3 Dimensional Vectors: Problems, Matrix, Angle & Components

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What are 3 Dimensional Vectors?

3-dimensional or 3D vectors are vectors that are represented on a three-dimensional plane or space to have three coordinates such as the x, y and z.

If we imagine a 3D plane with axis i, j and k, (which represents the x, y, and z-axis respectively) we can write a 3D vector as the sum of its i, j and k component.

Imagine a vector $\vec{A}$ which travels from the origin (0,0,0) and goes to the coordinates (3,2,5). We could write that vector as

$\vec{A} = (3 i + 2 j + 5 k)$

For this vector, the i component would be 3, the j component would be 2 and the k component would be 5.

What are the coordinates of a 3D vector?

The three-dimensional vector has three coordinates which are represented in the x, y and z-axis. Recall that in a two-dimensional plane, you have coordinates only on the x and y-axis. Thus, in a 2D vector coordinates are given in the form (x, y). However, the coordinates of 3D vectors are given in the form (x, y, z)

How do you plot a 3D vector?

Begin by drawing a set of axis. Firstly, draw the vertical z-axis. Perpendicular to that, draw a y-axis. In between the z and y-axis, draw the x-axis. Note that all 3 axes are perpendicular to each other.

3-Dimensional vectors, three dimensional axis, StudySmarter

3-Dimensional axis (math.brown.edu)

After that, place a scale on each axis and mark the point where the head of the vector arrives. Then draw an arrow between the origin and the head of the vector. Finally, mark the coordinates of the head of the arrow.

3-Dimensional Vectors, 3D vector, StudySmarter 3D vector

3D Vector Matrix

Vector can also be written in matrix form. In this form, we can write the vector as three rows by one column matrix. The first row is the i component, the second row is the j component and the third row is the k component.

We do not write the x, y, and z terms in matrix form.

If we use the vector $\vec{A}$ above as an example, we get:

$\vec{A} = [\begin{matrix} 3 \\ 2 \\ 5 \end{matrix}]$

We can combine two vectors to find the dot product of these vectors.

Suppose we have vector $\vec{A} = [\begin{matrix} a \\ b \\ c \end{matrix}]$ and vector $\vec{B} = [\begin{matrix} d \\ e \\ f \end{matrix}]$ , the dot product $\vec{A} \cdot \vec{B}$ can be found by following the method below:

Step 1: Transpose vector $\vec{A}$ , that is, convert it from a 3 rows by 1 column vector to a 1 row by 3 column vector.

For vector $\vec{A} = [\begin{matrix} a \\ b \\ c \end{matrix}]$ , vector $\overset{\to t}{A} = [\begin{matrix} a & b & c \end{matrix}]$

Step 2: Write the dot product of both vectors as the multiplication of both matrices.

$\vec{A} \cdot \vec{B} = [\begin{matrix} a & b & c \end{matrix}] [\begin{matrix} d \\ e \\ f \end{matrix}]$

Step 3: Perform the matrix multiplication:

$\vec{A} \cdot \vec{B} = [\begin{matrix} a d + b e + c f \end{matrix}]$

Step 4: Simplify the matrix. You should end up with a 1-by-1 matrix.

Let vector $\vec{A} = 3 i + 2 j + 2 k$ , and vector $\vec{B} = i + 2 j + k$ . Find the dot product of vectors $\vec{A}$ and $\vec{B}$ .

Solution:

Writing both vectors in matrix form, we get:

$\vec{A} = [\begin{matrix} 3 & 2 & 2 \end{matrix}]$ and $\vec{B} = [\begin{matrix} 1 & 2 & 1 \end{matrix}]$

Step 1:

$\overset{\to t}{A} = [\begin{matrix} 3 \\ 2 \\ 2 \end{matrix}]$

Step 2:

$\vec{A} \cdot \vec{B} = [\begin{matrix} 3 \\ 2 \\ 2 \end{matrix}] [\begin{matrix} 1 & 2 & 1 \end{matrix}]$

Step 3:

$\vec{A} \cdot \vec{B} = [\begin{matrix} 3 + 4 + 2 \end{matrix}]$

Step 4:

$\vec{A} \cdot \vec{B} = 9$

What are the 3D vector equations?

Essentially, there are two main 3D equations. However, a third equation which is the angle between 3D vectors is derived from these two main equations. The two main equations are the dot product and the magnitude of a 3D vector equation.

Dot product of 3D vectors

For two certain 3D vectors A (x₁, y₁, z₁) and B (x₂, y₂, z₂) which are represented in the vector form

$x_{1} i + y_{1} j + z_{1} k$

and

$x_{2} i + y_{2} j + z_{2} k$

The dot product is

$\vec{A} \cdot \vec{B} = x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}$

Find the product of Vector G and K located (-1, 2, 3) and (0, 5, 1) of a plane.

Solution:

By applying the dot product formula

$\vec{A} \cdot \vec{B} = x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}$

Then,

$\vec{G} \cdot \vec{K} = (- 1 \times 0) + (2 \times 5) + (3 \times 1) \vec{G} \cdot \vec{K} = 0 + 10 + 3 \vec{G} \cdot \vec{K} = 13$

Magnitude of a 3D vector

The magnitude of a three-dimensional vector is derived using the extended Pythagoras theorem. Recall that the Pythagoras theorem is applied knowing the x and y-axis are perpendicular, note that the additional z-axis in 3D is perpendicular to both the x and y-axis. Hence, in order to calculate the magnitude of a certain 3D vector A (x₁, y₁, z₁) which is represented in the vector form.

$x_{1} i + y_{1} j + z_{1} k$

apply

$|A| = \sqrt{{x_{1}}^{2} + {y_{1}}^{2} + {z_{1}}^{2}}$

Find the magnitude of vector C given by $3 i - 2 j + k$

Solution:

Since the magnitude of a vector $x_{1} i + y_{1} j + z_{1} k$ is calculated as

$|A| = \sqrt{{x_{1}}^{2} + {y_{1}}^{2} + {z_{1}}^{2}}$

Then the magnitude of vector C is

$|C| = \sqrt{3^{2} + {(- 2)}^{2} + 1^{2}} |C| = \sqrt{9 + 4 + 1} |C| = \sqrt{14}$

How is the angle between 3D vectors calculated?

To find the angle between two corresponding 3D vectors, use the formula below:

$θ = \cos^{- 1} (\frac{a \cdot b}{|a| |b|})$

3-Dimensional Vectors An illustration of the angle between two vectors in 3D StudySmarter Originals An illustration of the angle between two vectors in 3D, StudySmarter Originals

Where $θ$ is the angle between vectors a and b, $a \cdot b$ is the dot product of vectors a and b, and where $|a|$ and $|b|$ are the magnitudes of vector a and vector b respectively.

Find the magnitude of the vector traveling from the origin to the coordinates (2,1,2).

Solution:

The vector can be written as

$\vec{A} = 2 i + j + 2 k$

Using the equation above:

$|a| = \sqrt{2^{2} + 1^{2} + 2^{2}}$

$|a| = \sqrt{4 + 1 + 4} = \sqrt{9}$

Therefore:

$|a| = 3$

The magnitude of the vector is 3 units.

We can now combine all that we have learned to find the angle between two vectors!

Find the angle between vectors $\vec{A} = 2 i + 3 j + k$ and vector $\vec{B} = i + 4 j + 5 k$ .

Solution:

Writing the matrix form of these vectors:

$\vec{A} = [\begin{matrix} 2 \\ 3 \\ 1 \end{matrix}]$

and

$\vec{B} = [\begin{matrix} 1 \\ 4 \\ 5 \end{matrix}]$

Writing vector $\vec{A}$ in transcript form:

$\overset{\to t}{A} = [\begin{matrix} 2 & 3 & 1 \end{matrix}]$

Therefore:

$\vec{A} \cdot \vec{B} = [\begin{matrix} 2 & 3 & 1 \end{matrix}] [\begin{matrix} 1 \\ 4 \\ 5 \end{matrix}]$

$\vec{A} \cdot \vec{B} = [\begin{matrix} 2 + 12 + 5 \end{matrix}] = 19$

The magnitude of vector $\vec{A}$ is:

$|A| = \sqrt{2^{2} + 3^{2} + 1^{2}} = \sqrt{4 + 9 + 1} = \sqrt{14}$

The magnitude of vector $\vec{B}$ is:

$|B| = \sqrt{1^{2} + 4^{2} + 5^{2}} = \sqrt{42}$

Since:

$θ = \cos^{- 1} \frac{\vec{A} \cdot \vec{B}}{|A| |B|}$

Hence:

$θ = \cos^{- 1} (\frac{19}{\sqrt{14} \sqrt{42}}) = 38.41 ° (2 d . p .)$

3-Dimensional Vectors - Key takeaways

3D vectors have values i, j, and k for their x, y, and z-axis respectively.
3D vectors can be written in matrix form.
In this form, we can find the dot product of two vectors by performing matrix multiplication.
By also finding the magnitude of those vectors through an extended version of Pythagoras' theorem, we can find the angle between those vectors.
Graphing vectors comprise of drawing the axes, the coordinates where the vector ends and begins, and sketching a line connecting both points.

Flashcards in 3 Dimensional Vectors 1

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Can vectors be written in matrix form?

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Frequently Asked Questions about 3 Dimensional Vectors

What is a 3-Dimensional vector?

It is a line segment in three-dimensional space from Point A to Point B.

What are the steps to find the dot product of vectors A and B?

1. Write vector A in transpose form.

2. Write the dot product as a multiplication of the transposed matrix and the matrix for vector B.

3. Perform the vector multiplication.

4. Simplify.

Where does the equation for the magnitude of a vector derive from?

Pythagoras' theorem.

How to graph 3 dimensional vectors?

Begin by drawing a set of axis. Firstly, draw the vertical z-axis. Perpendicular to that, draw a y-axis. In between the z and y axis, draw the x-axis. Note that all 3 axes are perpendicular to each other.

How to find slope of 3 dimensional vector?

Generally, slope is only applicable to 2D vectors and as such the slope of vectors in 3D is undefined.

How to draw a 3 dimensional vector?

What are some 3 dimensional vector problems?

3 dimensional vector problems are questions about 3D vectors which may require you to determine the magnitude of 3D vectors, angle between two 3D vectors or anything related to vectors in 3D.

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3 Dimensional Vectors

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What are 3 Dimensional Vectors?