3D-Lines (eActivity)

Command last updated 6/3/2024.

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Extends Charlie Watson’s eActivity. Modify the values of the input variables, then click the top line and hit EXE. The outputs are stored in the relevant variables.

Line 2 pts

Find vector and Cartesian equations of line given 2 points on line.

Input: A, B

A, B are point vectors on line

Output: A, AB, C

A is point vector on line
AB is direction vector along line
C is list representing Cartesian equation

C = {(x-a)/b, (y-c)/d, (z-e)/f} represents (x-a)/b = (y-c)/d = (z-e)/f

r = A + λAB

Example

Find the line passing through points \((-3, -1, -10)\) and \((5, -6, 4)\). (Sample Specialist Exam 2 Section A Question 4)

A = [-3, -1, -10]; B = [5, -6, 4] A = [-3, -1, -10]; AB = [5, -6, 4]; C = {(x+3)/8, -(y+1)/5, (z+10)/14}

Line 1 pt, 1 vector

Find Cartesian equation of line given 1 point and 1 direction vector on line.

Input: A, P

A is point vector on line
P is direction vector along line

Output: C

C is list representing Cartesian equation

C = {(x-a)/b, (y-c)/d, (z-e)/f} represents (x-a)/b = (y-c)/d = (z-e)/f

r = A + λP

Example

Find a Cartesian equation of the line \(\mathbf{r}(\lambda) = -\mathbf{i} + \mathbf{j} - 3\mathbf{k} + \lambda(2\mathbf{i} + 4\mathbf{j} - 7\mathbf{k})\).

A = [-1, 1, -3]; P = [2, 4, -7] C = {(x+1)/2, (y-1)/4, -(z+3)/7}

Line Cartesian

Find vector equation of line given Cartesian equation of line.

Input: C

C is list representing Cartesian equation

Output: A, P

A is point vector on line
P is direction vector along line

C = {(x-a)/b, (y-c)/d, (z-e)/f} represents (x-a)/b = (y-c)/d = (z-e)/f

r = A + λP

Example

Find a vector equation of the line \(\dfrac{x+3}{2} = \dfrac{2-y}{3} = \dfrac{z+1}{5}\).

C = {(x+3)/2, (2-y)/3, (z+1)/5} A = [-3, 2, -1]; P = [2, -3, 5]

Distance point-line

Find shortest distance between point and line.

Input: P, A, B

P is point vector of point
A is point vector on line
B is direction vector along line

Output: D, M

D is shortest distance
M is point vector on line closest to P

r = A + λB

Example

\(A = (1, 1, 2)\), \(B = (1, 2, 3)\) and \(C = (3, 2, 4)\). Find the shortest distance from \(B\) to line segment \(AC\). (2023 Specialist Exam 2 Section B Question 5b)

P = [1, 2, 3]; A = [1, 1, 2]; B = [3, 2, 4] - [1, 1, 2] D = 1; M = [5/3, 4/3, 8/3]

Distance line-line

Find shortest distance between two lines.

Input: rA, vA, rB, vB

rA is point vector on line A
vA is direction vector along line A
rB is point vector on line B
rB is direction vector along line B

Output: AB, D

AB is shortest vector between lines (perpendicular to both lines)
D is shortest distance

r = rA + λvA

r = rB + μvB

Example

Find the shortest distance between the lines \(\mathbf{r}(t) = 4\mathbf{i} + 2\mathbf{j} + \mathbf{k} + t(-\mathbf{i} + \mathbf{j} + 3\mathbf{k})\) and \(\mathbf{r}(s) = 5\mathbf{i} + 4\mathbf{j} - 2\mathbf{k} + s(-\mathbf{i} + \mathbf{j} + 3\mathbf{k})\). (Sample Specialist Exam 2 Section B Question 4a)

rA = [4, 2, 1]; vA = [-1, 1, 3]; rB = [5, 4, -2]; vB = [-1, 1, 3] AB = [(11mu-8)/11-mu+1, -(11mu-8)/11+mu+2, -3(11mu-8)/11+3mu-3]; D = 3 sqrt(110) / 11

Intersection line-line

Find intersection between two lines.

Input: rA, vA, rB, vB

rA is point vector on line A
vA is direction vector along line A
rB is point vector on line B
vB is direction vector along line B

Output: Soln1, Soln2, X

Check Soln1 and Soln2 are the same. If different, then lines do not intersect.
X is point vector of intersection

r = rA + λvA

r = rB + μvB

Example

Find \(a\) and the point of intersection between the intersecting lines \(\mathbf{r}(t) = \mathbf{i} - 3\mathbf{j} + 6\mathbf{k} + t(3\mathbf{i} + 5\mathbf{j} - a\mathbf{k})\) and \(\mathbf{r}(s) = -6\mathbf{i} + 2\mathbf{j} + \mathbf{k} + s(4\mathbf{i} - 10\mathbf{j} + 6\mathbf{k})\). (Sample Specialist Exam 2 Section B Question 4b)

rA = [1, -3, 6]; vA = [3, 5, -a]; rB = [-6, 2, 1]; vB = [4, -10, 6] Soln1 = {lambda=-1, mu=1}; Soln2 = {lambda=-11/(2a+9), mu=(7a+15)/(4a+18); X = [-2, -8, a+6]

Angle line-line

Find both angles between two lines.

Input: vA, vB

vA is direction vector along line A
vB is direction vector along line B

Output: θ1, θ2

θ1, θ2 are angles

r = rA + λvA

r = rB + μvB

Example

Find the angles between the lines \(\mathbf{r}(t) = \mathbf{i} + 2\mathbf{k} + t(2\mathbf{i} - \mathbf{j} + \mathbf{k})\) and \(\mathbf{r}(s) = -2\mathbf{i} + 2\mathbf{j} + \mathbf{k} + s(-3\mathbf{i} + 2\mathbf{j} - \mathbf{k})\).

vA = [2, -1, 1]; vB = [-3, 2, -1] theta1 = 169; theta2 = 188

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