In this segment we'll do a numerical example involving rigid body displacements. Consider Frame A with mutually orthogonal uni-vectors, a1, a2 and a3 with origin P attached to a rigid body. Let Q be a point on this rigid body. As we've seen in lecture the position vector PQ can be written as a linear combination of a1, a2 and a3 using components q1, q2 and q3. PQ represents the position of the point Q relative to frame A. Suppose the body undergoes a rigid body displacement g. Let Frame B represent Frame A after the rigid body displacement with mutually orthogonal unit vectors b1, b2, and b3 and origin P prime. The point Q after the rigid body displacement is denoted Q prime. Again, as we have seen in lecture, the position vector P prime, Q prime can be written as a linear combination of b1, b2, and b3, using the same components q1, q2, and q3. Here, P prime, Q prime is the position of the displaced point Q prime relative to the displaced Frame B. Finally, consider the vector PQ prime. This vector gives us the position of the displaced point Q prime relative to frame A. To find the rotation matrix R we need to find the relationship between q1, q2, q3 and q1 prime, q2 prime and q3 prime. In lecture, we only looked at rigid body transformations that involve rotations. However, in this example, you'll notice that there is a displacement between frame a and frame b as well. As a result, q1, q2, and q3 will be related to q1 prime, q2 prime, q3 prime, through a rotation matrix. And an additional displacement component. We can consider these two components separately. First, let's focus on the rotation between these two frames. For now, suppose the origin of frame B coincides with fame A. In other words, fame B is a result of displacing frame A by only the rotation portion of the rigid body displacement Again, PQ is a position of q in frame a. Let PQ double prime be the position of q double prime. It is the position vector of the point q after the rotation in frame b. We know that the components of this vector will also be q1, q2, and q3. Finally, like Q one double prime, Q two double prime, and Q three double prime be the components of PQ double prime in frame A. If we can relate Q one double prime, Q two double prime, Q three double prime to Q one, Q two, and Q three, we have our rotation matrix R. To find this relationship, we first find the relationship between a1, a2, a3 and b1, b2, b3. We can easily see that, in this example, the vectors a1 and b1 are the same. This means that the axes b2 and b3 are in the same plane as the axes a2 and a3. Let theta be the angle of rotation between a2 and b2. We can then geometrically derive the relationship between b2, a2, and a3. The magnitude of b2 times the cosine of theta gives us the component of b2 along a2. Similarly, the magnitude of b2 times the sine of theta gives us the component of b2 along a3. As our result, we can write b2 as a linear combination of a2 and a3 using sine data and cosine data. Here we have used the fact that b2 is a unit vector. Therefore, its magnitude is 1. We can repeat this process for the unit vector b3. The angle between a3 and b3 is again theta. Notice that this is the same data as the angle between B2 and A2. We can again find the components of B3 along A2 and A3 and use this to write B3 as a linear combination of A2 and A3. Notice that here we have a negative sign in front of the A2 term because the components of B3 along A2 points in the negative direction. Recall that we can express PQ double prime in either frame A or frame B. These representations must ultimately be equal. We can substitute in the expressions we found for b1, b2, and b3 in terms of a1, a2, a3 into the first equation. We can now regroup the terms as coefficients of a1, a2 and a3. Then, we can equate the coefficients in their true representations of PQ double prime. This will give us the following three equations relating the two sets of coefficients. Putting this equation in matrix form gives us an expression for r in terms of the angle of rotation theta. We can now see, that we have derived the result we saw in lecture, for a simple rotation of theta, about the x axis. Suppose we are given that this angle is pie over 4, then we know that the rotation from frame A to frame B can be represented with the following rotation matrix. Recall that rigid body displacements can include a translation as well. The translation between frame A and frame B can be represented by a single vector, d. Which is the position vector from the origin of frame A to the origin of frame B. Expressed as a linear combination of a1, a2, a3. Suppose we are given that this vector is d = 1a1- 3a2 + 1a3. We can then characterize the rigid-body displacement G using the rotation matrix R we previously derived and this translation vector d. Now consider again the vector PQ prime. If we know the vector of PQ and the rigid body transformation defined by R and d, we can solve for PQ prime using the following equation. Suppose we know that the position factor PQ is given by components 1, 2, 1. Substituting in the rotation and displacement defined previously, we can find the components of PQ prime in frame A. These components are 2, -2.29, 3.12. Qualitatively we can see that the point Q prime relative to frame A is in the positive A1, negative A2, and positive A3 directions. If we know the components of PQ prime, we can invert the previous equation to find the components of PQ.