---
title: "Unit vector from A toward B"
subject: "Electromagnetism"
chapter: 1
key: "semester_3/electromagnetism:unit-vector-toward-b-cartesian-cylindrical"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/unit-vector-toward-b-cartesian-cylindrical
site: BCT Engineering
---

# Unit vector from A toward B

Electromagnetism · Chapter 1: Introduction

- Asked in: 2076 Asoj, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/unit-vector-toward-b-cartesian-cylindrical
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-1

## Question

Given points A(ρ = 5, φ = 70°, z = −3) and B(ρ = 2, φ = −30°, z = 1), find (a) a unit vector in cartesian coordinates at A directed toward B; (b) a unit vector in cylindrical coordinates at A directed toward B.

## Short answer

Convert A(5,70°,−3) and B(2,−30°,1) to Cartesian, find R_AB = (0.022, −5.698, 4), magnitude 6.962. Unit vector: (a) 0.003 a_x − 0.819 a_y + 0.575 a_z; (b) at A, project to cylindrical: −0.768 a_ρ − 0.283 a_φ + 0.575 a_z.

## Answer

**Given:** A(ρ = 5, φ = 70°, z = −3), B(ρ = 2, φ = −30°, z = 1).

#### Step 1: Convert both points to Cartesian

```text
A: x = 5 cos70° = 1.710,  y = 5 sin70° = 4.698,  z = −3
B: x = 2 cos(−30°) = 1.732,  y = 2 sin(−30°) = −1.000,  z = 1
```

#### (a) Unit vector in Cartesian coordinates

```text
R_AB = B − A = (1.732 − 1.710) a_x + (−1.000 − 4.698) a_y + (1 + 3) a_z
     = 0.022 a_x − 5.698 a_y + 4 a_z

|R_AB| = √(0.022² + 5.698² + 4²) = √(0.0005 + 32.47 + 16) = 6.962

a_AB = R_AB / |R_AB| = 0.0032 a_x − 0.8185 a_y + 0.5745 a_z
```

#### (b) Unit vector in cylindrical coordinates at A

At A, φ = 70° (cos70° = 0.3420, sin70° = 0.9397). Project the Cartesian components of a_AB onto a_ρ and a_φ at A:

```text
a_ρ component = (x-component) cosφ + (y-component) sinφ
a_φ component = −(x-component) sinφ + (y-component) cosφ
```

```text
a_ρ component = 0.0032(0.3420) + (−0.8185)(0.9397) = 0.0011 − 0.7692 = −0.768
a_φ component = −0.0032(0.9397) + (−0.8185)(0.3420) = −0.0030 − 0.2799 = −0.283
a_z component = 0.5745
```

**Check:** 0.768² + 0.283² + 0.5745² = 0.590 + 0.080 + 0.330 = 1.000 ✓

**Answer:**
- (a) a_AB = 0.003 a_x − 0.819 a_y + 0.575 a_z
- (b) a_AB = −0.768 a_ρ − 0.283 a_φ + 0.575 a_z (at A)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the vector from A toward B found?
   A. R_AB = A - B
   B. R_AB = B - A, after converting both points to Cartesian
   C. R_AB = B + A in cylindrical coordinates
   D. Subtract rho, phi and z of A from B directly
   Answer: B. Both points are converted to Cartesian and R_AB = B - A is formed, which gives (0.022, -5.698, 4). (Key point: Direction vector)

2. What is done to R_AB to get the Cartesian unit vector?
   A. Multiply by its magnitude
   B. Divide by its magnitude 6.962
   C. Divide by 1
   D. Take only its largest component
   Answer: B. A unit vector is R_AB divided by |R_AB| = 6.962, giving 0.003 a_x - 0.819 a_y + 0.575 a_z. (Key point: Making it a unit vector)

3. To express a_AB in cylindrical components at A, which angle is used?
   A. phi = -30°, the angle of B
   B. phi = 70°, the angle of A
   C. phi = 40°, the average of the two
   D. phi = 100°, the sum of the two
   Answer: B. Cylindrical unit vectors depend on the point where the vector sits, here A, so phi = 70° is used. (Key point: Cylindrical components at A)

4. What is the cylindrical form of the unit vector at A?
   A. 0.768 a_ρ + 0.283 a_φ + 0.575 a_z
   B. -0.768 a_ρ - 0.283 a_φ + 0.575 a_z
   C. -0.283 a_ρ - 0.768 a_φ + 0.575 a_z
   D. 0.003 a_ρ - 0.819 a_φ + 0.575 a_z
   Answer: B. a_ρ = 0.0032cos70° - 0.8185sin70° = -0.768 and a_φ = -0.0032sin70° - 0.8185cos70° = -0.283; the z component stays 0.575. (Key point: Cylindrical result)

## Related in Electromagnetism

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