MAT1503 Assignment 4 2025 - Due 29 August 2025

MAT1503
Assignment 4
Unique No:
Due 29 August 2025

, MAT1503
ASSIGNMENT 04

Friday, 29 August 2025

Question 1

1.1 Plane through the origin parallel to −𝑥 + 3𝑦 − 2𝑧 = 6 . Parallel ⇒ same normal 𝐧 =
(−1,3, −2). Through (0,0,0):

−𝑥 + 3𝑦 − 2𝑧 = 0 .

1.2 Distance from 𝑃(−1, −2,0) to plane 3𝑥 − 𝑦 + 4𝑧 = −2 . Write 3𝑥 − 𝑦 + 4𝑧 + 2 = 0 .

|3(−1) − 1(−2) + 4(0) + 2| | − 3 + 2 + 0 + 2| 1
𝑑= = = .
√32 + (−1)2 + 42 √26 √26




Question 2

2.1 Angle between 𝐯1 = ⟨−1, 1, 0, −1⟩ and 𝐯2 = ⟨1, −1, 3, −2⟩. 𝐯1 ⋅ 𝐯2 = −1 − 1 + 0 +
2 = 0 ⇒ 𝜃 = 90 ∘ . ∥ 𝐯1 ∥= √3, ∥ 𝐯2 ∥= √15.

Angle = 90 ∘ (perpendicular; not parallel) .

2.2 Direction cosines/angles of 𝐫 = ⟨0, −1, −2, 3 4⁄ ⟩. ∥ 𝐫 ∥= √02 + 12 + 22 + (3/4)2 =

√89/4. Direction cosines = (0, −4/ √89, −8/ √89, 3/√89). Direction angles:

𝛼1 = arccos0 = 𝜋 ⁄
2 , 𝛼 2 = arccos(− 4 ⁄√89), 𝛼3 = arccos(− 8 ⁄√89), 𝛼4 = arccos(3 ⁄√89) .

𝑑
2.3 𝐫(𝑡) = ⟨𝑡, −1/𝑡, 𝑡2 − 2⟩. 𝐫 ′ (𝑡) = ⟨1, 1/𝑡 2 , 2𝑡⟩ ⇒ 𝐫 ′ (1) = ⟨1, 1, 2⟩ . For 𝑑𝑡 (𝐕 ⋅ 𝐫) = 𝐕 ′ ⋅

𝐫 + 𝐕 ⋅ 𝐫 ′ : 𝐫(1) = ⟨1, −1, −1⟩, 𝐕(1) = ⟨−1,1, −2⟩, 𝐕 ′ (1) = ⟨1, −2,2⟩. 𝐕 ′ ⋅ 𝐫 = 1 + 2 − 2 =
1, 𝐕 ⋅ 𝐫′ = −1 + 1 − 4 = −4 .

(𝐕 ⋅ 𝐫)′ (1) = −3 .