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2.1 Vectors in the plane  (Page 5/29)

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Vector w has initial point ( −4 , −5 ) and terminal point ( −1 , 2 ) . Express w in component form.

⟨ 3 , 7 ⟩

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To find the magnitude of a vector, we calculate the distance between its initial point and its terminal point. The magnitude    of vector v = ⟨ x , y ⟩ is denoted ‖ v ‖ , or | v | , and can be computed using the formula

‖ v ‖ = x 2 + y 2 .

Note that because this vector is written in component form, it is equivalent to a vector in standard position, with its initial point at the origin and terminal point ( x , y ) . Thus, it suffices to calculate the magnitude of the vector in standard position. Using the distance formula to calculate the distance between initial point ( 0 , 0 ) and terminal point ( x , y ) , we have

‖ v ‖ = ( x − 0 ) 2 + ( y − 0 ) 2 = x 2 + y 2 .

Based on this formula, it is clear that for any vector v , ‖ v ‖ ≥ 0 , and ‖ v ‖ = 0 if and only if v = 0 .

The magnitude of a vector can also be derived using the Pythagorean theorem, as in the following figure.

This figure is a right triangle. The two sides are labeled “x” and “y.” The hypotenuse is represented as a vector and is labeled “square root (x^2 + y^2).”
If you use the components of a vector to define a right triangle, the magnitude of the vector is the length of the triangle’s hypotenuse.

We have defined scalar multiplication and vector addition geometrically. Expressing vectors in component form allows us to perform these same operations algebraically.

Definition

Let v = ⟨ x 1 , y 1 ⟩ and w = ⟨ x 2 , y 2 ⟩ be vectors, and let k be a scalar.

Scalar multiplication: k v = ⟨ k x 1 , k y 1 ⟩

Vector addition: v + w = ⟨ x 1 , y 1 ⟩ + ⟨ x 2 , y 2 ⟩ = ⟨ x 1 + x 2 , y 1 + y 2 ⟩

Performing operations in component form

Let v be the vector with initial point ( 2 , 5 ) and terminal point ( 8 , 13 ) , and let w = ⟨ −2 , 4 ⟩ .

  1. Express v in component form and find ‖ v ‖ . Then, using algebra, find
  2. v + w ,
  3. 3 v , and
  4. v − 2 w .
  1. To place the initial point of v at the origin, we must translate the vector 2 units to the left and 5 units down ( [link] ). Using the algebraic method, we can express v as v = ⟨ 8 − 2 , 13 − 5 ⟩ = ⟨ 6 , 8 ⟩ :
    ‖ v ‖ = 6 2 + 8 2 = 36 + 64 = 100 = 10 .

    This figure is the first quadrant of a coordinate system. It has two vectors. The first vector has initial point at (2, 5) and terminal point (8, 13). The second vector has initial point at the origin and terminal point at (6, 8).
    In component form, v = ⟨ 6 , 8 ⟩ .
  2. To find v + w , add the x -components and the y -components separately:
    v + w = ⟨ 6 , 8 ⟩ + ⟨ −2 , 4 ⟩ = ⟨ 4 , 12 ⟩ .
  3. To find 3 v , multiply v by the scalar k = 3 :
    3 v = 3 · ⟨ 6 , 8 ⟩ = ⟨ 3 · 6 , 3 · 8 ⟩ = ⟨ 18 , 24 ⟩ .
  4. To find v − 2 w , find −2 w and add it to v :
    v − 2 w = ⟨ 6 , 8 ⟩ − 2 · ⟨ −2 , 4 ⟩ = ⟨ 6 , 8 ⟩ + ⟨ 4 , −8 ⟩ = ⟨ 10 , 0 ⟩ .
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Let a = ⟨ 7 , 1 ⟩ and let b be the vector with initial point ( 3 , 2 ) and terminal point ( −1 , −1 ) .

  1. Find ‖ a ‖ .
  2. Express b in component form.
  3. Find 3 a − 4 b .

a. ‖ a ‖ = 5 2 , b. b = ⟨ −4 , −3 ⟩ , c. 3 a − 4 b = ⟨ 37 , 15 ⟩

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Now that we have established the basic rules of vector arithmetic, we can state the properties of vector operations. We will prove two of these properties. The others can be proved in a similar manner.

Properties of vector operations

Let u , v , and w be vectors in a plane. Let r and s be scalars.

i. u + v = v + u Commutative property ii. ( u + v ) + w = u + ( v + w ) Associative property iii. u + 0 = u Additive identity property iv. u + ( − u ) = 0 Additive inverse property v. r ( s u ) = ( r s ) u Associativity of scalar multiplication vi. ( r + s ) u = r u + s u Distributive property vii. r ( u + v ) = r u + r v Distributive property viii. 1 u = u , 0 u = 0 Identity and zero properties

Proof of commutative property

Let u = ⟨ x 1 , y 1 ⟩ and v = ⟨ x 2 , y 2 ⟩ . Apply the commutative property for real numbers:

u + v = ⟨ x 1 + x 2 , y 1 + y 2 ⟩ = ⟨ x 2 + x 1 , y 2 + y 1 ⟩ = v + u .

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Proof of distributive property

Apply the distributive property for real numbers:

r ( u + v ) = r · ⟨ x 1 + x 2 , y 1 + y 2 ⟩ = ⟨ r ( x 1 + x 2 ) , r ( y 1 + y 2 ) ⟩ = ⟨ r x 1 + r x 2 , r y 1 + r y 2 ⟩ = ⟨ r x 1 , r y 1 ⟩ + ⟨ r x 2 , r y 2 ⟩ = r u + r v .
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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