PhysicsCBSE Class 11 · Motion in a Plane

Vector Addition, Resolution and Products Class 11

Updated 17 October 2026 · Fact-checked

Vectors add by the triangle or parallelogram law: R = √(A² + B² + 2AB cos θ). A vector resolves into components A cos θ and A sin θ. The dot product A·B = AB cos θ is a scalar (zero for perpendicular vectors); the cross product A×B has magnitude AB sin θ, direction by the right-hand rule, and is zero for parallel vectors.

Draw A and B from the same point; the diagonal of the parallelogram is R, largest when θ = 0° and smallest when θ = 180°.

Illustration drawn for QuestionBuddy.

Understand Vector Addition, Resolution and Products

A vector has magnitude and direction and obeys the vector laws of addition; a scalar has only magnitude.

Addition. By the parallelogram law, two vectors $A$ and $B$ with angle $θ$ between them have resultant
$ $R = A^{2} + B^{2} + 2 A B cos θ, tan α = \frac{B s i n θ}{A + B c o s θ},$ $
where $α$ is the angle of $R$ from $A$ . The maximum ( $A + B$ ) occurs at $θ = 0^{\circ}$ , the minimum $∣ A - B ∣$ at $18 0^{\circ}$ , and for $θ = 9 0^{\circ}$ , $R = A^{2} + B^{2}$ . The difference is $∣ A - B ∣ = A^{2} + B^{2} - 2 A B cos θ$ . A set of vectors adds to zero if they close into a polygon; three magnitudes can sum to zero only if each is at most the sum of the other two.

Resolution. A vector at angle $θ$ to the x-axis has components $A_{x} = A cos θ$ , $A_{y} = A sin θ$ , so $A = A_{x} \hat{i} + A_{y} \hat{j}$ , with $A = A_{x}^{2} + A_{y}^{2}$ and $tan θ = A_{y} / A_{x}$ . The unit vector along $A$ is $\hat{A} = A / A$ . To add vectors, add component by component.

Scalar (dot) product. $A \cdot B = A B cos θ = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$ . It is commutative, zero for perpendicular vectors, and $\hat{i} \cdot \hat{i} = 1$ , $\hat{i} \cdot \hat{j} = 0$ . Examples: work $W = F \cdot d$ , power $P = F \cdot v$ .

Vector (cross) product. $A \times B = A B sin θ \overset{n}{^}$ , with $\overset{n}{^}$ given by the right-hand rule, perpendicular to both. It is anticommutative ( $A \times B = - B \times A$ ), zero for parallel vectors, and equals the determinant of $\hat{i}, \hat{j}, \hat{k}$ over the components. Its magnitude is the area of the parallelogram on the two vectors. Examples: torque $τ = r \times F$ , angular momentum $L = r \times p$ .

Relative velocity. $v_{A B} = v_{A} - v_{B}$ . For a boat in a river, the velocity of the boat relative to the ground is the vector sum of its velocity relative to the water and the water's velocity.

Vector Addition, Resolution and Products: formulas and key facts

Resultant (parallelogram law): $R = A^{2} + B^{2} + 2 A B cos θ$; $θ$ between the vectors.
Components: $A_{x} = A cos θ, A_{y} = A sin θ$; From the x-axis.
Dot product: $A \cdot B = A B cos θ = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$; Scalar.
Cross product: $∣ A \times B ∣ = A B sin θ$; Right-hand rule.
Cross product by components: $A \times B = \hat{i} A_{x} B_{x} \hat{j} A_{y} B_{y} \hat{k} A_{z} B_{z}$; Determinant form.
Relative velocity: $v_{A B} = v_{A} - v_{B}$; Velocity of A as seen from B.

How to solve Vector Addition, Resolution and Products questions

Resolve into components when the vectors are given by direction; use the parallelogram formula when only magnitudes and the angle between are given.

1Draw the vectors tail to tail (or tail to head) and mark the angle between them.
2For two vectors with an angle given, use R² = A² + B² + 2AB cos θ and the angle formula.
3For three or more, resolve each along x and y, add the components and recombine.
4For products, convert to component form and use the i, j, k rules.
5Use the dot product for an angle (cos θ) and for a perpendicularity test; use the cross product for an area or a torque.
6For river or wind problems, write the velocity relative to the medium plus the velocity of the medium.

Vector Addition, Resolution and Products in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Pulling a trolley with two ropes

Suppose two children pull a trolley with ropes at 90° to each other, with forces 30 N and 40 N.

Find the resultant force and its angle from the 30 N rope.

Try it first, then show the solutionSolution

R = √(30² + 40²) = 50 N.
tan α = 40/30 = 4/3, so α ≈ 53°.

Answer: 50 N at about 53° from the 30 N rope.

Memory hook: Perpendicular forces form a 3-4-5 triangle.

Wind and a plane

Suppose a plane flies at 100 km/h relative to the air, heading due north, while a wind blows at 60 km/h due east.

Find its speed and direction relative to the ground.

Try it first, then show the solutionSolution

Speed = √(100² + 60²) = √13600 = 116.6 km/h.
tan θ = 60/100 = 0.6, so θ ≈ 31° east of north.

Answer: About 117 km/h, 31° east of north.

Memory hook: Ground velocity = air velocity + wind velocity.

Tricks to remember Vector Addition, Resolution and Products

Short memory aids. Say each one aloud once, then use it on a question the same day.

Dot is cos, cross is sin
Dot product: cos θ and a scalar. Cross product: sin θ and a vector.
Use it when: Choosing the product.
Extremes
Resultant is largest at 0° and smallest at 180°.
Use it when: Range of a resultant.
ijk cycle
i×j = k, j×k = i, k×i = j; reverse order gives a minus sign.
Use it when: Cross product calculations.
3-4-5 triangle
Use sin 37° = 0.6, cos 37° = 0.8 when numbers allow.
Use it when: Fast components.

Vector Addition, Resolution and Products: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Resolving a force · 2 marks · Very short answerBoard level

A force of 50 N acts at 37° above the horizontal. Find its horizontal and vertical components (sin 37° = 0.6, cos 37° = 0.8).

Try it first, then show the solutionSolution

Fx = 50 cos 37° = 50 × 0.8 = 40 N. (1)
Fy = 50 sin 37° = 50 × 0.6 = 30 N. (1)

Answer: Fx = 40 N, Fy = 30 N.

Example 2: Resultant of perpendicular vectors · 3 marks · Short answerBoard level

Two vectors of magnitudes 6 and 8 act at right angles. Find the magnitude of the resultant and its angle with the vector of magnitude 6.

Try it first, then show the solutionSolution

R = √(6² + 8²) = √100 = 10. (1)
tan α = 8/6 = 4/3. (1)
α = 53° from the vector of magnitude 6. (1)

Answer: 10 units at 53° from the 6-unit vector.

Example 3: Dot and cross products in component form · 3 marks · Short answerHarder board level

Given A = 3i − 2j + k and B = 2i + j − 4k, find A·B and A×B. What do the results tell you about the angle between them?

Try it first, then show the solutionSolution

A·B = 6 − 2 − 4 = 0. (1)
A×B = i[(−2)(−4) − (1)(1)] − j[(3)(−4) − (1)(2)] + k[(3)(1) − (−2)(2)] = 7i + 14j + 7k. (1)
A·B = 0 means they are perpendicular; |A×B| = √294 = |A||B| = √14 × √21, consistent with sin 90° = 1. (1)

Answer: A·B = 0; A×B = 7i + 14j + 7k; the vectors are perpendicular.

Example 4: Finding the angle from the resultant · 3 marks · Short answerHarder board level

Two vectors of magnitude 3 and 5 have a resultant of magnitude 7. Find the angle between them and the magnitude of their difference.

Try it first, then show the solutionSolution

49 = 9 + 25 + 30 cos θ. (1)
cos θ = 15/30 = 0.5, so θ = 60°. (1)
|A − B| = √(9 + 25 − 2 × 15 × 0.5) = √19 ≈ 4.36. (1)

Answer: 60°; difference about 4.36.

Example 5: Crossing a river · 4 marks · Long answerHOTS / competency

A river 120 m wide flows at 3 m/s. A boat has speed 5 m/s in still water. (a) If the boat heads perpendicular to the bank, find the time to cross, the drift downstream and the speed relative to the ground. (b) At what angle upstream from the perpendicular must it head to land directly opposite, and how long does the crossing take?

Try it first, then show the solutionSolution

(a) t = 120/5 = 24 s; drift = 3 × 24 = 72 m. (1)
Ground speed = √(5² + 3²) = √34 ≈ 5.83 m/s. (1)
(b) Cancel the drift: 5 sin θ = 3, so θ = sin⁻¹(0.6) ≈ 37° upstream. (1)
Effective speed across = √(25 − 9) = 4 m/s, so t = 120/4 = 30 s. (1)

Answer: (a) 24 s, 72 m, 5.83 m/s; (b) about 37° upstream, 30 s.

Where students lose marks in Vector Addition, Resolution and Products

The slips examiners see most often, and what to write instead.

Mistake: Adding magnitudes directly for vectors at an angle.
Treating vectors as scalars.
Do this: Use R = √(A² + B² + 2AB cos θ).
Usually costs: The final mark.
Mistake: Substituting the head-to-tail (exterior) angle in R² = A² + B² + 2AB cos θ.
Mixing the triangle law with the parallelogram law.
Do this: Use the angle between the vectors when both start at the same point; for head-to-tail drawing the angle between them is 180° minus the angle shown.
Usually costs: Method mark.
Mistake: Getting the sign of the j component wrong in a cross product.
Forgetting the minus sign in the determinant.
Do this: Expand: +i(…) − j(…) + k(…).
Usually costs: 1 to 2 marks.
Mistake: Calling the dot product a vector.
Confusing the two products.
Do this: Dot gives a scalar; cross gives a vector.
Usually costs: Concept mark.

Exam tips

Quote the formula for the resultant before substituting.
State the angle of the resultant with a named reference vector or axis.
In cross product answers, give the direction (along which axis, with sign) as well as the magnitude.
A dot product of zero for non-zero vectors means perpendicular; a cross product of zero means parallel.
For river problems draw the velocity triangle.

Practice questions on Vector Addition, Resolution and Products

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQHarder board level
The magnitude of A×B equals A·B when the angle between A and B is
(a) 0° (b) 30° (c) 45° (d) 90°
Hint
Set sin θ = cos θ.
Answer
(c) 45°.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Three vectors of magnitudes 3, 4 and 5 can add to zero.
Reason (R): Three vectors can add to zero if they form a closed triangle, which needs each magnitude to be at most the sum of the other two.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
The 3-4-5 triangle closes.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Find the unit vector along A = 6i − 8j.
Hint
Divide by the magnitude.
Answer
|A| = 10, so unit vector = 0.6i − 0.8j.
Q4 · 3 marks · Short answerHarder board level
A force F = 2i + 3j N acts on a body that moves through d = 4i − j m. Find the work done and the angle between F and d.
Hint
W = F·d.
Answer
W = 8 − 3 = 5 J; cos θ = 5/(√13 × √17) = 0.336, so θ ≈ 70.4°.
Q5 · 4 marks · Long answerHOTS / competency
A force F = 2i − 5j N acts at the point with position vector r = 3i + 4j m. Find the torque about the origin (magnitude and direction) and the angle between r and F.
Hint
τ = r × F; cos θ = r·F/(rF).
Answer
τ = (3)(−5) − (4)(2) = −23 along k, i.e. 23 N m in the −z direction. r·F = 6 − 20 = −14; |r| = 5, |F| = √29 = 5.385; cos θ = −14/26.9 = −0.52, so θ ≈ 121°.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Resultant of A and B at angle θ?Show answer
√(A² + B² + 2AB cos θ).
2.A·B for perpendicular vectors?Show answer
Zero.
3.A×B for parallel vectors?Show answer
Zero.
4.i × j?Show answer
k.
5.Magnitude of A×B represents?Show answer
Area of the parallelogram on A and B.
6.Condition for three vectors to add to zero?Show answer
They close into a triangle.

Vector Addition, Resolution and Products: frequently asked questions

Why is A×B = −B×A?

Reversing the order reverses the direction given by the right-hand rule, while the magnitude AB sin θ stays the same.

Is work a vector?

No. Work is the dot product of force and displacement and is a scalar.

When is the resultant equal to the difference of magnitudes?

When the vectors are antiparallel (θ = 180°).

What is a unit vector?

A vector of magnitude 1 that gives only direction, written with a hat.

What is the relative velocity of two bodies moving in the same direction?

The difference of their speeds.