From Components to a Spatial Sketch
A vector field assigns a vector to each point in a region of space. A sketch represents this assignment by drawing selected arrows: each arrow is attached to the point where the field is evaluated, and its direction and length represent the assigned vector.
In three dimensions, the main challenge is to keep two pieces of information distinct: where an arrow is based and which way the arrow points. The point is described by three coordinates. The vector is described by three components. Those coordinates and components play different roles in the sketch.
A three-dimensional drawing is usually shown on a flat page, so it is a projection of space rather than a literal view from every direction. Even so, a consistent set of coordinate axes and careful use of signs make the arrows readable. This tutorial focuses on sketching at selected points and describing what the arrows show locally. An arrow diagram alone does not specify the paths of moving particles.
Points, Vectors, and Arrow Endpoints
The components \(u\), \(v\), and \(w\) specify signed displacements in the \(x\)-, \(y\)-, and \(z\)-directions. Thus, if the arrow starts at \(P=(a,b,c)\), its endpoint is \(Q=(a+u,b+v,c+w)\). A positive component moves in the positive direction of its axis; a negative component moves in the negative direction; a zero component contributes no displacement along that axis.
The endpoint is a useful construction aid. It does not mean that the vector is located at \(Q\): the arrow still begins at \(P\), and \(Q\) is only its head. In a sketch, mark or mentally identify the base point first, then use the components to place the arrowhead relative to that base.
A Reliable Sketching Procedure
Use three mutually perpendicular coordinate axes and mark the selected points in space. On a flat page, a common convention is to draw the \(x\)- and \(y\)-axes at an angle and the \(z\)-axis vertically. The precise angles on the page are a drawing choice; the positive and negative directions of each labeled axis are what matter.
Plot the coordinates \((a,b,c)\) on the three-dimensional axes. Keep track of which coordinates are positive, negative, or zero.
Substitute \(a\), \(b\), and \(c\) into all three component functions to obtain \(\langle u,v,w\rangle\).
Determine the arrow’s displacement along each axis. A zero component means there is no displacement along that axis.
Use the three displacements to locate the arrowhead, then connect it to the base point. Repeat consistently for each selected point.
When an exact scale would make a sketch crowded, arrows can be shortened proportionally for legibility. If you do this, preserve the direction and communicate relative lengths as consistently as possible. The drawing is meant to show the assigned arrows, not to suggest extra behavior between the sampled points.
Worked Example: Evaluating and Plotting Two Arrows
Worked Example: Evaluating and Plotting Two Arrows
Let \(\mathbf{F}(x,y,z)=\langle 2x-y,\ x+z,\ 1-z\rangle\). Find the vectors at \(P=(1,2,0)\) and \(R=(-1,0,2)\), and determine their arrow endpoints.
At \(P=(1,2,0)\), substitute \(x=1\), \(y=2\), and \(z=0\) into each component:
The base is \((1,2,0)\), so the endpoint is found by adding the components coordinate by coordinate:
Thus the arrow starts at \((1,2,0)\), has no \(x\)-displacement, and moves in the positive \(y\)- and \(z\)-directions to \((1,3,1)\).
At \(R=(-1,0,2)\), evaluation gives:
Its endpoint is:
The second arrow points in the negative \(x\)-direction, positive \(y\)-direction, and negative \(z\)-direction. Checking the endpoint differences confirms the plotted components: \((-3)-(-1)=-2\), \(1-0=1\), and \(1-2=-1\).
Why Adding Components Gives the Endpoint
The endpoint rule follows directly from the coordinate definition of a displacement vector. If an arrow goes from \(P=(a,b,c)\) to \(Q=(q_1,q_2,q_3)\), its displacement is the endpoint minus the starting point, coordinate by coordinate. Consequently, if that displacement is \(\langle u,v,w\rangle\), then \(q_1-a=u\), \(q_2-b=v\), and \(q_3-c=w\). Solving these three equations gives the endpoint rule.
This also explains why the vector components are not generally the coordinates of the arrowhead. They are the differences between the endpoint and the base point. They match the endpoint coordinates only when the base happens to be the origin.
Worked Example: Sampling a Field at Several Points
Worked Example: Sampling a Field at Several Points
Consider \(\mathbf{G}(x,y,z)=\langle x+z,\ y-z,\ x-y\rangle\). Evaluate the field at \(A=(1,0,1)\), \(B=(0,1,0)\), and \(C=(0,0,0)\). For each, state the endpoint needed to draw the arrow.
At \(A=(1,0,1)\):
The arrow endpoint is \(A+\mathbf{G}(A)=(1+2,0-1,1+1)=(3,-1,2)\). It has positive \(x\)- and \(z\)-components and a negative \(y\)-component.
At \(B=(0,1,0)\):
The endpoint is \(B+\mathbf{G}(B)=(0+0,1+1,0-1)=(0,2,-1)\). This arrow has no \(x\)-displacement, moves positively along \(y\), and moves negatively along \(z\).
At \(C=(0,0,0)\):
The endpoint is \(C+\mathbf{G}(C)=(0+0,0+0,0+0)=(0,0,0)\). The assigned vector is zero, so the sketch at \(C\) has no arrow with a nonzero length.
A compact record before sketching is:
| Base point | Assigned vector | Endpoint |
|---|---|---|
| \(A=(1,0,1)\) | \(\langle 2,-1,1\rangle\) | \((3,-1,2)\) |
| \(B=(0,1,0)\) | \(\langle 0,1,-1\rangle\) | \((0,2,-1)\) |
| \(C=(0,0,0)\) | \(\langle 0,0,0\rangle\) | \((0,0,0)\) |
The table provides a check on the drawing: every arrow must start at the listed base point, and the coordinate difference from its base to its endpoint must equal the listed vector.
Reading a Simple Three-Dimensional Sketch
To interpret a sketch, read one arrow at a time. First identify its base point; then compare its direction with the labeled axes. An arrow with positive \(z\)-component has an upward displacement in the positive \(z\)-direction, while one with negative \(z\)-component has a downward displacement. The same sign rule applies to the \(x\)- and \(y\)-axes, even when those axes are slanted on the page.
Next compare arrows attached to different points. You can state, for example, that one sampled vector has a positive \(x\)-component while another has a negative \(x\)-component, or that a particular arrow lies in a coordinate-parallel plane because one component is zero. These are local descriptions of the displayed vectors. Do not connect arrowheads or interpret the set of arrows as a path of motion: paths require additional information beyond a static field sketch.
Worked Example: Recognizing Component Patterns
Worked Example: Recognizing Component Patterns
For \(\mathbf{H}(x,y,z)=\langle z,x,y\rangle\), sample the points \(P=(1,0,0)\), \(Q=(0,1,0)\), and \(R=(0,0,1)\). Describe the component directions and endpoints, then state what can be concluded from this small sketch.
At \(P=(1,0,0)\), the components are \(\langle z,x,y\rangle=\langle 0,1,0\rangle\), so the endpoint is \((1,0,0)+\langle 0,1,0\rangle=(1,1,0)\). The arrow points in the positive \(y\)-direction.
At \(Q=(0,1,0)\), the components are \(\langle 0,0,1\rangle\), so the endpoint is \((0,1,0)+(0,0,1)=(0,1,1)\). The arrow points in the positive \(z\)-direction.
At \(R=(0,0,1)\), the components are \(\langle 1,0,0\rangle\), so the endpoint is \((0,0,1)+(1,0,0)=(1,0,1)\). The arrow points in the positive \(x\)-direction.
Each endpoint check recovers the assigned vector: \((1,1,0)-(1,0,0)=\langle 0,1,0\rangle\), \((0,1,1)-(0,1,0)=\langle 0,0,1\rangle\), and \((1,0,1)-(0,0,1)=\langle 1,0,0\rangle\). The sketch shows these three sampled directions. It does not establish what the arrows do at every other point, nor does it determine any trajectory.
A Common Mistake: Treating Components as Endpoint Coordinates
Suppose an arrow based at \(P=(2,-1,3)\) is assigned the vector \(\langle 1,2,-1\rangle\). A common error is to draw it from \(P\) to \((1,2,-1)\), treating the three components as the endpoint coordinates. That would produce the displacement \((1-2,2-(-1),-1-3)=\langle -1,3,-4\rangle\), which is not the assigned vector.
The correction is to add the components to the base coordinates:
Now the displacement check gives \((3-2,1-(-1),2-3)=\langle 1,2,-1\rangle\), exactly as required. Another visual error is to ignore an axis because it is foreshortened in the page projection. A component can be nonzero even when its direction is difficult to see in a particular drawing; use the axis labels and signs rather than relying only on the apparent slope of the arrow.
Why Three-Dimensional Field Sketches Matter
A sketch turns component formulas into a spatial picture that can be inspected at selected locations. It helps catch sign errors, makes the base point and assigned vector visually distinct, and provides a qualitative view of how the assignment varies across the sample points. This is useful before attempting more involved calculations involving vector fields.
However, a sketch is only as informative as its sampling and scale. A few arrows show the field values at those chosen points; they do not by themselves prove a rule for all points. Keep the interpretation tied to what is actually shown: base locations, component directions, and any comparisons justified by the displayed arrows.
Check Your Understanding
For each question, distinguish the coordinates of a base point from the components of the vector based there.
- If an arrow starts at \((1,-2,0)\) and represents \(\langle 2,1,-3\rangle\), what is its endpoint?
- What does a zero second component tell you about an arrow’s displacement?
- For \(\mathbf{F}(x,y,z)=\langle y,z,x\rangle\), evaluate the vector at \((2,0,1)\) and find its endpoint.
- An arrow starts at \((0,1,2)\) and ends at \((-1,3,2)\). What vector does it represent?
- Why does a sketch of arrows at selected points not, by itself, determine trajectories?