Lecture 1: Vectors (المتجهات) PDF

# Lecture 1: Vectors (المتجهات) ## Page 1 **Date:** 5/10/2024 * Quantities in nature are classified into two types: * Scalar quantities: * Quantities that need only a magnitude to be defined. * Examples: Mass (الكتلة), Heat (الحرارة) * Vector quantities: * Quanti...

# Lecture 1: Vectors (المتجهات) ## Page 1 **Date:** 5/10/2024 * Quantities in nature are classified into two types: * Scalar quantities: * Quantities that need only a magnitude to be defined. * Examples: Mass (الكتلة), Heat (الحرارة) * Vector quantities: * Quantities that need two different types of units for their definition. * One of them is the magnitude (مقدار), and the other is the direction (اتجاه). * To define a vector, we need to use one of three methods: * Placing an arrow above the symbol representing the vector. * Placing an arrow below the symbol representing the vector. * Using a boldfaced line (*Blad*) above the symbol (in older references). * The vector's magnitude is its length, and it is represented by the symbol $|A|$. * The length of the vector is calculated as follows: * $ab = A$. <br> ## Page 2 **Date:** 5/10/2024 ### Examples 1. Write the following vectors in a different form: * $A=(4, \frac{\pi}{4})$. * **Solution:** * $A_x = A \cos\alpha=4 \cos \frac{\pi}{4}=2\sqrt{2}$. * $A_y = A \sin\alpha=4 \sin \frac{\pi}{4}=2\sqrt{2}$. * $\Rightarrow A = ( 2\sqrt{2}, 2\sqrt{2})$. * $B = (4, 3)$. * **Solution:** * $B = \sqrt{B_x^2 + B_y^2} = \sqrt{4^2 + 3^2} =5$. * $\tan\alpha=\frac{B_y}{B_x} = \frac{3}{4} \Rightarrow A = tan^{-1}\frac{3}{4}$. * $\Rightarrow B = (\frac {3}{5}, \tan^{-1}\frac {3}{4})$. <br> ## Page 3 **Date:** 5/10/2024 ## Space Vectors (المتجهات في الفراغ) * These vectors are defined by their magnitude and at least two angles. <br> ## Page 4 **Date:** 5/10/2024 ### Examples 1. $B=(-12, 12, 6)$. * **Solution:** * $B = \sqrt{12^2+12^2+6^2} = 18$. * $\cos\alpha = \frac{B_x}{B} = \frac{12}{18} = \frac{2}{3}$. * $\cos\beta = \frac{B_y}{B} = \frac{12}{18} = \frac{2}{3}$. * $\cos\gamma = \frac{B_z}{B} = \frac{6}{18} = \frac{1}{3}$. * $B=(\frac{2}{3}B, \frac{2}{3}B, \frac{1}{3}B)= (18, \frac{2}{3}, \frac{1}{3})$. 2. $C=(18, \frac{\pi}{4}, \frac{\pi}{3})$. * **Solution:** * $C_x = 18 \cos \frac{\pi}{4} = 9\sqrt{2}$. * $C_y =18 \cos \frac{\pi}{3} = 9$. * $C_z = 18 \times \frac{1}{2} =9$. * $\Rightarrow C = (C_x, C_y, C_z) = (9\sqrt{2}, 9, 9)$. <br> ## Page 5 **Date:** 5/10/2024 ### Notes: * Two vectors, A and B, are said to be equal if their magnitudes and directions are the same. * If two vectors only share the same length but differ in their direction, then $A = -B$, which means that A has the opposite direction of B. * If a vector is multiplied by a scalar value, the magnitude of the resulting vector is multiplied by the scalar value. If the scalar is positive, the direction remains the same. If the scalar is negative, the direction is reversed. * $A$ is a unit vector of $A$, which has a magnitude of 1 and the same direction as A. * The ratio between two parallel vectors is a scalar value. <br> ## Page 6 **Date:** 5/10/2024 ## Notes: * The unit vector in the following directions is: * $x$ direction: $i$. * $y$ direction: $j$. * $z$ direction: $k$. * $A = Ae$. * $A = Ax i + Ay j + Az k$. * $| A | = \sqrt{A^2 x + A^2 y + A^2 z}$. ## Page 7 **Date:** 5/10/2024 ## Notes: * If we write the unit vectors separately, we keep the (+) sign instead of the (,) sign. ## Addition of two Vectors (جمع متجهين) **Example:** Find the magnitude and direction of vector C, where: * $C = A + B$. * $A = (6, \frac{2}{3}, - 1)$. * $B = (12, -10, 0)$. * **Solution:** ## Page 8 **Date:** 5/10/2024 * $A = A[l i + m j + n k]$. * $A = [\frac{6}{3}i + \frac{2}{3}j + \frac{-3}{3}k]$. * $A = 4i + \frac{2}{3}j + 4k$. * $B = 12i - 10j$. * $C = A + B = 16i - 8j + 4k$. * $C = \sqrt{16^2 +(-8)^2 + 4^2} = 12$. * $\cos\alpha = \frac{C_x}{C} = \frac{16}{12}=\frac{4}{3}$. * $\cos\beta = \frac{C_y}{C} = \frac{-8}{12}=-\frac{2}{3}$. * $\cos\gamma = \frac{C_z}{C} = \frac{4}{12}=\frac{1}{3}$. * **Go to Page 13 in the book.**