# Vector Concepts for Fluid Mechanics

Steven A. JonesBIEN 501

March 8, 2006

**Geometric Interpretations of Vector Relationships**

**Projection into a Plane**

The equation for a plane is

,

where , , and are constants. Such a plane is shown in Figure 1. The plane intersects the ¸, and axes at specific points (marked with ) which can be found by setting two coordinates to zero. For example, the intersection with the x axis is obtained by setting y and z to zero. Then from Equation 1,

.

If a normal vector to a plane is given, then the equation for the plane can be readily obtained. Let be any vector in the plane, then for that vector to be in the plane, its dot product with the normal to the plane must be zero (because (1) the dot product of two vectors that are normal to one another is zero and (2) any vector in the plane must be normal to the plane’s normal vector). We can therefore write the equation for the plane in vector form as:

Specifically, if then or , as in Eq. 1.

Example: Find the equation for the x-y plane.

Solution: The normal vector to the x-y plane is . Therefore, the equation for the plane is . This result is intuitive since it simply says that the plane in question is the plane.

Example: You are standing up facing north. Your upper arm is extended such that it points northeast and downward by 10 degrees. Set up a coordinate system where your elbow is the origin, the x-axis points to the North, the y-axis points to the East, and the z-axis points upward. Find a direction vector that describes your upper arm.

Solution: Let represent the correct vector, where is a unit vector along the correct direction.

We can find the equation for a plane that passes through two vectors that intersect at the origin by first finding a normal vector to that plane. Since the cross product of two vectors is a vector normal to both vectors, then if and are known vectors in the plane, is normal to both and . Therefore, an equation for the plane is:

Example: Say you are standing up facing north. Your upper arm is extended such that it points northeast and downward by 10 degrees, and your forearm is pointed north northwest and upward by 20 degrees. Find an equation for the plane that includes your upper arm and forearm.

Solution: Your upper arm and forearm are attached at the elbow. Let the elbow be the origin of a Cartesian coordinate system. The vector describing your upper arm then goes upward by 10 degrees and southeast. It can be described as

If the plane passes through the origin, then. A vector within that plane, say must be such that its dot product with the plane’s normal vector is 0. Thus, if we know the normal vector, we can write , which means that .

- What is the magnitude of the vector ? Answer:
- If ,, and , what is ? Answer:
- If , , and are vectors (you may express answers in dot and cross products):
- What is the projection of on ?Answer: or , where is the unit vector in the direction of .
- What is the equation for the plane passing through the origin that is perpendicular to ? Answer:, where . This answer arises because any vector in the plane must be perpendicular to , so its dot product with must be zero.
- What is the equation for the plane passing through that is perpendicular to ? Answer:, where .
- What vector is normal to the plane that passes through and ?Answer: A vector that is normal to the plane is normal to both and , and thus can be represented by .
- What is the equation for the plane that passes through the origin and is parallel to both and ? Answer: The solution comes from a combination of b and d. I.e., if we know a vector that is normal to the plane (question b), then we can find the equation for the plane from . But the normal to the plane is , from d. Therefore the equation is .
- What is the equation for the line passing through the origin and directed along the vector ? Answer: A line in space is represented by a pair of simultaneous equations. This representation may not be obvious when one examines the well known line . However, even in that case one should note that two equations are required, the relationship between and , and the statement that the line is in the plane, i.e. . The line along must be parallel to. Thus, if is a vector along that line, its cross product with must be zero. Hence the equation for the line is . If this is written out in detail, then

.

which yields three simultaneous equations:

While it may appear at first that the problem is overspecified (there are three equations and we require only two), close inspection shows that one of these equations is automatically satisfied if the other two are satisfied. In mathematical terms, the third equation is a linear combination of the first two; if the first is multiplied by , the second is multiplied by , and the two are subtracted, the terms containing cancel and the result is:

,

so the third equation is satisfied, as promised.

- What is the equation for the line defined by the intersection of (1) the plane passing through and and (2) the plane passing through and . Answer: The line must slmultaneously obey the equations for the two planes. Thus, we must have and . More succinctly, a vector in the direction of the line is perpendicular to both the normal to the first plane, and the normal to the second plane, . Once we have this vector, the result of f can be applied. Therefore, the equation for the line is: .
- What is the angle between and . Answer: If the angle is , then from the dot product, , so . In terms of the cross product, .
- What is the projection of a vector on a plane normal to (assume that all three vectors begin at the origin)? Answer:The projection vector is in the plane. It is also in a plane that passes through the vector and is normal to the plane, i.e. a plane that passes through both and . The vector therefore must be perpendicular to both and . The only vector direction perpendicular to both of these vectors is, and the corresponding unit vector is , where Therefore, the projection of along this direction is .
- What is the angle between and the projection of on the plane passing through and . Answer: is the normal to the plane. Let , which is normal to the plane containing both and . The angle between and the projection is the complement of the angle between and . Since , , or .

- Find a unit vector in the same direction as . Answer: .
- Sketch both and the unit vector in the x, y, z coordinate system.

- Let be a vector. What geometric object is described by the equation . Answer: A sphere.
- Sketch the object in 6 above.

- If , find the gradient of . Answer: .

- Find an equation that describes all of the planes that are perpendicular to the gradient of at any location in space. Answer: The gradient of evaluated at some location is . A plane perpendicular to this vector is then . Note that it is necessary to first take the gradient, then find its value at the given point, and then find the equation for the plane.