6\sqrt{3} ∣ a × b ∣ = 6 3
Summary table
Item Dot product Cross product Result Scalar Vector Uses cos θ \cos\theta cos θ sin θ \sin\theta sin θ Zero when Perpendicular Parallel Order Commutative Anti-commutative Geometric meaning Projection, work Area, normal direction
Exam tips
Use the dot product for angles and perpendicularity; use the cross product for areas and parallelism.
Lagrange's identity links the two products and saves time.
Remember the cyclic order i ^ → j ^ → k ^ → i ^ \hat{i} \to \hat{j} \to \hat{k} \to \hat{i} i ^ → j ^ → k ^ → i ^
Next, see the scalar and vector triple product , or browse the NIMCET Coordinate Geometry and Vectors chapter .
Common mistakes
Writing a ⃗ × b ⃗ = b ⃗ × a ⃗ \vec{a} \times \vec{b} = \vec{b} \times \vec{a} a × b = b × a
Forgetting the minus sign on the j ^ \hat{j} j ^
Using the parallelogram side formula when the diagonals are given.
Treating the dot product as a vector.
Frequently Asked Questions
What is the dot product of two vectors?
a · b = |a||b| cos θ = a₁b₁ + a₂b₂ + a₃b₃. It is a scalar.
What is the cross product of two vectors?
a × b is a vector of magnitude |a||b| sin θ, perpendicular to both a and b.
When is the dot product zero?
When the vectors are perpendicular (or one of them is the zero vector).
How do you find the area of a triangle using vectors?
Area = ½ |AB × AC|.
What is the projection of a on b?
It is (a · b)/|b|.
