MAT3701 Assignment 1 solutions 2026

MAT3701 Assignment 1 Solutions 2026
All questions are answered clearly and in full.


MAT3701 - Linear Algebra III ASSIGNMENT 01
Opens: 2 March 2026 Due: Friday, 30 April 2026
Instructions for the Assignment
(1) Carefully explain all your arguments.
(2) Only hand written PDF files will be accepted.
(3) Late submissions will not be marked.
(4) Write your name, surname and student number on the first
page.

,MAT3701 – Assignment 01 Solutions

(Write these neatly by hand)



Question 1

1.1 Equation of the line through (2,4,0) and (−3,−6,0)

Find the direction vector

A line through two points has direction

𝑑⃗ = 𝑃2 − 𝑃1
𝑃1 = (2,4,0), 𝑃2 = (−3, −6,0)
𝑑⃗ = (−3 − 2, −6 − 4,0 − 0)
𝑑⃗ = (−5, −10,0)


Parametric equation of a line

The vector form is

𝑟 = 𝑟0 + 𝑡𝑑

where

𝑟0 = (2,4,0)

Thus

(𝑥, 𝑦, 𝑧) = (2,4,0) + 𝑡(−5, −10,0)




Component form

𝑥 = 2 − 5𝑡
𝑦 = 4 − 10𝑡
𝑧=0


Equation of the line

(𝑥, 𝑦, 𝑧) = (2,4,0) + 𝑡(−5, −10,0)


or

𝑥 = 2 − 5𝑡, 𝑦 = 4 − 10𝑡, 𝑧 = 0

, 1.2 Equation of the plane through three points

Points:

𝐴(3, −6,7), 𝐵(−2,0, −4), 𝐶(5, −9, −2)


Find vectors in the plane

⃗⃗⃗⃗⃗⃗ = 𝐵 − 𝐴
𝐴𝐵
= (−2 − 3,0 + 6, −4 − 7)
= (−5,6, −11)


⃗⃗⃗⃗⃗⃗ = 𝐶 − 𝐴
𝐴𝐶
= (5 − 3, −9 + 6, −2 − 7)
= (2, −3, −9)



Find the normal vector

𝑛 = 𝐴𝐵 × 𝐴𝐶
𝑖 𝑗 𝑘
𝑛 =∣ −5 6 −11 ∣
2 −3 −9


Compute:

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)


Plane equation

𝑎(𝑥 − 𝑥0 ) + 𝑏(𝑦 − 𝑦0 ) + 𝑐(𝑧 − 𝑧0 ) = 0


Using point 𝐴(3, −6,7)

−87(𝑥 − 3) − 67(𝑦 + 6) + 3(𝑧 − 7) = 0