Exam (elaborations)
mat1503 exam pack
- Course
- Linear Algebra I (MAT1503)
- Institution
- University Of South Africa (Unisa)
Exam of 70 pages for the course Linear Algebra I at Unisa (mat1503 exam pack)
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MAT1503 ASSIGNMENT 5 2023written by
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MAT1503
ASSIGNMENT 5
2023
Downloaded by: mokonekgadiyamaake | njmaake111@gmail.com Want to earn
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QUESTION 1
Solution:
1.1).
Let: ⃗⃗⃗⃗
n1 be the normal of plane U and ⃗⃗⃗⃗
n2 be the normal of plane V.
n1 = 〈λ, 5, −2λ〉 and ⃗⃗⃗⃗
⃗⃗⃗⃗ n2 = 〈−λ, 1,2〉
a).
Planes U and V are orthogonal if ⃗⃗⃗⃗
n1 ∙ ⃗⃗⃗⃗
n2 = 0
n1 ∙ ⃗⃗⃗⃗
⃗⃗⃗⃗ n2 = 0
〈λ, 5, −2λ〉 ∙ 〈−λ, 1,2〉 = 0
−λ2 + 5 − 4λ = 0
λ2 − 5 + 4λ = 0
(λ − 1)(λ + 5) = 0
λ = 1 and λ = −5
b).
Planes U and V are orthogonal if ⃗⃗⃗⃗
n1 × ⃗⃗⃗⃗
n2 = 0
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n1 × ⃗⃗⃗⃗
⃗⃗⃗⃗ n2 = 0
i j k
|λ 5 −2λ| = 0
−λ 1 2
5 −2λ λ −2λ λ 5
i| |− j| | + k| | = 〈0,0,0〉
1 2 −λ 2 −λ 1
(10 + 2λ)i − (2λ − 2λ2 )j + (λ + 5λ)k = 〈0,0,0〉
10 + 2λ = 0 1 ⇒ λ = −5
2λ − 2λ2 = 0 2 ⇒ λ(2 − 2λ) = 0 ⇒ λ = 0 or λ = 1
λ + 5λ = 0 3 ⇒ 6λ = 0 ⇒ λ=0
We are getting different values of λ from different equation , so no such values of
λ exist.
1.2).
The normal of plane − x + 3y − 2z = 6 is 〈−1,3, −2〉
Since the plane that passes through the origin is parallel to the plane − x + 3y − 2z = 6
then they have the same normal vector.
Let: n
⃗ be the normal of the plane that passes through the origin.
Equation of plane: 〈x − x0 , y − y0 , z − z0 〉 ∙ n
⃗ =0
〈x − x0 , y − y0 , z − z0 〉 ∙ n
⃗ =0 ∴ 〈x0 , y0 , z0 〉 = 〈0,0,0〉 and n
⃗ = 〈−1,3, −2〉
〈x − 0, y − 0, z − 0〉 ∙ 〈−1,3, −2〉 = 0
〈x, y, z〉 ∙ 〈−1,3, −2〉 = 0
−x + 3y − 2z = 0 (equation of the plane passing through the origin)
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