Section 1.2 Vectors 

1.2 Vectors 

This section is an introduction for the vectors in 2-dimensional space and 3-dimensional space. 

Definition:

(a) A vector is a quantity that has both magnitude and direction. We use the notation [latex]\overrightarrow{v}[/latex] to present a vector and [latex]||\overrightarrow{v}||[/latex] to present the magnitude. The zero vector is a vector with magnitude [latex]0[/latex].

(b) Vectors, [latex]\overrightarrow{u}[/latex] and [latex]\overrightarrow{v}[/latex], are said to be parallel if they have the same direction. We write [latex]\overrightarrow{u} \parallel \overrightarrow{v}[/latex].

Vectors, [latex]\overrightarrow{u}[/latex] and [latex]\overrightarrow{v}[/latex], are said to be equivalent vectors if they have the same magnitude and direction. We write [latex]\overrightarrow{u}=\overrightarrow{v}[/latex]. 

(c) The vector with initial point [latex](0,0)[/latex] or [latex](0,0,0)[/latex] and

terminal point [latex](x,y)[/latex] or [latex](x,y,z)[/latex] can be written in component form as [latex]\overrightarrow{v}=\lt x,y \gt[/latex]  or [latex]\overrightarrow{v}=\lt x,y,z \gt[/latex].

The scalars [latex]x,y,z[/latex] are called the components of v. The magnitude of [latex]\overrightarrow{v}[/latex] is [latex]||\overrightarrow{v}||=\sqrt{x^{2}+y^{2}}[/latex]

or [latex]||\overrightarrow{v}||=\sqrt{x^{2}+y^{2}+z^{2}}[/latex].

EQ1: The magnitude of the vector [latex]\overrightarrow{v}=\lt 1,-2,3 \gt[/latex] is 

A: [latex]1^{2}-2^{2}+3^{2}[/latex]; B: [latex]1^{2}+2^{2}+3^{2}[/latex];

C: [latex]\sqrt{1^{2}+2^{2}+3^{2}}[/latex]; D: [latex]\sqrt{1^{2}-2^{2}+3^{2}}[/latex]

Definition 

(a) Let [latex]k[/latex] be a real number. The product [latex]k\overrightarrow{v}[/latex] of a vector [latex]\overrightarrow{v}[/latex] and a scalar [latex]k[/latex] is a vector with a magnitude [latex]|k|||\overrightarrow{v}||[/latex] and with a direction that is the same as the direction of [latex]\overrightarrow{v}[/latex] if [latex] k\gt 0[/latex], and opposite the direction of [latex]\overrightarrow{v}[/latex] if [latex] k\lt 0[/latex]. This is called scalar multiplication. If [latex]k=0[/latex] or [latex]\overrightarrow{v}=0[/latex], then [latex]k\overrightarrow{v}=0[/latex]. Let [latex]\overrightarrow{v}=\lt x,y \gt[/latex]  or [latex]\overrightarrow{v}=\lt x,y,z \gt[/latex], then [latex]k\overrightarrow{v}=\lt kx,ky \gt[/latex]  or [latex]k\overrightarrow{v}=\lt kx,ky,kz \gt[/latex].

(b) The sum of two vectors [latex]\overrightarrow{u}[/latex] and [latex]\overrightarrow{v}[/latex], [latex]\overrightarrow{u}+\overrightarrow{v}[/latex], can be constructed graphically by placing the initial point of [latex]\overrightarrow{v}[/latex] at the terminal point of [latex]\overrightarrow{u}[/latex] or vice versa. Let  [latex]\overrightarrow{u}=\lt x_1,y_1 \gt[/latex]  or [latex]\overrightarrow{u}=\lt x_1,y_1,z_1 \gt[/latex],  and  [latex]\overrightarrow{v}=\lt x_2,y_2 \gt[/latex]  or [latex]\overrightarrow{v}=\lt x_2,y_2,z_2 \gt[/latex], then  [latex]\overrightarrow{u}+\overrightarrow{v}=\lt x_1+x_2,y_2+y_2 \gt[/latex]  or [latex]\overrightarrow{u}+\overrightarrow{v}=\lt x_1+x_2,y_1+y_2,z_1+z_2 \gt[/latex].

Example 1: In 2-dimensional space, a vector with starting point [latex]A=(x_{1},y_{1})[/latex] and ending point [latex]B=(x_{2},y_{2})[/latex], we write vector [latex]\overrightarrow{v}=\overrightarrow{AB}=\lt x_2-x_1,y_2-y_1 \gt[/latex]  to present the vector. The magnitude of the vector [latex]\overrightarrow{AB}[/latex] is the length of the line segment [latex]\overline{AB}=||\overrightarrow{v}||=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/latex]. This makes sense because if we let [latex]O=(0,0)[/latex] then [latex]\overrightarrow{OA}=\lt x_1,y_1 \gt[/latex]  and [latex]\overrightarrow{OB}=\lt x_2,y_2 \gt[/latex]  then [latex] \overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{OB}+(-\overrightarrow{OA})=\lt x_2,y_2 \gt +\lt -x_1,-y_1 \gt =\lt x_2-x_1,y_2-y_1 \gt .[/latex]

Example 2: In 3-dimensional space, a vector with starting point [latex]A=(x_{1},y_{1},z_{1})[/latex] and ending point [latex]B=(x_{2},y_{2},z_{2})[/latex], we write vector [latex]\overrightarrow{v}=\overrightarrow{AB}=\lt x_2-x_1,y_2-y_1,z_2-z_1 \gt[/latex]  to present the vector. The magnitude of the vector [latex]\overrightarrow{AB}[/latex] is the length of the line segment [latex]\overline{AB}=||\overrightarrow{v}||=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}[/latex].

EQ2: Which one presents a vector starting from the point [latex]A=(2,1,-1)[/latex] and ending at the point [latex]B=(-2,1,3)[/latex]?

A: [latex]\lt 0,2,2\gt[/latex]; B: [latex]\lt-4,0,4\gt[/latex]; C: [latex]\lt 4,0,-4\gt[/latex]; D: [latex]\lt -4,1,2\gt[/latex]

Definition: (a) A unit vector is a vector with magnitude [latex]1[/latex]. For any nonzero vector [latex]\overrightarrow{v}[/latex], we can use scalar multiplication to find a unit vector [latex]\overrightarrow{u}[/latex] that has the same direction as [latex]\overrightarrow{v}[/latex]. To do this, we multiply the vector [latex]\overrightarrow{v}[/latex] by the reciprocal of its magnitude, [latex]\frac{1}{||\overrightarrow{v}||}[/latex]:  \[\overrightarrow{u}=\frac{1}{||\overrightarrow{v}||}\overrightarrow{v}.\]

(b) Standard basis vectors are vectors of [latex]\overrightarrow{i}= \lt 1,0\gt[/latex] and [latex]\overrightarrow{j}= \lt 0,1\gt[/latex] in 2-dimensional space and [latex]\overrightarrow{i}= \lt 1,0,0\gt[/latex], [latex]\overrightarrow{j}= \lt 0,1,0\gt[/latex] and [latex]\overrightarrow{k}= \lt 0,0,1\gt[/latex]. They are all unit vectors and any vector can be written as combination of standard basis vectors: [latex]\overrightarrow{v}= \lt x,y\gt =x \lt 1,0\gt +y \lt 0,1\gt =x\overrightarrow{i}+y\overrightarrow{j}[/latex] or [latex]\overrightarrow{v}= \lt x,y,z\gt =x \lt 1,0,0\gt +y \lt 0,1,0\gt +z \lt 0,0,1\gt   =x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}[/latex].

EQ3: The unit vector that has the same direction of [latex]\overrightarrow{v}= \lt -2,0,3\gt[/latex] is A: [latex]13 \lt -2,0,3\gt[/latex]; B: [latex]\sqrt{13} \lt -2,0,3\gt ; C: [latex]\frac{1}{13} \lt -2,0,3\gt[/latex] D: [latex]\frac{1}{\sqrt{13}} \lt -2,0,3\gt[/latex]

Example 3: (a) Express the vector [latex]\overrightarrow{w}= \lt 5,-12\gt[/latex] in terms of standard basis vectors.  (b)Vector [latex]\overrightarrow{u}[/latex] is a unit vector that forms an angle of [latex]\frac{\pi}{6}[/latex] with the positive [latex]x[/latex]-axis. Use standard basis vectors to describe [latex]\overrightarrow{u}[/latex].

Exercise 1: (a) Express the vector [latex]\overrightarrow{w}= \lt -3,4\gt[/latex] in terms of standard basis vectors.  (b)Vector [latex]\overrightarrow{u}[/latex] is a unit vector that forms an angle of [latex]\frac{\pi}{3}[/latex] with the positive [latex]x[/latex]-axis. Use standard basis vectors to describe [latex]\overrightarrow{u}[/latex].

Theorem: Properties of Vector Operations

Let [latex]\overrightarrow{u},\overrightarrow{v},[/latex] and [latex]\overrightarrow{w}[/latex] be vectors. Let [latex]r[/latex] and [latex]s[/latex] be scalars. }

i. [latex]\overrightarrow{u}+\overrightarrow{v}=\overrightarrow{v}+\overrightarrow{u}[/latex], Commutative property

ii. [latex](\overrightarrow{u}+\overrightarrow{v})+\overrightarrow{w}=\overrightarrow{u}+(\overrightarrow{v}+\overrightarrow{w})[/latex], Associative property

iii. [latex]\overrightarrow{u}+\overrightarrow{0}=\overrightarrow{u}[/latex], Additive identity property

iv. [latex]\overrightarrow{u}+(-\overrightarrow{u})=\overrightarrow{0}[/latex], Additive inverse property

v. [latex]r(s\overrightarrow{u})=(rs)\overrightarrow{u}[/latex], Associativity of scalar multiplication

vi. [latex](r+s)\overrightarrow{u}=r\overrightarrow{u}+s\overrightarrow{u}[/latex], Distributive property

vii. [latex]r(\overrightarrow{u}+\overrightarrow{v})=r\overrightarrow{u}+r\overrightarrow{v}[/latex], Distributive property

viii. [latex]1\overrightarrow{u}=\overrightarrow{u},\text{ }0\overrightarrow{u}=0[/latex] Identity and zero properties

EQ4: Let [latex]\overrightarrow{u}= \lt 1,-2,3\gt[/latex], [latex]\overrightarrow{v}= \lt -1,1,-2\gt[/latex] and [latex]\overrightarrow{w}= \lt 2,-1,-1\gt[/latex]. Which of the following is wrong?

A: [latex]||\overrightarrow{u}||=\sqrt{1+4+9}[/latex];

B: [latex]3\overrightarrow{v}+4\overrightarrow{w}= \lt 5,-1,-10\gt[/latex];C: [latex]\overrightarrow{v}-\overrightarrow{u}= \lt -2,3,-5\gt[/latex];

D: [latex]-\overrightarrow{w}= \lt -2,-1,-1\gt[/latex] .

A force is represented by a vector because it has both a magnitude and a direction. If several forces are acting on an object, the resultant force experienced by the object is the vector sum of these forces.

Example 4: A football thrown by a quarterback has an initial speed of 70 mph and an angle of elevation of 30°. Determine the velocity vector in mph and express it in component form. 

Example 5: Jane's car is stuck in the mud. Lisa and Jed come along in a truck to help pull her out. They attach one end of a tow strap to the front of the car and the other end to the truck's trailer hitch, and the truck starts to pull. Meanwhile, Jane and Jed get behind the car and push. The truck generates a horizontal force of 300 lb on the car. Jane and Jed are pushing at a slight upward angle [latex]\frac{\pi}{6}[/latex] and generate a force of 150 lb on the car. Find the resultant force, and give its magnitude, and its direction angle from the positive [latex]x[/latex]-axis.

Group work:

1. A bullet is fired with an initial velocity of 1000 ft/sec at an angle of [latex]\frac{\pi}{3}[/latex] with the horizontal. Find the horizontal and vertical components of the velocity vector of the bullet.

2. Two forces, a horizontal force of 40 lb and another of 50 lb, act on the same object. The angle between these forces is 45°. Find the magnitude and direction angle from the positive [latex]x[/latex]-axis of the resultant force that acts on the object. 

3. Two forces, a vertical force of 20 lb and another of 45 lb, act on the same object. The angle between these forces is 60°. Find the magnitude and direction angle from the positive [latex]x[/latex]-axis of the resultant force that acts on the object. 

4. Three forces with magnitudes 80 lb, 120 lb, and 60 lb act on an object at angles of 45°, 60° and 30°, respectively, with the positive [latex]x[/latex]-axis. Find the magnitude and direction angle from the positive x-axis of the resultant force.