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MAT1503 Assignment 8 2021

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  • MAT1503 - Linear Algebra (MAT1503)
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  • University Of South Africa (Unisa)

UNISA MAT1503 Linear Algebra Assignment EIGHT of 2021 solutions. Topics covered: Vectors. Planes. Parallel and Perpendicular planes. Finding equations of planes. Unit vectors. Complex numbers. Polar forms Standard form Operations on complex numbers

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  • July 25, 2021
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  • 2021/2022
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Subjects

  • mat1503
  • linear algebra
  • vectors
  • planes
  • complex numbers

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  • University of South Africa (Unisa)
  • MAT1503 - Linear Algebra (MAT1503)
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MAT1503 ASSIGNMENT 8 2021



Question 1



(1.1)



𝑈: 𝜆𝑥 + 5𝑦 − 2𝜆𝑧 − 3 = 0
⃗⃗⃗⃗1 = 〈𝜆, 5, −2𝜆〉
𝑉𝑒𝑐𝑡𝑜𝑟 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑡𝑜 𝑝𝑙𝑎𝑛𝑒 𝑖𝑠 𝑛



𝑉: −𝜆𝑥 + 𝑦 + 2𝜆𝑧 + 1 = 0
⃗⃗⃗⃗2 = 〈−𝜆, 1, 2𝜆〉
𝑉𝑒𝑐𝑡𝑜𝑟 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑡𝑜 𝑝𝑙𝑎𝑛𝑒 𝑖𝑠 𝑛


(𝑎) 𝑈 𝑎𝑛𝑑 𝑉 𝑎𝑟𝑒 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙



𝑇ℎ𝑒 𝑡𝑤𝑜 𝑝𝑙𝑎𝑛𝑒𝑠 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑖𝑓 𝑡ℎ𝑒𝑖𝑟 𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑎𝑟𝑒 𝑎𝑙𝑠𝑜 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙.
𝑛
⃗⃗⃗⃗1 • 𝑛
⃗⃗⃗⃗2 = 0
〈𝜆, 5, −2𝜆〉 • 〈−𝜆, 1, 2𝜆〉 = 0

𝜆 × −𝜆 + 5 × 1 − 2𝜆 × 2𝜆 = 0

−𝜆2 + 5 − 4𝜆2 = 0
−𝜆2 − 4𝜆2 + 5 = 0

−5𝜆2 + 5 = 0

−5𝜆2 5 0
+ =
−5 −5 −5
𝜆2 − 1 = 0
(𝜆 + 1)(𝜆 − 1) = 0

𝜆 = −1 𝑜𝑟 𝜆 = 1 𝐵𝑜𝑡ℎ 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝜆 𝑤𝑖𝑙𝑙 𝑙𝑒𝑎𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑝𝑙𝑎𝑛𝑒𝑠 𝑡𝑜 𝑏𝑒 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙.

,𝑈: 𝜆𝑥 + 5𝑦 − 2𝜆𝑧 − 3 = 0
⃗⃗⃗⃗1 = 〈𝜆, 5, −2𝜆〉
𝑉𝑒𝑐𝑡𝑜𝑟 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑡𝑜 𝑝𝑙𝑎𝑛𝑒 𝑖𝑠 𝑛



𝑉: −𝜆𝑥 + 𝑦 + 2𝜆𝑧 + 1 = 0
⃗⃗⃗⃗2 = 〈−𝜆, 1, 2𝜆〉
𝑉𝑒𝑐𝑡𝑜𝑟 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑡𝑜 𝑝𝑙𝑎𝑛𝑒 𝑖𝑠 𝑛


(𝑏) 𝑈 𝑎𝑛𝑑 𝑉 𝑎𝑟𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙



𝑇ℎ𝑒 𝑡𝑤𝑜 𝑝𝑙𝑎𝑛𝑒𝑠 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑖𝑓 𝑡ℎ𝑒𝑖𝑟 𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑎𝑟𝑒 𝑎𝑙𝑠𝑜 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙.



𝑛
⃗⃗⃗⃗1 = 𝑘𝑛
⃗⃗⃗⃗2 𝑘∈ℝ
〈𝜆, 5, −2𝜆〉 = 𝑘〈−𝜆, 1, 2𝜆〉 𝑘∈ℝ
〈𝜆, 5, −2𝜆〉 = 〈−𝜆𝑘, 𝑘, 2𝜆𝑘〉 𝑘∈ℝ



𝜆 = −𝜆𝑘 𝑎𝑛𝑑 5 = 𝑘 𝑎𝑛𝑑 − 2𝜆 = 2𝜆𝑘
𝜆 = −5𝜆 𝑎𝑛𝑑 𝑘 = 5 𝑎𝑛𝑑 − 2𝜆 = 2(5)𝜆
𝜆 = −5𝜆 𝑎𝑛𝑑 𝑘 = 5 𝑎𝑛𝑑 − 2𝜆 = 10𝜆
𝜆 + 5𝜆 = 0 𝑎𝑛𝑑 𝑘 = 5 𝑎𝑛𝑑 − 2𝜆 − 10𝜆 = 0
6𝜆 = 0 𝑎𝑛𝑑 𝑘 = 5 𝑎𝑛𝑑 − 12𝜆 = 0
𝜆 = 0 𝑎𝑛𝑑 𝑘 = 5 𝑎𝑛𝑑 𝜆 = 0
𝜆 = 0 𝑤𝑖𝑙𝑙 𝑙𝑒𝑎𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑝𝑙𝑎𝑛𝑒𝑠 𝑡𝑜 𝑏𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙.


(1.2)



𝑇ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑝𝑎𝑠𝑠𝑒𝑠 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 (0, 0, 0) 𝑎𝑛𝑑 𝑖𝑠 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒
−𝑥 + 3𝑦 − 2𝑧 = 6 𝑤ℎ𝑜𝑠𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑠 〈−1, 3, −2〉



𝑇ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 〈−1, 3, −2〉 𝑤𝑖𝑙𝑙 𝑎𝑙𝑠𝑜 𝑏𝑒 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑤ℎ𝑜𝑠𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑤𝑒 𝑚𝑢𝑠𝑡 𝑓𝑖𝑛𝑑.



𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒:
𝑎(𝑥 − 𝑥0 ) + 𝑏(𝑦 − 𝑦0 ) + 𝑐(𝑧 − 𝑧0 ) = 0
〈𝑎, 𝑏, 𝑐〉 = 𝑎 𝑣𝑒𝑐𝑡𝑜𝑟 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒
(𝑥0 , 𝑦0 , 𝑧0 ) = 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒

,〈𝑎, 𝑏, 𝑐〉 = 〈−1, 3, −2〉
(𝑥0 , 𝑦0 , 𝑧0 ) = (0, 0, 0)



𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒:
𝑎(𝑥 − 𝑥0 ) + 𝑏(𝑦 − 𝑦0 ) + 𝑐(𝑧 − 𝑧0 ) = 0
−1(𝑥 − 0) + 3(𝑦 − 0) − 2(𝑧 − 0) = 0
−1𝑥 + 0 + 3𝑦 − 0 − 2𝑧 + 0 = 0
−𝑥 + 3𝑦 − 2𝑧 = 0


(1.3)

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑝𝑜𝑖𝑛𝑡 (−1, −2, 0) 𝑎𝑛𝑑 𝑝𝑙𝑎𝑛𝑒: 3𝑥 − 𝑦 + 4𝑧 = −2



𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑎 𝑝𝑜𝑖𝑛𝑡 (𝑢, 𝑣, 𝑤) 𝑎𝑛𝑑 𝑎 𝑝𝑙𝑎𝑛𝑒 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑑 = 0
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒(𝑎𝑢 + 𝑏𝑣 + 𝑐𝑤 + 𝑑)
𝑑=
√𝑎2 + 𝑏 2 + 𝑐 2


𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑝𝑜𝑖𝑛𝑡 (−1, −2, 0) 𝑎𝑛𝑑 𝑝𝑙𝑎𝑛𝑒: 3𝑥 − 𝑦 + 4𝑧 + 2 = 0
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒(3𝑥 − 𝑦 + 4𝑧 + 2)
𝑑=
√32 + (−1)2 + 42
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒(3(−1) − (−2) + 4(0) + 2)
𝑑=
√9 + 1 + 16
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒(−3 + 2 + 0 + 2)
𝑑=
√26
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒(1)
𝑑=
√26
1
𝑑=
√26

, Question 2



(2.1)

𝑈𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑣 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 𝑣 • 〈3, −1〉 = 0



𝐿𝑒𝑡 𝑣 = 〈𝑥, 𝑦〉 𝑥, 𝑦 ∈ ℝ
〈𝑥, 𝑦〉 • 〈3, −1〉 = 0

𝑥 × 3 + 𝑦 × −1 = 0
3𝑥 − 𝑦 = 0
3𝑥 = 𝑦
𝑦 = 3𝑥



𝑣 𝑖𝑠 𝑎 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟.
‖𝑣 ‖ = 1
‖〈𝑥, 𝑦〉‖ = 1

√𝑥 2 + 𝑦 2 = 1
2
(√𝑥 2 + 𝑦 2 ) = 12

𝑥2 + 𝑦2 = 1
𝑅𝑒𝑐𝑎𝑙𝑙 𝑡ℎ𝑎𝑡 𝑦 = 3𝑥

𝑥 2 + (3𝑥)2 = 1

𝑥 2 + 9𝑥 2 = 1

10𝑥 2 = 1
1
𝑥2 =
10

1
𝑥 = ±√
10

1
𝑥=±
√10

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