Step-by-Step Solution

Topic: Vector Addition & Products

(I) MOVEMENT PLAN (TRIANGLE LAW)

Using the Triangle Law of Vector Addition:

• 1. Path from A to C: OA + AC = OC   ⇒   AC = OC – OA
• 2. Path from B to C: OB + BC = OC   ⇒   BC = OC – OB

This describes the displacement vectors required to travel from destinations A and B to the final meeting point C.

(II) FINDING VECTORS AC AND BC

Given: OA = a, OB = b, OC = 5a – 2b.

Vector AC:

AC = (5a – 2b) – a = 4a – 2b

Vector BC:

BC = (5a – 2b) – b = 5a – 3b

(III)(A) ANGLE & CROSS PRODUCT MAGNITUDE

1. Find the Angle (θ):

Formula: cos θ = (a · b) / (|a| |b|)

Substitute values: a · b = 1, |a| = 1, |b| = 2.

cos θ = 1 / (1 × 2) = 1/2

2. Find |a × b|:

Formula: |a × b| = |a| |b| sin θ

|a × b| = 1 × 2 × sin(60°)

|a × b| = 2 × (√3 / 2)

(III)(B) UNIT VECTOR PERPENDICULAR TO (a+b) AND (a-b)

Given: a = 2î – ĵ + 4k̂ and b = 0î + ĵ – k̂

Step 1: Find Sum and Difference Vectors

p = a + b = 2î + 0ĵ + 3k̂

q = a – b = 2î – 2ĵ + 5k̂

Step 2: Cross Product (p × q) using Determinant

n = î(0 – (-6)) – ĵ(10 – 6) + k̂(-4 – 0)

n = 6î – 4ĵ – 4k̂

Step 3: Calculate Unit Vector

Magnitude |n| = √(6² + (-4)² + (-4)²) = √(36 + 16 + 16) = √68 = 2√17

Unit Vector = n / |n| = (6î – 4ĵ – 4k̂) / 2√17