Notes on Position Vectors, Right-Hand 3D Coordinate System, and Force Projections

Position Vectors in 3D and the Right-Hand Coordinate System

In three-dimensional space, we utilize a right-hand coordinate system defined by three mutually perpendicular axes: x, y, and z. This system is called "right-hand" because if you curl the fingers of your right hand from the positive x-axis towards the positive y-axis, your thumb will point in the direction of the positive z-axis. A position vector $\mathbf{r}$ describes the unique location of a point with respect to the origin (the intersection of all three axes) and is expressed in components along the Cartesian basis vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ as:

$\mathbf{r} = x\,\hat{i} + y\,\hat{j} + z\,\hat{k}.$

Here, $x$, $y$, and $z$ represent the scalar components of the vector along the respective x, y, and z axes. These components are the coordinates of the point in space. The basis vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ are unit vectors (magnitude of 1) pointing in the positive x, y, and z directions, respectively, and are orthogonal to each other.

The vector from the origin to a specific point A with coordinates $(x<em>A, y</em>A, z<em>A)$ is denoted as $\mathbf{r}</em>A = x<em>A\hat{i} + y</em>A\hat{j} + z<em>A\hat{k}$. If we consider two distinct points A and B in space, with their corresponding position vectors $\mathbf{r}</em>A$ and $\mathbf{r}*B$ from the origin, the vector from point A to point B (representing the displacement from A to B) is determined by subtracting the initial position vector from the final position vector:

$\mathbf{r}<em>{AB} = \mathbf{r}</em>B - \mathbf{r}<em>A = (x</em>B - x<em>A)\hat{i} + (y</em>B - y<em>A)\hat{j} + (z</em>B - z*A)\hat{k}.$

The resulting components $(x<em>B - x</em>A)$, $(y<em>B - y</em>A)$, and $(z<em>B - z</em>A)$ are the scalar components of the vector $\mathbf{r}<em>{AB}$ along the x, y, and z axes, respectively. This means the vector $\mathbf{r}</em>{AB}$ starts at A and ends at B.

AB as a numerical example

To illustrate this, let's take point A with coordinates $(1,-2,-4)$ and point B with coordinates $(-1,4,5)$. The vector from A to B, or $\mathbf{r}*{AB}$, is calculated by subtracting the coordinates of A from those of B:

$\mathbf{r}*{AB} = (-1-1)\hat{i} + (4-(-2))\hat{j} + (5-(-4))\hat{k} = (-2,\;6,\;9).$

This vector indicates a displacement of -2 units along the x-axis, 6 units along the y-axis, and 9 units along the z-axis when moving from A to B. The length, or magnitude, of this displacement vector is found using the Pythagorean theorem in 3D:

$|\mathbf{r}*{AB}| = \sqrt{(-2)^2 + 6^2 + 9^2} = \sqrt{4+36+81} = \sqrt{121} = 11.$

The unit vector along the direction of AB, denoted as $\hat{\mathbf{u}}<em>{AB}$, represents the direction of the vector $\mathbf{r}</em>{AB}$ without its magnitude. It is calculated by dividing the vector by its magnitude:

$\hat{\mathbf{u}}<em>{AB} = \frac{\mathbf{r}</em>{AB}}{|\mathbf{r}*{AB}|} = \left(-\frac{2}{11},\; \frac{6}{11},\; \frac{9}{11}\right).$

This unit vector is crucial for defining the direction of forces. For instance, if a force of magnitude 200 acts along the line AB, its vector form, $\mathbf{F}*{AB}$, is simply the magnitude multiplied by the unit vector:

$\mathbf{F}*{AB} = 200\,\hat{\mathbf{u}}_{AB} = \left(-\frac{400}{11},\;\frac{1200}{11},\;\frac{1800}{11}\right)\ \text{(units of force)} \approx (-36.36,\;109.09,\;163.64).$

This vector form of the force explicitly gives its components in each Cartesian direction.

Two random points and the AB/AC setup

The principle of defining a vector between two points extends independently of the origin. If you have two points, $A=(xA,yA,zA)$ and $B=(xB,yB,zB)$, the position vector of B relative to A, or $\mathbf{r}*{AB}$, is directly obtained by component-wise subtraction:

$\mathbf{r}<em>{AB} = (x</em>B - x<em>A)\hat{i} + (y</em>B - y<em>A)\hat{j} + (z</em>B - z_A)\hat{k}.$

This intuitive method, often referred to as "far point minus near point" (or "head minus tail"), consistently yields the vector pointing from the first specified point to the second across each axis. It's a fundamental operation in vector mechanics and geometry.

Forces in space and the line of action

A force in three-dimensional space is a vector quantity, meaning it has both magnitude and direction. This force acts along a specific path in space called its line of action. To uniquely define this line and thus the direction of the force, two points through which the force acts are required. If a force, with a scalar magnitude $F$, acts along the line extending from point A to point B, the force vector $\mathbf{F}<em>{AB}$ is precisely defined by this magnitude and the direction specified by the unit vector $\hat{\mathbf{u}}</em>{AB}$, which points from A to B:

$\mathbf{F}<em>{AB} = F \, \hat{\mathbf{u}}</em>{AB}.$

The unit vector $\hat{\mathbf{u}}*{AB}$ effectively encapsulates the orientation of the line of action in space. The magnitude $F$ is always a positive scalar value, representing the strength of the force.

How to construct a force in Cartesian form

Converting a conceptual force (magnitude and line of action) into its Cartesian vector components is a systematic process:

1. Compute the position vector from A to B: First, determine the displacement vector $\mathbf{r}<em>{AB} = \mathbf{r}</em>B - \mathbf{r}*A$. This vector establishes the geometric path from A to B.

2. Compute its magnitude: Calculate the length of the position vector, \u_003cbr/ >|\mathbf{r}{AB}| = \sqrt{(xB - xA)^2 + (yB - yA)^2 + (zB - z*A)^2}. This scalar value represents the distance between points A and B.

3. Form the unit vector: Normalize the position vector by dividing it by its magnitude to obtain the unit vector $\hat{\mathbf{u}}<em>{AB} = \mathbf{r}</em>{AB} / |\mathbf{r}*{AB}|$\u_003cbr/ >. This unit vector has a magnitude of one and points purely in the direction from A to B.

4. Multiply by the given magnitude: Finally, multiply the scalar magnitude of the force ($F$) by the unit vector $\hat{\mathbf{u}}<em>{AB}$ to get the force vector in Cartesian form: $\mathbf{F}</em>{AB} = F \hat{\mathbf{u}}_ {AB}$. This step scales the directional information by the actual strength of the force.

Worked example: AB and AC with forces along lines AB and AC

Let's reuse the coordinates for A and B from our previous example: $A=(1,-2,-4)$ and $B=(-1,4,5)$.

• From our prior calculations: $\mathbf{r}<em>{AB} = (-2, \, 6, \, 9)$ and $|\mathbf{r}</em>{AB}| = 11$.

• The unit vector for the line AB is: $\hat{\mathbf{u}}_ {AB} = (-2/11, \, 6/11, \, 9/11)$.

• If the force $F*{AB}$ has a magnitude of 200 units, then its vector form is:

  $\mathbf{F}*{AB} = 200 \hat{\mathbf{u}}_ {AB} = \left(-\frac{400}{11}, \, \frac{1200}{11}, \, \frac{1800}{11}\right) \approx (-36.36, \, 109.09, \, 163.64).$

Now, let's consider a second line of action, AC, which depends on the coordinates of point C. The provided transcript contains two different data sets for point C, leading to different numerical outcomes.

Case A: Point C as absolute coordinates

Let's assume $C = (-3,-5,-10)$ (as used in one part of the discussion, implying C is an absolute point). Then:

• The position vector from A to C is $\mathbf{r}<em>{AC} = \mathbf{r}</em>C - \mathbf{r}<em>A = (-3-1, -5-(-2), -10-(-4)) = (-4, -3, -6)$. Its magnitude is $|\mathbf{r}</em>{AC}| = \sqrt{(-4)^2 + (-3)^2 + (-6)^2} = \sqrt{16+9+36} = \sqrt{61} \approx 7.810$.

• The unit vector along AC is $\hat{\mathbf{u}}_ {AC} = (-4/\sqrt{61},\; -3/\sqrt{61},\; -6/\sqrt{61}) \approx (-0.512,\; -0.384,\; -0.769).$

• If a force $F*{AC}$ has a magnitude of 100 units, then its vector form is:

  $\mathbf{F}*{AC} = 100 \hat{\mathbf{u}}_ {AC} \approx (-51.2,\; -38.4,\; -76.9).$

• The component of force $\mathbf{F}<em>{AB}$ along the line of action of $\mathbf{F}</em>{AC}$ (i.e., the scalar projection of $\mathbf{F}<em>{AB}$ onto $\hat{\mathbf{u}}</em>{AC}$) tells us how much of $\mathbf{F}<em>{AB}$ acts in the direction of $\mathbf{F}</em>{AC}$. This is calculated using the dot product:

  $\text{Comp}<em>{AB\parallel AC} = \mathbf{F}</em>{AB} \cdot \hat{\mathbf{u}}<em>{AC} = (200 \hat{\mathbf{u}}</em>{AB}) \cdot \hat{\mathbf{u}}<em>{AC} = 200\, (\hat{\mathbf{u}}</em>{AB} \cdot \hat{\mathbf{u}}*{AC}).$

  First, calculate the dot product of the unit vectors:

  $\hat{\mathbf{u}}<em>{AB} \cdot \hat{\mathbf{u}}</em>{AC} = \frac{(-2)(-4) + 6(-3) + 9(-6)}{11\sqrt{61}} = \frac{8 - 18 - 54}{11\sqrt{61}} = \frac{-64}{11\sqrt{61}} \approx -0.745,$

  Then, compute the scalar projection:

  $\text{Comp}*{AB\parallel AC} \approx 200 \times (-0.745) \approx -149.0 \text{ pounds}.$

  The negative sign indicates that the component of $\mathbf{F}*{AB}$ along the line AC acts in the direction *opposite* to $\hat{\mathbf{u}}*{AC}$.

• The angle $\theta$ between the two force vectors, $\mathbf{F}<em>{AB}$ and $\mathbf{F}</em>{AC}$, is found using the dot product formula, which simplifies when using unit vectors:

  $\cos\theta = \frac{\mathbf{F}<em>{AB} \cdot \mathbf{F}</em>{AC}}{|\mathbf{F}<em>{AB}|\,|\mathbf{F}</em>{AC}|} = \hat{\mathbf{u}}<em>{AB} \cdot \hat{\mathbf{u}}</em>{AC} = \frac{-64}{11\sqrt{61}} \approx -0.745,$

  Therefore, the angle is:

  $\theta \approx \arccos(-0.745) \approx 138.2^{\circ}.$

Case B: Vector AC provided directly

Alternatively, consider the scenario where the vector from A to C is given directly as $\mathbf{r}_ {AC} = (-3, -5, -10)$ (meaning this vector is already defined, and its magnitude is \sqrt{(-3)^2 + (-5)^2 + (-10)^2} = \sqrt{9+25+100} = \sqrt{134} \approx 11.58$). Then:

• The unit vector \hat{\mathbf{u}}*{AC} = (-3/\sqrt{134},\; -5/\sqrt{134},\; -10/\sqrt{134}\) \approx (-0.259, -0.431, -0.862).

• If $F{AC}=100$, then \mathbf{F}{AC} = 100 \hat{\mathbf{u}}_ {AC} \approx (-25.9, -43.1, -86.2)$.</p></li><li><p>The component of$\mathbf{F}*{AB} along $AC$ (scalar projection) becomes:

  \text{Comp}{AB\parallel AC} = \mathbf{F}{AB} \cdot \hat{\mathbf{u}}*{AC} \approx -178.6\ \text{pounds},

  (calculated as $200 \times ((-2)(-0.259) + (6)(-0.431) + (9)(-0.862)) = 200 \times (-0.893)$)

• And the angle between the forces satisfies:

  \cos\theta = \hat{\mathbf{u}}{AB} \cdot \hat{\mathbf{u}}{AC} \approx -0.894,$</p><p>So, the angle is:</p><p>$\theta \approx 153.2^{\circ}.$</p></li></ul><p>Notes on these results:</p><ul><li><p>The numerical outcomes for projections and angles are sensitive to the initial coordinate data for point C. It highlights the critical importance of using consistent and correctly defined point coordinates throughout a problem. Discrepancies often arise from misinterpreting a given vector as coordinate points or vice-versa.</p></li><li><p>Regardless of the specific data, the underlying mathematical principles (dot product, unit vector definitions) remain consistent. These versatile tools can be applied to compute projections and angles using either the full vector forms or their corresponding unit vectors.</p></li></ul><h5 id="e5df6087-deb1-4fff-9c8b-fd77c16bb9bb" data-toc-id="e5df6087-deb1-4fff-9c8b-fd77c16bb9bb" collapsed="false" seolevelmigrated="true">Key definitions and formulas (summary)</h5><ul><li><p><strong>Position vector of a point</strong>: Defines a point's location relative to the origin.</p><p>$\mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k}.$</p></li><li><p><strong>Vector between two points A and B</strong>: Represents the displacement from A to B.</p><p>$\mathbf{r}{AB} = \mathbf{r}B - \mathbf{r}A = (xB - xA)\hat{i} + (yB - yA)\hat{j} + (zB - z_A)\hat{k}.$</p></li><li><p><strong>Magnitude of a vector</strong>: The scalar length of a vector.</p><p>$|\mathbf{v}| = \sqrt{vx^2 + vy^2 + v_z^2}.$</p></li><li><p><strong>Unit vector</strong>: A vector with a magnitude of one, indicating only direction.</p><p>$\hat{\mathbf{u}} = \frac{\mathbf{v}}{|\mathbf{v}|}.$</p></li><li><p><strong>Direction cosines (angles with axes)</strong>: The cosines of the angles a vector makes with the positive x, y, and z axes. If$\hat{\mathbf{u}} = l\hat{i} + m\hat{j} + n\hat{k}$, then$l = \cos \alpha$(angle with x-axis),$m = \cos \beta$(angle with y-axis),$n = \cos \gamma$(angle with z-axis). A key property is$l^2 + m^2 + n^2 = 1$.</p></li><li><p><strong>Direction cosine form of AB</strong>: Directly from the unit vector components:</p><p>$l = (xB - xA)/|\mathbf{r}{AB}|$,$m = (yB - yA)/|\mathbf{r}{AB}|\u_003cbr/ >, n = (zB - zA)/|\mathbf{r}_ {AB}|$</p></li><li><p><strong>Force along a line of action</strong>: A vector representing a force with magnitude F acting along a specific direction.</p><p>$\mathbf{F}{AB} = F \hat{\mathbf{u}}{AB},$</p><p>where$\hat{\mathbf{u}}{AB} = \frac{\mathbf{r}{AB}}{|\mathbf{r}_ {AB}|}.$</p></li><li><p><strong>Component of one vector along the line of action of another (scalar projection)</strong>: The scalar value indicating how much of one vector's effect acts in the direction of another. This is found by the dot product of the first vector with the unit vector of the second line.</p><p>$\text{Comp}{\mathbf{F}1 \parallel \mathbf{F}2} = \mathbf{F}1 \cdot \hat{\mathbf{u}}_2.$</p></li><li><p><strong>Angle between two vectors</strong>: The spatial angle separating the directions of two vectors. It can be found using the dot product formula, which provides the cosine of the angle.</p><p>$\cos\theta = \frac{\mathbf{F}1 \cdot \mathbf{F}2}{|\mathbf{F}1|\,|\mathbf{F}2|}.$</p></li><li><p><strong>Special case for unit vectors</strong>: When working with unit vectors, the formula simplifies significantly because their magnitudes are 1:</p><p>$\cos\theta = \hat{\mathbf{u}}1 \cdot \hat{\mathbf{u}}2.$</p></li></ul><h5 id="23325efe-8c06-48cf-be6a-ddfe67f09466" data-toc-id="23325efe-8c06-48cf-be6a-ddfe67f09466" collapsed="false" seolevelmigrated="true">Practical and exam-oriented notes</h5><ul><li><p><strong>Accuracy with Cartesian Form</strong>: Always write vectors in Cartesian form ($x\hat{i} + y\hat{j} + z\hat{k}) and diligently keep track of signs. The signs of the components are critical as they immediately convey both the magnitude contribution and the precise direction along each axis.

• Effective Projection: When determining the component of a force along another line, it is most efficient and geometrically sound to use the dot product of the force vector with the unit vector along the target line. This ensures you isolate only the directional component without altering the magnitude by an extraneous factor.

• Angle Interpretation: The angle between two forces is derived from their dot product divided by the product of their magnitudes. The sign of the dot product offers quick insight: a positive dot product means an acute angle ($0^{\circ} \le \theta < 90^{\circ}$), while a negative dot product indicates an obtuse angle ($90^{\circ} < \theta \le 180^{\circ}$). If the dot product is zero, the vectors are perpendicular.

• Dimensional Reduction: If a problem's geometry naturally confines forces or objects to a two-dimensional plane (e.g., only x and y components are non-zero), you can simplify the analysis by dropping the z-component. All the same vector projection and angle calculation principles apply, just within the 2D plane, making calculations simpler.

• Real-World Relevance: These foundational vector concepts are indispensable in a wide array of engineering and physics applications. They form the basis for statics (analyzing forces in equilibrium), dynamics (analyzing forces causing motion), finite element analysis, cable tension calculations in structures, predicting forces at joints of linkages, and understanding fluid flow, where precise direction and magnitude of forces are paramount for design and safety.

Quick recap for studying

• Vector Formation: Be proficient in computing a vector \mathbf{r}*{AB}$$ from the coordinates of points A and B, and subsequently determining its magnitude and unit vector.

• Force Vector Construction: Be able to accurately form a force vector in Cartesian notation when given its total magnitude and the line of action (defined by two points).

• Force Projection Skill: Practice calculating the scalar projection of one force onto the line of action of another using the dot product with the relevant unit vector.

• Angle Calculation: Ensure you can compute the angle between any two force vectors using their dot product.

• Data Consistency: Understand and actively distinguish between different data sets for points (like the two cases for point C discussed in the example) and recognize how different coordinate choices directly lead to different numerical results in vector calculations. This highlights the importance of precise problem interpretation and data management.