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Subsection 2.9.5 Manipulating Vectors

Manipulating vectors involves performing operations such as addition, scalar multiplication, and translation. These operations allow us to combine and transform vectors in various ways. Some of the key methods for manipulating vectors are graphical addition and subtraction, algebraic addition and subtraction, scalar multiplication, and translation.

Subsubsection 2.9.5.1 Adding Vectors

Learner Experience 2.9.53.

Work in groups
What you require: Graph paper,ruler

(a)

Draw the \(x\) and \(y\) axis on the graph paper.

(b)

Draw vector \(\textbf{AB}\) from point \(A(0,0)\) to point \(B(2,2)\text{.}\)

(c)

Draw vector \(\textbf{BC}\) from point \(B(2,2)\) to point \(C(5,2)\text{.}\)

(d)

Count the number of units moved horizontally (along the \(x\) axis) from the starting point \(A\) to the final point \(C\text{.}\)

(e)

Similarly, count the number of units moved vertically (along the \(y\) axis) from point \(A\) to point \(C\text{.}\)

(f)

Write the resultant displacement in coordinate form \(\begin{pmatrix} x \\ y \end{pmatrix}\text{,}\) where \(x\) represents displacement along the \(x\) axis and \(y\) represents displacement along the \(y\) axis.

(g)

Discuss and share your findings with the rest of the class.

Key Takeaway 2.9.54.

Consider a displacement from point \(P\) to point \(Q\text{,}\) followed by another displacement from point \(Q\) to point \(N\text{.}\) The total resultant displacement from \(P\) to \(N\) is obtained by adding the two vectors sequentially.
This can be expressed as:
\begin{align*} \mathbf{PN} \amp = \mathbf{PQ} + \mathbf{QN} \end{align*}
where:
  1. \(\mathbf{PQ} = \mathbf{r}\) represents the first displacement.
  2. \(\mathbf{QN} = \mathbf{s}\) represents the second displacement.
Therefore, \(\mathbf{PN} = \mathbf{r} + \mathbf{s}\)
Figure 2.9.55.
We can also re-arrange the two vectors and add them together as shown in Figure 2.9.56 below.
Figure 2.9.56.

Example 2.9.57.

In Figure 2.9.58 below, find vector \(\mathbf{AD}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\text{.}\)
Figure 2.9.58.
Solution.
\begin{align*} \overrightarrow{AD} \amp = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD}\\ \amp = \mathbf{a} + \mathbf{b} + \mathbf{c} \end{align*}
Thus, \(\overrightarrow{AD} = \mathbf{a} + \mathbf{b} + \mathbf{c} \)

Exercises Exercises

1.

\(PQNM\) is a square with vectors \(\mathbf{PQ}\) and \(\mathbf{PM}\) given as \(\mathbf{a} \text{ and } \mathbf{b}\) respectively, as shown in Figure 2.9.59 . Express the \(\mathbf{PN}\) and \(\mathbf{MQ}\) vectors in terms of \(\mathbf{a}\) and \(\mathbf{b}\)
Figure 2.9.59.
Answer.
\(\mathbf{PN} = \mathbf{a} + \mathbf{b}\)
\(\mathbf{MQ} = \mathbf{a} - \mathbf{b}\) or \(-\mathbf{b} + \mathbf{a}\)

2.

Use Figure 2.9.60 to answer the questions below:
Figure 2.9.60.
  1. List pairs of equal vectors.
  2. Name pairs of vectors with equal magnitude but opposite directions.
Answer.
  1. Pairs of equal vectors
    • \(\mathbf{e}\) and \(\mathbf{d}\) (both are \(2\) units right, \(1\) unit up).
    • \(\mathbf{g}\) and \(\mathbf{h}\) (both are \(3\) units right, \(2\) units up).
  2. Equal magnitude, opposite direction
    • \(\mathbf{f} \) and \(\mathbf{b}\) both have same magnitude but different directions.

3.

Given the vectors \(\mathbf{a} = \binom{2}{3}\) and \(\mathbf{b} = \binom{4}{-1}\text{,}\) find \(\mathbf{a} + \mathbf{b}\) and illustrate the solution graphically.
Answer.
\(\mathbf{a} + \mathbf{b} = \binom{6}{2}\)
Figure 2.9.61.

Checkpoint 2.9.62.

Checkpoint 2.9.63.

Checkpoint 2.9.64.

Subsection 2.9.5.1 Multiplying Vectors by Scalars

Learner Experience 2.9.66.

Work in groups

(a)

On a graph paper, draw the \(x\) axis and \(y\) axis.

(b)

Draw a directed line passing through point \(A(0,2)\) and \(B(2,2)\text{.}\)

(c)

From point \(B(2,2)\text{,}\) draw directed line to point \(C(4,2)\text{.}\)

(d)

Determine the coordinate representations of \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\text{..}\)

(e)

How does \(\overrightarrow{AB}\) relate to \(\overrightarrow{AC}\text{?}\)

(f)

Discuss and share your findings with the rest of the class.
\(\textbf{Key Takeaway}\)
Positive Scalar
In Figure 2.9.67, the vector \(\mathbf{PQ}\) is represented as \(\mathbf{a}\text{.}\) When we multiply \(\mathbf{a}\) by a positive scalar, say \(2\text{,}\) the length of the vector doubles, making it \(\mathbf{2a}\) as shown in Figure 2.9.68. The direction of the vector remains unchanged, but its magnitude increases.
Figure 2.9.67.
In Figure 2.9.68, the vector \(\mathbf{PN}\) is given by: \(\mathbf{PN} = \mathbf{a} + \mathbf{a} = 2\mathbf{a}\text{.}\) This means \(\mathbf{PN}\) has the same direction as \(\mathbf{PQ}\text{,}\) but its magnitude twice that of \(\mathbf{PQ}.\)
Figure 2.9.68.
Negative Scalar
Consider the vector \(\mathbf{AB}\text{,}\)denoted as \(\mathbf{a}\text{,}\) in Figure 2.9.69 The vector points to the right and has a magnitude of \(\mathbf{a}\text{.}\)
Figure 2.9.69.
In Figure 2.9.70, the vector \(\mathbf{AB}\) is obtained by multiplying \(\mathbf{AB}\) by \(-2\text{,}\) giving:
\begin{equation*} \mathbf{AC} = \mathbf{-2} \times \mathbf{a} = \mathbf{-2a} \end{equation*}
This means that \(\mathbf{AC}\) has twice the magnitude of \(\mathbf{AB}\text{,}\) but its direction is reversed.
Figure 2.9.70.
Multiplying a vector by a negative scalar reverses its direction, making it point in the opposite direction.
Zero Scalar
When a vector \(\mathbf{a}\text{,}\) as shown in Figure 2.9.71, is multiplied by \(0\text{,}\) its magnitude becomes \(0\text{,}\) resulting in a zero vector.
Figure 2.9.71.
\begin{equation*} \mathbf{a} \cdot 0 = 0 \end{equation*}

Example 2.9.72.

Consider the vector \(\vec{v} = \langle 3, -4, 5 \rangle\text{.}\)
  1. Multiply the vector by the scalar \(k = 2\text{.}\)
  2. Multiply the vector by the scalar \(k = -\frac{1}{2}\text{.}\)
  3. Explain the geometric effect of multiplying a vector by a scalar.
Solution.
  1. Multiplying by \(k = 2\text{:}\)
    \(2 \vec{v} = 2 \langle 3, -4, 5 \rangle = \langle 6, -8, 10 \rangle\)
  2. Multiplying by \(k = -\frac{1}{2}\text{:}\)
    \(-\frac{1}{2} \vec{v} = -\frac{1}{2} \langle 3, -4, 5 \rangle = \langle -\frac{3}{2}, 2, -\frac{5}{2} \rangle\)
  3. Geometric effect:
    1. Multiplying a vector by a positive scalar \(k > 1\) stretches the vector in the same direction.
    2. Multiplying by a scalar \(0 < k < 1\) shrinks the vector.
    3. Multiplying by a negative scalar \(k < 0\) reverses the direction of the vector and scales its magnitude by the absolute value of the scalar.

Example 2.9.73.

Given the vectors \(u = 2p + 5q\) and \(v = p - 3q\text{,}\) express \(3u + 2v \) in terms of \(p\) and \(q:\)
Solution.
Substitute the given expressions for u and v.
\begin{align*} 3u + 2v \amp = 3(2p + 5q) + 2(p - 3q) \\ \amp = 6p + 15q + 2p - 6q \end{align*}
Combine like terms (terms with p and terms with q).
\begin{align*} 3u + 2v \amp = (6p + 2p) + (15q - 6q) \\ \amp = 8p + 9q \end{align*}

Exercises Exercises

1.

Simplify the following:
  1. \(\displaystyle 5x + 3y - z + 2(3x - z) + (8x - 6y)\)
  2. \(\displaystyle (\mathbf{a} - \mathbf{b}) + (c - \mathbf{a}) + (\mathbf{b} - c)\)
  3. \(\displaystyle 4m - 2n + 5(k - m) + 2(m + n)\)
Answer.
  1. \(\displaystyle 19x - 3y - 3z\)
  2. \(\displaystyle 0\)
  3. \(\displaystyle m + 5k\)

2.

Given that \(x = 3m - n\) and \(y = n +4m\text{,}\) express the following vectors in terms of \(m\) and \(n:\)
  1. \(\displaystyle 3x\)
  2. \(\displaystyle \frac{2}{3} y\)
  3. \(\displaystyle 6x - 9y\)
  4. \(\displaystyle 3y - x\)
Answer.
  1. \(\displaystyle 9m - 3n\)
  2. \(\displaystyle \frac{8}{3}m + \frac{2}{3}n\)
  3. \(\displaystyle -18m - 15n\)
  4. \(\displaystyle 9m + 4n\)

3.

A pentagon \(ABCDE\) with \(\overrightarrow{AB} = m\text{,}\) \(\overrightarrow{BC} = n\text{,}\) and \(\overrightarrow{CD} = k\text{.}\) Express the following vectors in terms of \(m\text{,}\) \(n\text{,}\) and \(k:\)
  1. \(\displaystyle \overrightarrow{AC}\)
  2. \(\displaystyle \overrightarrow{AD}\)
Answer.
  1. \(\displaystyle m+n\)
  2. \(\displaystyle m+n+k\)

Checkpoint 2.9.74.

Subsubsection 2.9.5.2 Translating Vectors

Learner Experience 2.9.76.

Work in groups
What you require: Graph paper

(a)

Draw the \(x\) axis and \(y\) axis on the graph paper.

(b)

Plot the triangle with vertices \(A(-3,1)\text{,}\) \(B(-1,1)\text{,}\) and \(C(-2,3)\text{.}\)

(c)

Translate each point by moving \(2\) units to the right parallel to the \(x\) axis and \(3\) units up in the \(y\) axis. Label the new points as \(A', B',\) and \(C'\text{.}\)

(d)

Draw the new triangle \(A'B'C'\) on the graph.

(e)

Use dotted lines to connect each original point to its corresponding translated point \((A \text{ to } A', B \text{ to } B', C \text{ to } C')\text{,}\) add arrows to indicate the direction.

(f)

Observe and describe any similarities between triangle \(ABC\) and triangle \(A'B'C'.\)

(g)

Analyze the distance each point moved.

(h)

Discuss and share your findings with your classmates in the class.

Key Takeaway 2.9.77.

A square \(\mathbf{ABCD}\) undergoes a translation when each of its vertices (\(\mathbf{A}\text{,}\)\(\mathbf{B}\text{,}\)\(\mathbf{C}\) and \(\mathbf{D}\)) is moved the same distance and in the same direction. A translation vector, denoted by \(\mathbf{T}\text{,}\) describes this movement.
Using \(\mathbf{T}\) to represent a translation, the notation \(\mathbf{T}(\mathbf{P})\) indicates the application of the translation \(\mathbf{T}\) on \(\mathbf{P}\text{.}\) In Figure 2.9.78, shows \(\mathbf{A'B'C'D'}\) is the image of \(\mathbf{ABCD}\) under a translation.
Figure 2.9.78.

Example 2.9.79.

The position vector of point \(P\) is \(\vec{OP} = \langle 2, 3 \rangle\) in the plane. A translation is represented by the vector \(\vec{v} = \langle 5, -2 \rangle\text{.}\)
Find the position vector of the new point \(Q\) after translating \(P\) by \(\vec{v}\text{.}\)
Solution.
Translation of a point by a vector is performed by adding the translation vector to the original position vector:
\(\vec{OQ} = \vec{OP} + \vec{v}\)
Substitute the values:
\begin{align*} \vec{OQ} = \langle 2, 3 \rangle + \langle 5, -2 \rangle = \langle 2+5, 3+(-2) \rangle \amp \\ = \amp \langle 7, 1 \rangle \end{align*}
Therefore, the new point \(Q\) has coordinates \((7,1)\text{.}\)
Geometric interpretation: The point moves \(5\) units to the right and \(2\) units down along the plane.

Example 2.9.80.

Triangle \(\mathbf{ABC}\) has vertices \(\mathbf{A}(1,3)\text{,}\) \(\mathbf{B}(3,0)\) and \(\mathbf{C}(4,4)\text{.}\)The triangle undergoes a translation \(\mathbf{T}\) defined by the vector \(\begin{pmatrix} 4 \\ 3 \end{pmatrix}\text{.}\)
  1. Determine the coordinates of the translated vertices \(\mathbf{A'}\text{,}\) \(\mathbf{B'}\) , and \(\mathbf{C'}\) ’.
  2. Draw the triangle \(\mathbf{ABC}\) and its image under \(\mathbf{T}\text{.}\)
Solution.
To find the coordinates of the translated vertices, we apply the translation \(\mathbf{T}\) to each original vertex.
\begin{align*} \mathbf{OA'} \amp = \mathbf{OA} + \mathbf{T}\\ \mathbf{OB'} \amp = \mathbf{OB} + \mathbf{T}\\ \mathbf{OC'} \amp = \mathbf{OC} + \mathbf{T} \end{align*}
\begin{align*} \mathbf{OA'} \amp = \binom{1}{3} + \binom{4}{3} = \binom{1+4}{3+3} = \binom{5}{6} \\ \mathbf{OB'} \amp = \binom{3}{0} + \binom{4}{3} = \binom{3+4}{0+3} = \binom{7}{3} \\ \mathbf{OC'} \amp = \binom{4}{4} + \binom{4}{3} = \binom{4+4}{4+3} = \binom{8}{7} \end{align*}
Figure 2.9.81.

Exercises Exercises

1.

Draw triangle \(\mathbf{XYZ}\) with vertices \(\mathbf{X(1, 4)}\text{,}\) \(\mathbf{Y(6, 2)}\text{,}\) and \(\mathbf{Z(5, 3)}\text{.}\) On the same axes, plot \(\mathbf{X'Y'Z'}\text{,}\) the image of triangle \(\mathbf{XYZ}\) under a translation given by \(\binom{4}{9}.\)
Answer.
Figure 2.9.82.

2.

The following points have been translated using the given vectors. Determine their original positions:
  1. \(\displaystyle (4,-1); \, \, \binom{2}{3}\)
  2. \(\displaystyle (0,-3); \, \, \binom{1}{3}\)
  3. \(\displaystyle (-3,8); \, \, \binom{-2}{7}\)
  4. \(\displaystyle (11,-5); \,\, \binom{5}{-1}\)
  5. \(\displaystyle (-12,5); \,\, \binom{3}{-10}\)
  6. \(\displaystyle (2,-7); \,\, \binom{-10}{15}\)
Answer.
  1. \(\displaystyle (2,-4)\)
  2. \(\displaystyle (-1,-6)\)
  3. \(\displaystyle (-1,1)\)
  4. \(\displaystyle (6,-4)\)
  5. \(\displaystyle (-15,15)\)
  6. \(\displaystyle (12,-22)\)

3.

A point \(\mathbf{P(5, -3)}\) is mapped to a new position after a translation. If the new coordinates are \(\mathbf{(9, 1)}\text{,}\) determine the translation vector used.
Answer.
\(\binom{4}{4}\)

4.

A point \(\mathbf{M(1, -4)}\) undergoes a translation by \(\binom{3}{5}.\) Determine the coordinates of \(\mathbf{M'}\text{,}\) the transformed point. If \(\mathbf{M'}\) is then translated by \(\binom{-4}{2},\) find the final position \(\mathbf{M''}\text{.}\) What is the single translation vector that maps \(\mathbf{M'}\) to \(\mathbf{M''}\) directly?
Answer.
\(\binom{-1}{7}\)

5.

Translate each of the following points using the given vector:
  1. \(\displaystyle (10,1); \, \, \binom{-11}{2}\)
  2. \(\displaystyle (-2,-5); \, \, \binom{6}{14}\)
  3. \(\displaystyle (3,-15); \, \, \binom{-16}{11}\)
  4. \(\displaystyle (-11,4); \, \, \binom{-15}{10}\)
  5. \(\displaystyle (1,10); \, \, \binom{7}{2}\)
  6. \(\displaystyle (4,-9); \, \, \binom{5}{-8}\)
  7. \(\displaystyle (-2,13); \, \, \binom{-3}{1}\)
  8. \(\displaystyle (-6,5); \, \, \binom{-1}{2}\)
Answer.
  1. \(\displaystyle (-1, 3)\)
  2. \(\displaystyle (4, 9)\)
  3. \(\displaystyle (-13, -4)\)
  4. \(\displaystyle (-26, 14)\)
  5. \(\displaystyle (8, 12)\)
  6. \(\displaystyle (9, -17)\)
  7. \(\displaystyle (-5, 14)\)
  8. \(\displaystyle (-7, 7)\)

Checkpoint 2.9.83.

Checkpoint 2.9.84.

Subsubsection 2.9.5.3 Equivalent Vectors

Now that you understand how to represent vectors using magnitude and direction, we can explore how to compare them. Is it possible for two vectors to be considered “equal” even if they are located in different places on a graph? In this section, you will examine the specific conditions regarding length and direction that must be met for two vectors to be called equivalent.

Learner Experience 2.9.86.

Work in groups
What you require: Graph paper,ruler

(a)

Draw the \(x\) and \(y\) axis on the graph paper.

(b)

Plot the points \(A(0,4), B(3,4), C(0,2)\) and \(D(3,2)\text{.}\)

(c)

Draw a line to connect point \(A\) and \(B,\) add an arrow pointing to point \(B.\)

(d)

Draw a line to connect point \(C\) and \(D,\) add an arrow pointing to point \(D.\)

(e)

Look at the two arrows you have drawn. Do they look like "clones" (exact copies) of each other?

(f)

Imagine sliding the vector \(\mathbf{AB}\) straight down without turning it.
  • If you move point \(A\) so it sits exactly on top of point \(C\text{,}\) where does point \(B\) land?
  • Does it land exactly on point \(D\text{?}\)

(g)

Calculate the length of both vectors and compare the direction they are pointing. What two properties do vector \(\mathbf{AB}\) and vector \(\mathbf{CD}\) have in common?

(h)

Since these vectors share the exact same properties, discuss with your group what relationship exists between them.

Key Takeaway 2.9.87.

Two or more vectors are said to be equivalent if they satisfy the following conditions:
  1. They have same magnitude.
  2. They point in the same direction.
Figure 2.9.88.
In Figure 2.9.88 above, \(\textbf{MN} = \textbf{PQ}\) since they have same direction and equal magnitude.

Example 2.9.89.

Using Figure 2.9.90 below, determine whether vector \(\mathbf{AB}\) and \(\mathbf{DC}\) are equivalent.
Figure 2.9.90.
Solution.
Vector \(\mathbf{AB} \text{ and } \mathbf{DC}\) are equivalent because they have the same magnitude, \(|\mathbf{AB}| = |\mathbf{DC}| \text{,}\) and they point in the same direction

Exercises Exercises

1.

Is it possible for two vectors to have the same direction but not to be equivalent? Explain your answer.
Answer.
Yes. Two vectors are only equivalent if they have both the same direction and the same magnitude (length).

2.

In Figure 2.9.91, identify pairs of equivalent vectors and non-equivalent vectors.
Figure 2.9.91.
Answer.
  • Equivalent Pairs
    \(\overrightarrow{DA}\) and \(\overrightarrow{CB}\text{:}\) Both point directly upward and have the same length (sides of the square).
    \(\overrightarrow{DC}\) and \(\overrightarrow{AB}\text{:}\) Both point directly to the right and have the same length.
    \(\overrightarrow{DO}\) and \(\overrightarrow{OB}\) Both point in the same direction along the diagonal and have the same length.
  • Non-Equivalent Pairs
    \(\overrightarrow{DA}\) and \(\overrightarrow{BC}\text{:}\) While they have the same length, they point in opposite directions (up vs. down).
    \(\overrightarrow{DC}\) and \(\overrightarrow{DA}\text{:}\) These have the same length but different directions (horizontal vs. vertical).
    \(\overrightarrow{DB}\) and \(\overrightarrow{AC}\text{:}\) These are the diagonals; they have the same length but different directions.

3.

Draw two vectors that have the same magnitude and direction but start at different points.
Answer.
Figure 2.9.92.

Checkpoint 2.9.93.

Checkpoint 2.9.94.

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