MAT3701 Assignment 1 2026
DUE 30 April 2026
Question 1 (10 marks) - Lines and Planes in Space
1.1 Equation of a line through two points
Points (2,4,0) and (-3,-6,0)
Find the direction vector
Direction vector = point2 - point1 = (-3-2, -6-4, 0-0) = (-5, -10, 0)
You can simplify by dividing by -5: (1, 2, 0) - this is parallel, so acceptable
Write parametric equations
Using point (2,4,0) and direction (-5,-10,0):
x = 2 - 5t
y = 4 - 10t
z=0
Convert to symmetric form (optional)
Since z = 0 constant, the line lies in the xy-plane:
(x-2)/(-5) = (y-4)/(-10), z = 0
1.2 Equation of a plane through three points
(3,-6,7), (-2,0,-4), (5,-9,-2)
Find two direction vectors in the plane
v₁ = point2 - point1 = (-2-3, 0-(-6), -4-7) = (-5, 6, -11)
v₂ = point3 - point1 = (5-3, -9-(-6), -2-7) = (2, -3, -9)
Find normal vector using cross product
n = v₁ × v₂
Using the determinant method from Friedberg Section 1.1:
n=|i j k |
| -5 6 -11 |
|2 -3 -9 |
Calculate
DUE 30 April 2026
Question 1 (10 marks) - Lines and Planes in Space
1.1 Equation of a line through two points
Points (2,4,0) and (-3,-6,0)
Find the direction vector
Direction vector = point2 - point1 = (-3-2, -6-4, 0-0) = (-5, -10, 0)
You can simplify by dividing by -5: (1, 2, 0) - this is parallel, so acceptable
Write parametric equations
Using point (2,4,0) and direction (-5,-10,0):
x = 2 - 5t
y = 4 - 10t
z=0
Convert to symmetric form (optional)
Since z = 0 constant, the line lies in the xy-plane:
(x-2)/(-5) = (y-4)/(-10), z = 0
1.2 Equation of a plane through three points
(3,-6,7), (-2,0,-4), (5,-9,-2)
Find two direction vectors in the plane
v₁ = point2 - point1 = (-2-3, 0-(-6), -4-7) = (-5, 6, -11)
v₂ = point3 - point1 = (5-3, -9-(-6), -2-7) = (2, -3, -9)
Find normal vector using cross product
n = v₁ × v₂
Using the determinant method from Friedberg Section 1.1:
n=|i j k |
| -5 6 -11 |
|2 -3 -9 |
Calculate
