MAT3701 Assignment 1 Complete Solutions UNISA 2026 LINEAR ALGEBRA III

MAT3701
ASSIGNMENT 1
Linear Algebra III

FULL SOLUTIONS
COMPLETE SOLUTIONS

MEMORANDUM
UNISA 2026

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,SOLUTIONS:
Question 1

(1.1) The line passes through (2, 4, 0) and (−3, −6,0). A direction vector is

𝐝 = (−3 − 2, −6 − 4,0 − 0) = (−5, −10,0).



Since (2, 4, 0) lies on the line, the vector equation is

(𝑥, 𝑦, 𝑧) = (2,4,0) + 𝑡(−5, −10,0), 𝑡 ∈ ℝ.



The parametric equations are 𝑥 = 2 − 5𝑡, 𝑦 = 4 − 10𝑡, 𝑧 = 0.

(1.2) Let the points be 𝐴(3, −6,7), 𝐵(−2,0, −4), and 𝐶(5, −9, −2). Vectors in the plane are

⃗⃗⃗⃗⃗
𝐴𝐵 = (−5,6, −11), ⃗⃗⃗⃗⃗
𝐴𝐶 = (2, −3, −9).



A normal vector 𝐧 is ⃗⃗⃗⃗⃗ 𝐴𝐶 :
𝐴𝐵 × ⃗⃗⃗⃗⃗



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, 𝐢 𝐣 𝐤
𝐧 =∣ −5 6 −11 ∣
2 −3 −9
= 𝐢(6(−9) − (−11)(−3)) − 𝐣((−5)(−9) − (−11)(2)) + 𝐤((−5)(−3) − 6(2)).
= 𝐢(−54 − 33) − 𝐣(45 + 22) + 𝐤(15 − 12) = (−87, −67,3).



Using point 𝐴(3, −6,7), the plane equation is

−87(𝑥 − 3) − 67(𝑦 + 6) + 3(𝑧 − 7) = 0.



Simplifying:

−87𝑥 + 261 − 67𝑦 − 402 + 3𝑧 − 21 = 0 ⟹ − 87𝑥 − 67𝑦 + 3𝑧 − 162 = 0.



So the plane is 87𝑥 + 67𝑦 − 3𝑧 = −162.




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