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MAT1503 Assignment 7 2021

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UNISA MAT1503 Linear Algebra Assignment SEVEN of 2021 solutions. Topics covered are: Points in 3D. Vectors in 3D. Straight lines in 3D. Planes.

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MAT1503 ASSIGNMENT 7 2021



Question 1



(1.1)

(𝑎) 𝑥 = 1 + 𝑡, 𝑦 = −1 − 𝑡, 𝑧 = −2𝑡 𝑎𝑛𝑑 𝑥 + 2𝑦 + 3𝑧 − 9 = 0



𝑥 = 1 + 𝑡, 𝑦 = −1 − 𝑡, 𝑧 = 0 − 2𝑡 𝑎𝑛𝑑 𝑥 + 2𝑦 + 3𝑧 − 9 = 0
𝑉𝑒𝑐𝑡𝑜𝑟 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑙𝑖𝑛𝑒: 〈1, −1, −2〉 𝑉𝑒𝑐𝑡𝑜𝑟 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑝𝑙𝑎𝑛𝑒 〈1, 2, 3〉


〈1, −1, −2〉 • 〈1, 2, 3〉 = 1 × 1 − 1 × 2 − 2 × 3
〈1, −1, −2〉 • 〈1, 2, 3〉 = 1 − 2 − 6
〈1, −1, −2〉 • 〈1, 2, 3〉 = −7
〈1, −1, −2〉 • 〈1, 2, 3〉 ≠ 0
〈1, −1, −2〉 𝑎𝑛𝑑 〈1, 2, 3〉 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟

𝑉𝑒𝑐𝑡𝑜𝑟 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑛𝑜𝑡 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑝𝑙𝑎𝑛𝑒.
𝑇ℎ𝑒 𝑙𝑖𝑛𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟.


(𝑏) 〈0, 1, 2〉 + 𝑡〈3, 2, −1〉 𝑎𝑛𝑑 4𝑥 − 𝑦 + 2𝑧 + 1 = 0



𝑉𝑒𝑐𝑡𝑜𝑟 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑙𝑖𝑛𝑒: 〈3, 2, −1〉 𝑉𝑒𝑐𝑡𝑜𝑟 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑝𝑙𝑎𝑛𝑒 〈4, −1, 2〉


〈3, 2, −1〉 • 〈4, −1, 2〉 = 3 × 4 + 2 × −1 − 1 × 2
〈3, 2, −1〉 • 〈4, −1, 2〉 = 12 − 2 − 2
〈3, 2, −1〉 • 〈4, −1, 2〉 = 8
〈3, 2, −1〉 • 〈4, −1, 2〉 ≠ 0
〈3, 2, −1〉 𝑎𝑛𝑑 〈4, −1, 2〉 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟

𝑉𝑒𝑐𝑡𝑜𝑟 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑛𝑜𝑡 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑝𝑙𝑎𝑛𝑒.
𝑇ℎ𝑒 𝑙𝑖𝑛𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟.

, Question 2



(2.1)

𝑃 = (2, 0, −1) 𝑎𝑛𝑑 𝑛⃗ = 〈2, 1, 3〉



𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑙𝑖𝑛𝑒:
(𝑥, 𝑦, 𝑧) = 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑙𝑖𝑛𝑒 + 𝑡(𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑙𝑖𝑛𝑒) 𝑡∈ℝ


(𝑥, 𝑦, 𝑧) = (2, 0, −1) + 𝑡(2, 1, 3) 𝑡∈ℝ
(𝑥, 𝑦, 𝑧) = (2, 0, −1) + (2𝑡, 1𝑡, 3𝑡) 𝑡∈ℝ
(𝑥, 𝑦, 𝑧) = (2 + 2𝑡, 0 + 𝑡, −1 + 3𝑡) 𝑡 ∈ ℝ
(𝑥, 𝑦, 𝑧) = (2 + 2𝑡, 𝑡, −1 + 3𝑡) 𝑡∈ℝ



𝑥 = 2 + 2𝑡, 𝑦 = 𝑡, 𝑧 = −1 + 3𝑡 𝑡∈ℝ



(2.2)

𝐴 = (1, 2, −3) 𝑎𝑛𝑑 𝐵 = (7, 2, −4)
⃗⃗⃗⃗⃗ 𝑖𝑠 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒.
𝑇ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝐴𝐵



⃗⃗⃗⃗⃗
𝐴𝐵 = 〈7 − 1, 2 − 2, −4 + 3〉
⃗⃗⃗⃗⃗
𝐴𝐵 = 〈8, 0, −1〉



𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑙𝑖𝑛𝑒:
(𝑥, 𝑦, 𝑧) = 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑙𝑖𝑛𝑒 + 𝑡(𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑙𝑖𝑛𝑒) 𝑡∈ℝ


(𝑥, 𝑦, 𝑧) = (1, 2, −3) + 𝑡(8, 0, −1) 𝑡∈ℝ
(𝑥, 𝑦, 𝑧) = (1, 2, −3) + (8𝑡, 0𝑡, −1𝑡) 𝑡 ∈ ℝ
(𝑥, 𝑦, 𝑧) = (1, 2, −3) + (8𝑡, 0, −𝑡) 𝑡∈ℝ
(𝑥, 𝑦, 𝑧) = (1 + 8𝑡, 2 + 0, −3 − 𝑡) 𝑡∈ℝ
(𝑥, 𝑦, 𝑧) = (1 + 8𝑡, 2, −3 − 𝑡) 𝑡∈ℝ



𝑥 = 1 + 8𝑡, 𝑦 = 2, 𝑧 = −3 − 𝑡 𝑡∈ℝ

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