Calculating Projections (university physics only)
The dot product of vectors `A and `B is `A dot `B = || `A || * || `B || * cos(theta), where theta is the angle between the vectors.
The magnitude of the projection of `A on `B is easily seen to be equal to || `A || * | cos(theta) |, which by the above is equal to | `A dot `B | / || `B ||.
The expression `A dot `B / || `B || is equal either to the magnitude of the projection, or the negative of the magnitude of the projection, depending on whether the projection is in the direction of `B or opposite the direction of `B. It this quantity is multiplied by a unit vector in the direction of `B, the result will be the projection vector. We might refer to this quantity as the 'signed magnitude'.
A unit vector in the direction of `B is the vector `B / || `B ||.
It follows that the projection of `A on `B is
This form shows how the projection is constructed and a 'signed magnitude' multiplied by a unit vector. This results in the formula
It is recommended that you use the previous expression ( `A dot `B / || `B || ) * `B / || `B || for now, in order to remind yourself of the construction. You should also sketch the corresponding diagram, for the same reason.
Calculating cross product
Use the determinant form to represent the cross product.
Alternatively, note that `i X `j = `k, `i X `k = - `j, and `j X `k = `i. The distributive law works for cross products. Thus
(a1 `i + a2 `j + a3 `k) X (b1 `i + b2 `j + b3 `k)
= a1 `i X (b1 `i + b2 `j + b3 `k) + a2 `i X (b1 `i + b2 `j + b3 `k) + a3 `i X (b1 `i + b2 `j + b3 `k)
= a1 b1 `i X `i + a1 b2 `i X `j + a1 b3 `i X `k +
a2 b1 `j X `i + a2 b2 `j X `j + a2 b3 `j X `k +
a3 b1 `k X `i + a3 b2 `k X `j + a3 b3 `k X `k
= (a1 b2 - a2 b1) `i X `j + (a1 b3 - a3 b1) `i X `k + (a2 b3 - a3 b2) `j X `k
= (a1 b2 - a2 b1) `k + (-a1 b3 + a3 b1) `j + (a2 b3 - a3 b2) `i
In the above we have made use of the fact that for any two vectors `A and `B we have `A X `B = - `B X `A (the reason should be obvious by the right-hand rule); so for example since `i X `k = -`j, `k X `i = `j.
The last form is identical to the form you get by evaluating the determinant form.
Applications of dot and cross product
This is one answer to a question about how you integrate in 3-dimensional space, the answer most immediately relevant to the dynamics class being taken by several students:
If `F and ``dx represent a force vector and a displacement through which the force vector acts, then the expression
`F dot ``dx
represents the product of || `F || cos(theta) with ``dx, where theta is the angle between ``dx and `F.
|| `F || cos(theta) is the component of `F in the direction of ``dx, i.e., the component of the force vector in the direction of the displacement.
It is this component that determines the work done by the force. The more of the force component in the direction of motion, the more work the force does, and the greater the displacement ``dx the greater the work.
The expression `F dot ``dx is the work done by the force `F if exerted through displacement ``dx.
Now if you have a force field `F and a curved path through that force field, you can divide that path up into a large number of nearly-straight segments of form ``dx. At the location of a typical segment you have a force `F, which does work `dW = `F ``dx over that part of the path. If we add all these `dW contributions up, we get a good approximation to the total work done over the entire path. The smaller the ``dx segments, i.e., the more finely we partition the curved path, the more accurate our approximation. In the limit as the lengths of the segments approach zero, we get the exact work done. This is clearly an integration process. The details depend on the force function `F, which is usually a function of position, e.g., `F(x, y, z), and the parameterization of the curve (for example a set of three continuous functions x(t), y(t), z(t), over some t interval, are said to parameterize the path in terms of the parameter t; that is, if you evaluate the three functions at some value of t, you get the corresponding position on the curve). So starting with force as a function of position, and a parameterization that gives us position (x, y, z) as a function of some parameter like t, we can integrate `F over the corresponding path. This requires more details and some practice, but this is an outline of how we might integrate over a curve in 3-dimensional space. Glad you asked? Maybe soon...
An example of cross product:
The torque of a force exerted perpendicular to a rod constrained to rotate about some axis (think of a wrench turning a bolt, with the axis along the bolt) is equal to the distance from the axis at which the force is exerted, and the force. There is a big advantage to regarding the torque as a cross product of the `r vector (from the axis to the point of applications) and the force vector. The direction of the torque is therefore perpendicular to the `r vector and the force vector, by the right-hand rule. The direction of the torque is therefore the direction in which it would advance a right-hand screw.
If the force vector is parallel to the `r vector no torque is exerted (think of trying to turn a nut by pushing on the end of the wrench, pushing it toward the nut; you will have no turning effect. Your force will be in the direction of the `r vector, the angle between the two being 180 degrees, and the cross product will be zero since the sine of 180 degrees is zero).
If the force is neither perpendicular nor parallel to the `r vector, then the force vector will have a component parallel to the `r vector, and a component perpendicular to the `r vector. It is the perpendicular component that turns the nut or bolt. The perpendicular component is the force is || `F || sin(theta), where theta is the angle between the `r vector and the force vector.