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MAT2615 ASSIGNMENT 1 2022
- Institution
- University Of South Africa (Unisa)
This document contains MAT2615 ASSIGNMENT 1 2022 solutions. Clearly step by step calculations and explanations are provided.
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MAT2615ASSIGNMENT 1
SEMESTER 1
2022
,Solution:
̅ 𝑎𝑛𝑑 𝒗
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 𝑙𝑖𝑛𝑒𝑠 𝒍𝟏 𝑎𝑛𝑑 𝒍𝟐 𝑏𝑒 𝒖 ̅ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦
𝑢̅ = (1,1,1) 𝑎𝑛𝑑 𝑣̅ = (1, −1,0)
𝑛̅ = 𝑛𝑜𝑟𝑚𝑎𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒
𝑛̅ = 𝑢̅ × 𝑣̅
𝑖 𝑗 𝑘
= |1 1 1|
1 −1 0
1 1 1 1 1 1
= 𝑖| |−𝑗| |+𝑘| |
−1 0 1 0 1 −1
= 𝑖(0 + 1) − 𝑗(0 − 1) + 𝑘(−1 − 1)
= 𝑖 + 𝑗 − 2𝑘
= (1,1, −2)
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝑝𝑙𝑎𝑛𝑒 ∶ 𝑟 ∙ 𝑛̅ = 𝑟𝑜 ∙ 𝑛̅
𝑤ℎ𝑒𝑟𝑒 ∶ 𝑟 = (𝑥, 𝑦, 𝑧)
𝑛̅ = 𝑛𝑜𝑟𝑚𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 = (1,1, −2)
𝑟𝑜 = 𝑝𝑜𝑖𝑛𝑡 𝑡ℎ𝑎𝑡 𝑙𝑖𝑒𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒
𝑊𝑒 𝑐𝑎𝑛 𝑐ℎ𝑜𝑜𝑠𝑒 𝑟𝑜 = (1,0, −1) 𝑜𝑟 𝑟𝑜 = (2,1,0) 𝑏𝑢𝑡 𝑖𝑛 𝑡ℎ𝑖𝑠 𝑐𝑎𝑠𝑒 𝑖𝑚 𝑔𝑜𝑖𝑛𝑔 𝑡𝑜 𝑐ℎ𝑜𝑜𝑠𝑒
𝑟𝑜 = (1,0, −1)
, 𝑟 ∙ 𝑛̅ = 𝑟𝑜 ∙ 𝑛̅
(𝑥, 𝑦, 𝑧) ∙ (1,1, −2) = (1,0, −1) ∙ (1,1, −2)
𝑥 + 𝑦 − 2𝑧 = 1 + 0 + 2
𝑥 + 𝑦 − 2𝑧 = 3
𝑥 + 𝑦 − 2𝑧 − 3 = 0
𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒂 𝒑𝒍𝒂𝒏𝒆: 𝒙 + 𝒚 − 𝟐𝒛 − 𝟑 = 𝟎
Solution:
𝑥 − 𝑦 + 2𝑧 − 1 = 0
𝑥 − 𝑦 + 2𝑧 = 1 1
3𝑥 + 2𝑦 − 6𝑧 + 4 = 0
3𝑥 + 2𝑦 − 6𝑧 = −4 2
3× 1 − 2 ∶
𝑥 − 𝑦 + 2𝑧 = 1 ×3
− 3𝑥 + 2𝑦 − 6𝑧 = −4
0 − 5𝑦 + 12𝑧 = 7
−5𝑦 + 12𝑧 = 7
12𝑧 = 7 + 5𝑦
7 + 5𝑦
𝑧=
12
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