Vector Basics: Learn It 3

Vector Components

Working with vectors becomes much more straightforward when you use a coordinate system. Instead of relying purely on geometric methods, you can use the coordinates of initial and terminal points to perform precise calculations.

Since vectors can be positioned anywhere in the plane, calculations become easier when we place the vector’s initial point at the origin. A vector positioned this way is called a standard-position vector.

For any standard-position vector, the initial point is [latex](0,0)[/latex], so you only need the terminal point coordinates to describe the vector completely. If vector [latex]\mathbf{v}[/latex] has terminal point [latex](x,y)[/latex], we write:

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

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

component form

A vector [latex]\mathbf{v}[/latex] with initial point at the origin [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].

Use angle brackets [latex]\langle x,y \rangle[/latex] for vectors in component form, not parentheses [latex](x,y)[/latex]. Parentheses describe points in the plane, while angle brackets describe vectors. The vector [latex]\langle x,y \rangle[/latex] starts at [latex](0,0)[/latex] and ends at [latex](x,y)[/latex].

Converting Any Vector to Component Form

When you have a vector that’s not already in standard position, you can find its component form using two approaches:

Geometric Approach: Sketch the vector in the coordinate plane, then draw an equivalent standard-position vector.

Algebraic Approach: Use the coordinates directly. If a vector has initial point [latex](x_1, y_1)[/latex] and terminal point [latex](x_2, y_2)[/latex], then:

  • [latex]x[/latex]-component = [latex]x_2 - x_1[/latex]
  • [latex]y[/latex]-component = [latex]y_2 - y_1[/latex]

The vector in component form is [latex]\langle x_2 - x_1, y_2 - y_1 \rangle[/latex].

converting to component form

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

Vector Magnitude

To find the magnitude of a vector, you calculate the distance between its initial point and terminal point. For a vector [latex]\mathbf{v} = \langle x,y \rangle[/latex], the magnitude is:

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

This formula comes from the distance formula. Since any vector in component form is equivalent to a standard-position vector with initial point [latex](0,0)[/latex] and terminal point [latex](x,y)[/latex], we calculate:

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

Note that [latex]||\mathbf{v}|| \geq 0[/latex] for any vector [latex]\mathbf{v}[/latex], and [latex]||\mathbf{v}|| = 0[/latex] if and only if [latex]\mathbf{v} = \mathbf{0}[/latex].

You can also think of this using the Pythagorean theorem—the vector’s components form the legs of a right triangle, and the magnitude is the hypotenuse.

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.

Operations with Component Form

Expressing vectors in component form makes scalar multiplication and vector addition much simpler—you can perform these operations algebraically instead of geometrically.

vector operations in component form

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

  • Scalar multiplication: [latex]k\mathbf{v} = \langle kx_1, ky_1 \rangle[/latex]
  • Vector addition: [latex]\mathbf{v} + \mathbf{w} = \langle x_1 + x_2, y_1 + y_2 \rangle[/latex]
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].

Properties of Vector Operations

Vector operations follow familiar algebraic properties. These properties make vector calculations predictable and allow you to manipulate vector expressions just like algebraic expressions.

properties of vector operations

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

Addition Properties:

  • [latex]\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}[/latex] (Commutative)
  • [latex](\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})[/latex] (Associative)
  • [latex]\mathbf{u} + \mathbf{0} = \mathbf{u}[/latex] (Identity)
  • [latex]\mathbf{u} + (-\mathbf{u}) = \mathbf{0}[/latex] (Inverse)

Scalar Multiplication Properties:

  • [latex]r(s\mathbf{u}) = (rs)\mathbf{u}[/latex] (Associative)
  • [latex](r+s)\mathbf{u} = r\mathbf{u} + s\mathbf{u}[/latex] (Distributive)
  • [latex]r(\mathbf{u} + \mathbf{v}) = r\mathbf{u} + r\mathbf{v}[/latex] (Distributive)
  • [latex]1\mathbf{u} = \mathbf{u}[/latex] and [latex]0\mathbf{u} = \mathbf{0}[/latex] (Identity and Zero)

Proof

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]

Prove the additive inverse property.

Finding Components from Magnitude and Direction

Sometimes you know a vector's magnitude and direction angle, but not specific coordinate points. You can use trigonometry to find the components.

Consider a vector [latex]\mathbf{v}[/latex] that makes angle [latex]\theta[/latex] with the positive [latex]x[/latex]-axis. The vector's components form the legs of a right triangle, with the vector itself as the hypotenuse.

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.

Using basic trigonometry:

  • [latex]x[/latex]-component = [latex]||\mathbf{v}|| \cos \theta[/latex]
  • [latex]y[/latex]-component = [latex]||\mathbf{v}|| \sin \theta[/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.

[latex]\mathbf{v} = \langle ||\mathbf{v}|| \cos \theta, ||\mathbf{v}|| \sin \theta \rangle[/latex]

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.