MAT1503
ASSIGNMENT 7 2021
QUESTION 1
QUESTION 1.1
a).
𝑥 = 1 + 𝑡 , 𝑦 = −1 − 𝑡 , 𝑧 = −2𝑡
(𝑥, 𝑦, 𝑧) = (1 + 𝑡 , −1 − 𝑡 , −2𝑡)
(𝑥, 𝑦, 𝑧) = (1, −1,0) + 𝑡(1, −1, −2)
𝐿𝑒𝑡 ∶ 𝑣⃗ = 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑙𝑖𝑛𝑒 = (1, −1, −2)
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒 ∶ 𝑥 + 2𝑦 + 3𝑧 − 9 = 0
𝑛⃗⃗ = 𝑛𝑜𝑟𝑚𝑎𝑙 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒
𝑛⃗⃗ = (1,2,3)
𝑛⃗⃗ ∙ 𝑣⃗ = (1,2,3) ∙ (1, −1, −2)
=1−2−6
= −7
𝑆𝑖𝑛𝑐𝑒 𝑛⃗⃗ ∙ 𝑣⃗ ≠ 0 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑎𝑛𝑑 𝑝𝑙𝑎𝑛𝑒 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙.
, b).
𝐿𝑒𝑡 ∶ 𝑣⃗ = 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑙𝑖𝑛𝑒 = (3,2, −1)
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒 ∶ 4𝑥 − 𝑦 + 2𝑧 + 1 = 0
𝑛⃗⃗ = 𝑛𝑜𝑟𝑚𝑎𝑙 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒
𝑛⃗⃗ = (4, −1,2)
𝑛⃗⃗ ∙ 𝑣⃗ = (4, −1,2) ∙ (3,2, −1)
= 12 − 2 − 2
=8
𝑆𝑖𝑛𝑐𝑒 𝑛⃗⃗ ∙ 𝑣⃗ ≠ 0 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑎𝑛𝑑 𝑝𝑙𝑎𝑛𝑒 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙.
QUESTION 2
QUESTION 2.1
𝑃 = (2,0, −1) , 𝑛⃗⃗ = 〈2,1,3〉
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑙𝑖𝑛𝑒: 𝑟 = 𝑟𝑜 + 𝑡𝑣
𝑟 = 〈𝑥, 𝑦, 𝑧〉 = 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑝𝑜𝑖𝑛𝑡
𝑟𝑜 = 𝑝𝑜𝑖𝑛𝑡 𝑡ℎ𝑎𝑡 𝑙𝑖𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 = 〈2,0, −1〉
𝑣 = 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑙𝑖𝑛𝑒 = 𝑛⃗⃗
𝑣 = 〈2,1,3〉
〈𝑥, 𝑦, 𝑧〉 = 〈2,0, −1〉 + 𝑡〈2,1,3〉
〈𝑥, 𝑦, 𝑧〉 = 〈2 + 2𝑡 , 𝑡 , −1 + 3𝑡〉
𝑃𝑎𝑟𝑎𝑚𝑒𝑡𝑟𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛:
𝑥 = 2 + 2𝑡 , 𝑦 = 𝑡 , 𝑧 = −1 + 3𝑡
ASSIGNMENT 7 2021
QUESTION 1
QUESTION 1.1
a).
𝑥 = 1 + 𝑡 , 𝑦 = −1 − 𝑡 , 𝑧 = −2𝑡
(𝑥, 𝑦, 𝑧) = (1 + 𝑡 , −1 − 𝑡 , −2𝑡)
(𝑥, 𝑦, 𝑧) = (1, −1,0) + 𝑡(1, −1, −2)
𝐿𝑒𝑡 ∶ 𝑣⃗ = 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑙𝑖𝑛𝑒 = (1, −1, −2)
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒 ∶ 𝑥 + 2𝑦 + 3𝑧 − 9 = 0
𝑛⃗⃗ = 𝑛𝑜𝑟𝑚𝑎𝑙 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒
𝑛⃗⃗ = (1,2,3)
𝑛⃗⃗ ∙ 𝑣⃗ = (1,2,3) ∙ (1, −1, −2)
=1−2−6
= −7
𝑆𝑖𝑛𝑐𝑒 𝑛⃗⃗ ∙ 𝑣⃗ ≠ 0 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑎𝑛𝑑 𝑝𝑙𝑎𝑛𝑒 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙.
, b).
𝐿𝑒𝑡 ∶ 𝑣⃗ = 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑙𝑖𝑛𝑒 = (3,2, −1)
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒 ∶ 4𝑥 − 𝑦 + 2𝑧 + 1 = 0
𝑛⃗⃗ = 𝑛𝑜𝑟𝑚𝑎𝑙 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒
𝑛⃗⃗ = (4, −1,2)
𝑛⃗⃗ ∙ 𝑣⃗ = (4, −1,2) ∙ (3,2, −1)
= 12 − 2 − 2
=8
𝑆𝑖𝑛𝑐𝑒 𝑛⃗⃗ ∙ 𝑣⃗ ≠ 0 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑎𝑛𝑑 𝑝𝑙𝑎𝑛𝑒 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙.
QUESTION 2
QUESTION 2.1
𝑃 = (2,0, −1) , 𝑛⃗⃗ = 〈2,1,3〉
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑙𝑖𝑛𝑒: 𝑟 = 𝑟𝑜 + 𝑡𝑣
𝑟 = 〈𝑥, 𝑦, 𝑧〉 = 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑝𝑜𝑖𝑛𝑡
𝑟𝑜 = 𝑝𝑜𝑖𝑛𝑡 𝑡ℎ𝑎𝑡 𝑙𝑖𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 = 〈2,0, −1〉
𝑣 = 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑙𝑖𝑛𝑒 = 𝑛⃗⃗
𝑣 = 〈2,1,3〉
〈𝑥, 𝑦, 𝑧〉 = 〈2,0, −1〉 + 𝑡〈2,1,3〉
〈𝑥, 𝑦, 𝑧〉 = 〈2 + 2𝑡 , 𝑡 , −1 + 3𝑡〉
𝑃𝑎𝑟𝑎𝑚𝑒𝑡𝑟𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛:
𝑥 = 2 + 2𝑡 , 𝑦 = 𝑡 , 𝑧 = −1 + 3𝑡