Synopsis of the Lecture
Using the given source material as a guide, the following is a synopsis of the lecture that highlights the subjects and subtopics covered:
An Overview of 3D Geometry Topics
The equation of a line perpendicular to two given lines and the perpendicular distance of a point from a line will be covered in the video. Although these are considered minor subjects, it is crucial to understand how to approach problems of this nature. Vector knowledge is deemed extremely important for this chapter.
1. Determining a Point's Perpendicular Distance from a Line
This is the first subject covered. A given line (defined by a point A vector and a parallel vector B vector) and a point P (P vector) are involved in the problem. The aim is to find the distance between point P and the line perpendicularly. Although this problem type is now addressed in 3D, it may be remembered from 2D geometry (straight lines).
- Assuming a point C (C vector) on the line is the method. Given that C is on the line, its vector can be written as follows: A vector plus a scalar lambda (λ) multiplied by B vector (A vector + λB vector).
- The most important requirement is that the parallel vector B vector and the vector PC (C - P) be perpendicular.
- The PC vector dot B vector must equal zero in order to be considered perpendicular.
- When the PC vector is substituted with the C vector, the result is (C vector - P vector) · B vector = 0.
- The formula λ = (P vector - A vector) · B vector / |B vector|² can be obtained by solving this equation for lambda (λ).
- Calculating the point C (A vector + λB vector) is possible once lambda has been determined.
- The magnitude of the vector PC (C vector - P vector) is the perpendicular distance.
An example of a problem is: An example is given where the line is (x-6)/3 = (y-7)/2 = (z-7)/-2 and point P is (1, 2, 3).
- I + 2j + 3k is the vector of P.
- 6i + 7j + 7k is a vector that represents a point on the line and is taken from the numerator of the line equation.
- The parallel vector, denoted by the B vector (from the denominator), is 3i + 2j - 2k.
- A vector yields -5i - 5j - 4k when the P vector is calculated.
- Dot product calculation (P vector - A vector) results in -17.
- The B vector's magnitude squared is 17.
- With λ = -1, the point C is A vector + (-1) using the lambda formula.
- The magnitude of (3i + 5j + 9k) is the distance PC.
**Homework Issue:** Locate a point on the y-axis such that the distance between it and the line x = y = 0 is unity perpendicularly. One can assume that the point on the y-axis is (0, k, 0). By equating the calculated distance to 1, the same method can be used to determine the value of k.
2. Determining a Line's Equation Perpendicular to Two Given Lines
- This is the second subject covered. The task is to determine the equation of a third line (L3) that passes through a particular point and is perpendicular to two given lines (L1 and L2).
- You need a line's parallel vector and the point it passes through, which is provided, in order to determine the line's equation.
- The parallel vector for L3 is unique in that it needs to be perpendicular to both L1 and L2.
- The parallel vector of L3 (B3 vector) is perpendicular to both B1 and B2.
- It should be clear from this that taking the cross product of two vectors is necessary to find a vector perpendicular to them.
Consequently, the parallel vector B3 for the needed line is equal to the cross product of the parallel vectors of the two provided lines: B3 vector = B1 vector × B2 vector.
An example of a problem is: Considering L1: (x-1)/1 = (y-2)/2 = (z-3)/3, L2: (x-2)/4 = (y-3)/5 = (z-1)/6, and the point (1, 1, 1) that L3 passes through.
- i + 2j + 3k is the B1 vector (from L1).
- The B2 vector is 4i + 5j + 6k (from L2).
- Finding the cross product between B1 and B2.
| i j k |
| 1 2 3 |
| 4 5 6 |
= i(2*6 - 3*5) - j(1*6 - 3*4) + k(1*5 - 2*4)
= i(12 - 15) - j(6 - 12) + k(5 - 8)
= -3i - (-6)j - 3k
= -3i + 6j - 3k.
Thus, the B3 vector is equal to -3i + 6j - 3k.
Confirming Perpendicularity
The dot product of B3 with B1 and B2 must be zero:
- B3 · B1 = (-3)(1) + (6)(2) + (-3)(3) = -3 + 12 - 9 = 0.
- B3 · B2 = (-3)(4) + (6)(5) + (-3)(6) = -12 + 30 - 18 = 0.
This demonstrates that B3 is perpendicular to B1 and B2.
Line L3
The necessary line L3 has a parallel vector (-3, 6, -3) and passes through (1, 1, 1).
In symmetric form, the equation of line L3 is:
(x - 1)/-3 = (y - 1)/6 = (z - 1)/-3.
The procedure is to take the cross product of the parallel vectors of the two lines to find the parallel vector for the perpendicular line, and then use the given point to write the equation. The point (1, 1, 1) is in the numerator, and the components of the parallel vector (-3, 6, -3) are in the denominator.
Conclusion
Two approaches were presented:
- Determining a line perpendicular to two others (using the cross product of parallel vectors).
- Determining a perpendicular distance (using a formula for lambda, calculating the point C, and then determining the distance).
To become comfortable with these problem types, practice is advised, including with the homework problem.
