MAT1501
Assignment 04
Friday, 29 August 2025
, MAT1503 Assignment 04 — Due: 29 August 2025
MAT1503 Assignment 04
Due Date: Friday, 29 August 2025
Question 1
Problem 1.1
Problem Statement: Find an equation for the plane that passes through the origin
(0, 0, 0) and is parallel to the plane −x + 3y − 2z = 6.
Step 1: Identify the normal vector. A plane of the form ax + by + cz = d has
normal vector n = (a, b, c). For −x + 3y − 2z = 6, the normal is n = (−1, 3, −2).
Step 2: Use “parallel” and “through the origin”. Planes that are parallel share
the same normal vector. A plane with normal n = (−1, 3, −2) passing through the
origin has equation
−x + 3y − 2z = 0
because substituting (0, 0, 0) must satisfy the equation.
Step 3: Check. Any vector lying in the original plane is orthogonal to (−1, 3, −2), so
both planes are parallel.
Final Answer:
−x + 3y − 2z = 0
Page 1
, MAT1503 Assignment 04 — Due: 29 August 2025
Problem 1.2
Problem Statement: Find the distance between the point (−1, −2, 0) and the plane
3x − y + 4z = −2.
Step 1: Put the plane in standard form.
3x − y + 4z + 2 = 0 ⇒ a = 3, b = −1, c = 4, d = 2.
Step 2: Recall the distance formula. For point P (x0 , y0 , z0 ) and plane ax + by +
cz + d = 0,
|ax0 + by0 + cz0 + d|
dist(P, Π) = √ .
a2 + b 2 + c 2
Step 3: Substitute (−1, −2, 0).
|ax0 + by0 + cz0 + d| = |3(−1) + (−1)(−2) + 4(0) + 2|
= | − 3 + 2 + 0 + 2| = |1| = 1,
√ p √ √
a2 + b2 + c2 = 32 + (−1)2 + 42 = 9 + 1 + 16 = 26.
Final Answer:
1
√
26
Page 2
Assignment 04
Friday, 29 August 2025
, MAT1503 Assignment 04 — Due: 29 August 2025
MAT1503 Assignment 04
Due Date: Friday, 29 August 2025
Question 1
Problem 1.1
Problem Statement: Find an equation for the plane that passes through the origin
(0, 0, 0) and is parallel to the plane −x + 3y − 2z = 6.
Step 1: Identify the normal vector. A plane of the form ax + by + cz = d has
normal vector n = (a, b, c). For −x + 3y − 2z = 6, the normal is n = (−1, 3, −2).
Step 2: Use “parallel” and “through the origin”. Planes that are parallel share
the same normal vector. A plane with normal n = (−1, 3, −2) passing through the
origin has equation
−x + 3y − 2z = 0
because substituting (0, 0, 0) must satisfy the equation.
Step 3: Check. Any vector lying in the original plane is orthogonal to (−1, 3, −2), so
both planes are parallel.
Final Answer:
−x + 3y − 2z = 0
Page 1
, MAT1503 Assignment 04 — Due: 29 August 2025
Problem 1.2
Problem Statement: Find the distance between the point (−1, −2, 0) and the plane
3x − y + 4z = −2.
Step 1: Put the plane in standard form.
3x − y + 4z + 2 = 0 ⇒ a = 3, b = −1, c = 4, d = 2.
Step 2: Recall the distance formula. For point P (x0 , y0 , z0 ) and plane ax + by +
cz + d = 0,
|ax0 + by0 + cz0 + d|
dist(P, Π) = √ .
a2 + b 2 + c 2
Step 3: Substitute (−1, −2, 0).
|ax0 + by0 + cz0 + d| = |3(−1) + (−1)(−2) + 4(0) + 2|
= | − 3 + 2 + 0 + 2| = |1| = 1,
√ p √ √
a2 + b2 + c2 = 32 + (−1)2 + 42 = 9 + 1 + 16 = 26.
Final Answer:
1
√
26
Page 2
