Class 11 Mathematics Vectors Notes

Class 11 Mathematics Vectors Notes | Scalar Product, Projection & Vector Method
Vectors – Original PDF

Class 11 Mathematics Vectors Notes

Complete typed notes based on the supplied handwritten Vectors PDF, including unit vectors, displacement, vector addition, scalar product, projection, angle between vectors and solved Exercise 9.

1. Unit Vector

A vector whose magnitude is unity is called a unit vector.

â = a⃗ / |a⃗|

2. Displacement Along an Axis or on a Line

Let O be the origin. If the position vectors of points P and Q are OP⃗ = x₁ and OQ⃗ = x₂, then:

PQ⃗ = x₂ – x₁

In three dimensions:

OP⃗ = (x₁, y₁, z₁)
OQ⃗ = (x₂, y₂, z₂)
PQ⃗ = (x₂-x₁, y₂-y₁, z₂-z₁)

3. Addition of Two Vectors

AB⃗ = a⃗ + b⃗

4. Triangle Law of Vectors

If vectors are represented by two sides of a triangle taken in order, then their sum is represented by the third side taken in the same order.

AB⃗ = AO⃗ + OB⃗

5. Midpoint Formula

If C is the midpoint of AB, then:

OA⃗ + OB⃗ = 2OC⃗
OC⃗ = (OA⃗ + OB⃗)/2

6. Magnitude and Direction Cosines of a Vector

For a vector with components x, y and z, the direction cosines are:

l = x/√(x²+y²+z²)
m = y/√(x²+y²+z²)
n = z/√(x²+y²+z²)

The notes prove:

l² + m² + n² = 1

Parallelogram Example

If ABCD is a parallelogram and O is the point of intersection of its diagonals, the notes use the midpoint property of both diagonals to obtain the corresponding vector relation among OA⃗, OB⃗, OC⃗ and OD⃗.

Collinearity Example

The source checks whether three points are collinear by comparing the direction vectors between them. If one vector is a scalar multiple of another, the three points are collinear.

7. Scalar Product of Vectors

If a⃗ = (a₁, a₂) and b⃗ = (b₁, b₂), then:

a⃗ · b⃗ = a₁b₁ + a₂b₂

In terms of the angle θ between them:

a⃗ · b⃗ = |a⃗||b⃗| cosθ

Properties of Scalar Product

  1. Commutative: a⃗ · b⃗ = b⃗ · a⃗
  2. Distributive: a⃗ · (b⃗ + c⃗) = a⃗ · b⃗ + a⃗ · c⃗
  3. Self product: a⃗ · a⃗ = |a⃗|²

Useful Identities

|a⃗ + b⃗|² = |a⃗|² + 2a⃗·b⃗ + |b⃗|²
|a⃗ – b⃗|² = |a⃗|² – 2a⃗·b⃗ + |b⃗|²
(a⃗+b⃗)·(a⃗-b⃗) = |a⃗|² – |b⃗|²

8. Angle and Projection

Angle Between Two Vectors

cosθ = (a⃗·b⃗)/(|a⃗||b⃗|)
θ = cos-1((a⃗·b⃗)/(|a⃗||b⃗|))

Projection

Projection of a⃗ on b⃗ = (a⃗·b⃗)/|b⃗|
Projection of b⃗ on a⃗ = (a⃗·b⃗)/|a⃗|

Parallel Vectors

If two vectors are parallel, θ = 0°, so:

a⃗·b⃗ = |a⃗||b⃗|

Perpendicular Vectors

If two vectors are perpendicular, θ = 90°, so:

a⃗·b⃗ = 0

9. Exercise 9

1. Find the scalar product

(a)

a⃗ = i⃗ + 2j⃗ + 3k⃗
b⃗ = 2i⃗ – 5j⃗ + k⃗
a⃗·b⃗ = 2 – 10 + 3 = -5

(b)

a⃗ = 2j⃗ + 5k⃗
b⃗ = 7i⃗ – 5j⃗ + 2k⃗
a⃗·b⃗ = 0 – 10 + 10 = 0

2. Evaluate Vector Expressions

The notes use identities such as:

(a⃗-b⃗)·(a⃗+b⃗) = |a⃗|² – |b⃗|²

and scalar multiplication to simplify the given expressions.

3. Find the Angle Between Pairs of Vectors

The source uses:

cosθ = (a⃗·b⃗)/(|a⃗||b⃗|)

One worked example gives:

cosθ = √(2/3)
θ = cos-1(√(2/3))

Another example has dot product 0, giving θ = π/2.

4. Find p if Two Vectors are Perpendicular

Using a⃗·b⃗ = 0, the handwritten solution obtains:

p = 3

5. Find the Projection

The notes first calculate the scalar product and magnitude, then use:

Projection of b⃗ on a⃗ = (a⃗·b⃗)/|a⃗|

6. Find the Angles of a Triangle Using Vectors

The source forms AB⃗, BC⃗ and CA⃗ from the coordinates of the three vertices, then applies the scalar-product formula to determine each angle.

7. If |a⃗+b⃗|² = |a⃗-b⃗|², prove a⃗ ⟂ b⃗

|a⃗|² + 2a⃗·b⃗ + |b⃗|² = |a⃗|² – 2a⃗·b⃗ + |b⃗|²
4a⃗·b⃗ = 0
a⃗·b⃗ = 0

Therefore a⃗ is perpendicular to b⃗.

8. If (a⃗+b⃗)·(a⃗-b⃗)=0, prove |a⃗|=|b⃗|

|a⃗|² – |b⃗|² = 0
|a⃗|² = |b⃗|²
|a⃗| = |b⃗|

9. Prove a Dot-Product Identity

a⃗·b⃗ = 1/4 [|a⃗+b⃗|² – |a⃗-b⃗|²]

The source proves this by expanding both squared magnitudes.

10. Parallelogram Law Using Vectors

If AC and BD are diagonals of a parallelogram ABCD, the notes derive vector expressions for the diagonals and prove identities relating their squared lengths to the sides.

10. Vector Method in a Triangle

11. Prove Relations in Triangle ABC by Vector Method

The source derives cosine-rule type identities including:

a² = b² + c² – 2bc cos A
b² = c² + a² – 2ca cos B

12. Prove Trigonometric Identities Using Vectors

Using position vectors and the scalar product, the notes prove:

cos(A-B) = cosA cosB + sinA sinB

and similarly:

cos(A+B) = cosA cosB – sinA sinB

Note: This typed version follows the uploaded 21-page Vectors PDF closely. The Google Drive PDF supplied in your message is embedded at the top.

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