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MAT1503 Assignment 04 Solutions ( 20 August 2024 ) R50,00   Add to cart
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MAT1503 Assignment 04 Solutions ( 20 August 2024 )
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The document provided contains accurate and comprehensive solutions for Assignment 04 of Linear Algebra 1 (MAT1503) Module offered by the University of South Africa in 2024.

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  • August 7, 2024
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  • 2024/2025
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  • mat1503 linear algebra 1 assignment 04 solutions

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ASSIGNMENT 04
SOLUTIONS 2024


DUE DATE : 20 AUGUST 2024

TOTAL MARKS : 90 (10 QUESTIONS)




This Document has been compiled by The SST Academy -Science Department Team.

,QUESTION 1




Solution:


𝑛⃗ ∙ 𝑢
⃗ = 𝑥 + 𝑦 − 2𝑧
𝑥 + 𝑦 − 2𝑧 = 0 “orthogonal”


𝑛⃗ ∙ 𝑣 = −𝑥 + 2𝑦
1
−𝑥 + 2𝑦 = 0 → 𝑥 = 𝑦 “orthogonal”
2



𝑛⃗ ∙ 𝑤
⃗⃗ = −𝑥 + 𝑧
−𝑥 + 𝑧 = 0 → 𝑥 = 𝑧 “orthogonal”


1
𝑥 + 𝑦 − 2𝑧 = 0 → 𝑥 + 𝑥 − 2𝑥 = 0 → 𝑥 = 0 ✓
2
∴𝑦=0 𝑧=0
𝑛⃗ =< 0,0,0 > ✓ (2)




Solution:


Let orthogonal vector = 𝑤
⃗⃗ =< 𝑎, 𝑏, 𝑐 >
Using Cross Product:
𝑖 𝑗 𝑘
𝑤 ⃗ × 𝑣 = |0 −1 −1| = (1 − 0)𝑖 − (0 + 1)𝑗 + (0 + 1)𝑘 = 1𝑖 − 1𝑗 + 1𝑘
⃗⃗ = 𝑢
1 0 −1
→𝑤 ⃗ × 𝑣 =< 1, −1,1 > ✓
⃗⃗ = 𝑢
Unit vector :
1
𝑤
̂= 𝑢
⃗ ×𝑣
‖𝑢
⃗ × 𝑣‖
1

, 1 1 1 1 1
𝑤
̂= < 1, −1,1 > = < 1, −1,1 > = < ,− , > ✓
√(1)2 + (−1)2 + (1)2 √3 √3 √3 √3

(2)

QUESTION 2




Solution:

𝑖 𝑗 𝑘
(𝑣 × 𝑤
⃗⃗ ) = |−3 2 −1| = (10 + 3)𝑖 − (−15 + 1)𝑗 + (−9 − 2)𝑘 = 13𝑖 + 14𝑗 − 11𝑘
1 3 5

⃗⃗ ) = < 13,14, −11 > ✓
→ (𝑣 × 𝑤

𝑖 𝑗 𝑘
⃗ × (𝑣 × 𝑤
𝑢 ⃗⃗ ) = |−2 1 −1 | = (−11 + 14)𝑖 − (22 + 13)𝑗 + (−28 − 13)𝑘✓
13 14 −11

= 3𝑖 − 35𝑗 − 41𝑘

⃗⃗ ) = < 3, −35, −41 > ✓
⃗ × (𝑣 × 𝑤
∴𝑢

𝑖 𝑗 𝑘
(𝑢 ⃗⃗ ) = |−2 1
⃗ ×𝑤 −1| = (5 + 3)𝑖 − (−10 + 1)𝑗 + (−6 − 1)𝑘 = 8𝑖 + 9𝑗 − 7𝑘
1 3 5

→ (𝑢 ⃗⃗ ) = < 8,9, −7 > ✓
⃗ ×𝑤

𝑖 𝑗 𝑘
(𝑢 ⃗⃗ ) × 𝑣 = | 8
⃗ ×𝑤 9 −7| = (−9 + 14)𝑖 − (−8 − 21)𝑗 + (16 + 27)𝑘✓
−3 2 −1

= 5𝑖 + 29𝑗 + 43𝑘
1
∴ (𝑢 ⃗⃗ ) × 𝑣 = < 5,29,43 > ✓
⃗ ×𝑤 6 × = (3)
2




2

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