Vector Components

Learning Outcomes

  • Express a vector in component form.
  • Explain the formula for the magnitude of a vector.

Working with vectors in a plane is easier when we are working in a coordinate system. When the initial points and terminal points of vectors are given in Cartesian coordinates, computations become straightforward.

Example: Comparing Vectors

Are [latex]\bf{v}[/latex] and [latex]\bf{w}[/latex] equivalent vectors?

Try It

Which of the following vectors are equivalent?

alt="This figure is a coordinate system with 6 vectors, each labeled a through f. Three of the vectors, “a,” “b,” and “e” have the same length and are pointing in the same direction."

Figure 3. Determine which vectors are equivalent.

We have seen how to plot a vector when we are given an initial point and a terminal point. However, because a vector can be placed anywhere in a plane, it may be easier to perform calculations with a vector when its initial point coincides with the origin. We call a vector with its initial point at the origin a standard-position vector. Because the initial point of any vector in standard position is known to be [latex](0,0)[/latex], we can describe the vector by looking at the coordinates of its terminal point. Thus, if vector [latex]\bf{v}[/latex] has its initial point at the origin and its terminal point at [latex](x,y)[/latex], we write the vector in component form as

[latex]\bf{v}[/latex] [latex]= \langle x,y \rangle[/latex].

When a vector is written in component form like this, the scalars [latex]x[/latex] and [latex]y[/latex] are called the components of [latex]\bf{v}[/latex].

Definition

The vector with initial point [latex](0,0)[/latex] and terminal point [latex](x,y)[/latex] can be written in component form as

[latex]\bf{v}[/latex] [latex]= \langle x,y \rangle[/latex].

The scalars [latex]x[/latex] and [latex]y[/latex] are called the components of [latex]\bf{v}[/latex].

Recall that vectors are named with lowercase letters in bold type or by drawing an arrow over their name. We have also learned that we can name a vector by its component form, with the coordinates of its terminal point in angle brackets. However, when writing the component form of a vector, it is important to distinguish between [latex]\langle x,y \rangle[/latex] and [latex](x,y)[/latex]. The first ordered pair uses angle brackets to describe a vector, whereas the second uses parentheses to describe a point in a plane. The initial point of [latex]\langle x,y \rangle[/latex] is [latex](0,0)[/latex]; the terminal point of [latex]\langle x,y \rangle[/latex] is [latex](x,y)[/latex].

When we have a vector not already in standard position, we can determine its component form in one of two ways. We can use a geometric approach, in which we sketch the vector in the coordinate plane, and then sketch an equivalent standard-position vector. Alternatively, we can find it algebraically, using the coordinates of the initial point and the terminal point. To find it algebraically, we subtract the [latex]x[/latex]-coordinate of the initial point from the [latex]x[/latex]-coordinate of the terminal point to get the [latex]x[/latex] component, and we subtract the [latex]y[/latex]-coordinate of the initial point from the [latex]y[/latex]-coordinate of the terminal point to get the [latex]y[/latex] component.

Component Form of a Vector

Let [latex]\bf{v}[/latex] be a vector with initial point [latex](x_i,y_i)[/latex] and terminal point [latex](x_t,y_t)[/latex]. Then we can express [latex]\bf{v}[/latex] in component form as
[latex]\bf{v}[/latex] [latex]=\langle x_t − x_i,y_t − y_i\rangle[/latex].

Example: Expressing Vectors in Component Form

Express vector [latex]\bf{v}[/latex] with initial point [latex](−3,4)[/latex] and terminal point [latex](1,2)[/latex] in component form.

Try It

Vector [latex]\bf{w}[/latex] has initial point [latex](−4,−5)[/latex] and terminal point [latex](−1,2)[/latex]. Express [latex]\bf{w}[/latex] in component form.

To find the magnitude of a vector, we calculate the distance between its initial point and its terminal point. The magnitude of vector
[latex]\bf{v}[/latex][latex]=\langle x,y\rangle[/latex] is denoted [latex]\bf{||v||}[/latex], or [latex]\bf{|v|}[/latex], and can be computed using the formula

[latex]\bf{||v||}[/latex] [latex]=\sqrt{x^2 + y^2}[/latex]

.

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 [latex](x,y)[/latex]. Thus, it suffices to calculate the magnitude of the vector in standard position. Using the distance formula to calculate the distance between initial point [latex](0,0)[/latex] and terminal point [latex](x,y)[/latex], we have

[latex]\bf{||v||}[/latex][latex]=\sqrt{(x - 0)^2 + (y - 0)^2}[/latex]

 

[latex]=\sqrt{x^2 + y^2}[/latex]

 

Based on this formula, it is clear that for any vector [latex]\bf{v}[/latex], [latex]\bf{||v||}[/latex][latex]\geq 0[/latex], and [latex]\bf{||v||}[/latex][latex]= 0[/latex] if and only if [latex]\bf{v}[/latex][latex]= 0[/latex].

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).”

Figure 5. 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 [latex]\bf{v}[/latex][latex]= \langle x_1, y_1\rangle.[/latex] and [latex]\bf{w}[/latex][latex]= \langle x_2, y_2\rangle[/latex] be vectors, and let [latex]k[/latex] be a scalar.

Scalar multiplication: [latex]k\bf{v}[/latex][latex]=\langle kx_1, ky_1\rangle[/latex]

Vector addition: [latex]\bf{v + w}[/latex][latex]=\langle x_1, y_1\rangle + \langle x_2, y_2\rangle = \langle x_1 + x_2, y_1 + y_2\rangle[/latex]

Example: Performing Operations in Component Form

Let [latex]\bf{v}[/latex] be the vector with initial point [latex](2,5)[/latex] and terminal point [latex](8,13)[/latex], and let [latex]\bf{w}[/latex][latex]= \langle−2,4\rangle[/latex].

Try It

Let [latex]\bf{a}[/latex][latex]=\langle7,1\rangle[/latex] and let [latex]\bf{b}[/latex] be the vector with initial point [latex](3,2)[/latex] and terminal point [latex](−1,−1)[/latex].

Watch the following video to see the worked solution to the above Try IT.

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 Theorem

Let [latex]\bf{u}[/latex],[latex]\bf{v}[/latex], and [latex]\bf{w}[/latex] be vectors in a plane. Let [latex]r[/latex] and [latex]s[/latex] be scalars.

\[ \begin{array}{lrl}
\mbox{i.} & \textbf{u} + \textbf{v} &= \textbf{v} +\textbf{u} & \mbox{Commutative property} \\
\mbox{ii.} & (\textbf{u} + \textbf{v}) + \textbf{w} &= \textbf{u} + (\textbf{v} + \textbf{w}) & \mbox{Associative property} \\
\mbox{iii.} & \textbf{u} + 0 &= \textbf{u} & \mbox{Additive identity property} \\
\mbox{iv.} & \textbf{u} + (-\textbf{u}) &= 0 & \mbox{Additive inverse property} \\
\mbox{v.} & r(s\textbf{u}) &= (rs)\textbf{u} & \mbox{Associativity of scalar multiplication} \\
\mbox{vi.} & (r+s)\textbf{u} &= r\textbf{u} +s\textbf{u} & \mbox{Distributive property} \\
\mbox{vii.} & r(\textbf{u} + \textbf{v}) &= r\textbf{u} + r\textbf{v} & \mbox{Distributive property} \\
\mbox{viii.} & 1\textbf{u} &= \textbf{u} , 0\textbf{u} = 0 & \mbox{Identity and zero properties}\end{array}\]

Proof of Cummutative Property

Let [latex]{\bf{u}}=\langle x_1, y_1\rangle[/latex] and [latex]{\bf{v}}=\langle x_2, y_2\rangle[/latex]. Apply the commutative property for real numbers:

[latex]{\bf{u}} + {\bf{v}} = \langle x_1 + x_2, y_1 + y_2\rangle = \langle x_2 + x_1, y_2 + y_1\rangle = {\bf{u}} + {\bf{v}}[/latex].

[latex]_\blacksquare[/latex]

Proof of Distrubutive Property

Apply the distributive property for real numbers:

[latex]\begin{array}{ccc} \hfill r({\bf{u}} + {\bf{v}}) & =\hfill & r \cdot \langle x_1 + x_2,y_1 + y_2\rangle \hfill \\ \hfill & =\hfill & \langle r(x_1 + x_2), r(y_1 + y_2)\rangle \hfill \\ \hfill & =\hfill & \langle rx_1 + rx_2, ry_1 + ry_2\rangle \hfill \\ \hfill & =\hfill & \langle rx_1,ry_1\rangle \langle rx_2, ry_2\rangle \hfill \\ \hfill & =\hfill & r{\bf{u}} + r{\bf{v}}\end{array}[/latex]

[latex]_\blacksquare[/latex]

Try It

Prove the additive inverse property.

We have found the components of a vector given its initial and terminal points. In some cases, we may only have the magnitude and direction of a vector, not the points. For these vectors, we can identify the horizontal and vertical components using trigonometry (Figure 16).

This figure is a right triangle. There is an angle labeled theta. The two sides are labeled “magnitude of v times cosine theta” and “magnitude of v times sine theta.” The hypotenuse is labeled “magnitude of v.”

Figure 7. The components of a vector form the legs of a right triangle, with the vector as the hypotenuse.

Consider the angle [latex]\theta[/latex] formed by the vector [latex]{\bf{v}}[/latex] and the positive [latex]x[/latex]-axis. We can see from the triangle that the components of vector [latex]{\bf{v}}[/latex] are [latex]\langle ||{\bf{v}}||\cos{\theta},||{\bf{v}}||\sin{\theta} \rangle[/latex]. Therefore, given an angle and the magnitude of a vector, we can use the cosine and sine of the angle to find the components of the vector.

Example: Finding the Component Form of a Vector Using Trigonometry

Find the component form of a vector with magnitude [latex]4[/latex] that forms an angle of [latex]−45°[/latex] with the [latex]x[/latex]-axis.

Try It

Find the component form of vector [latex]{\bf{v}}[/latex] with magnitude [latex]10[/latex] that forms an angle of [latex]120°[/latex] with the positive [latex]x[/latex]-axis.

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