MAT1503 Assignment 4 Solutions 2026

MAT1503 Assignment 4 Solutions 2026
UNISA
ASSIGNMENT 04
Due date: Sunday, 31 August 2025
Total Marks: 40

,Question 1(a)

Given

𝑎⃗ = ⟨2,2,1⟩
𝑏⃗⃗ = ⟨1,0, −2⟩


The projection of 𝑎⃗onto 𝑏⃗⃗is

𝑎⃗ ⋅ 𝑏⃗⃗
proj⁡𝑏⃗⃗ 𝑎⃗ = 𝑏⃗⃗
∥ 𝑏⃗⃗ ∥2


Calculate the dot product:

𝑎⃗ ⋅ 𝑏⃗⃗ = (2)(1) + (2)(0) + (1)(−2)
= 2+0−2
=0


Therefore
0
proj⁡𝑏⃗⃗ 𝑎⃗ = 𝑏⃗⃗
12 + 02 + (−2)2
0
= 𝑏⃗⃗
5
= 0⃗
⃗


Hence

proj⁡𝑏⃗⃗ 𝑎⃗ = ⃗0⃗


Since 𝑤
⃗⃗⃗is supposed to have magnitude √12and be in the direction of this projection,
the statement is impossible because the zero vector has no direction.

, Question 1(b)

Plane 𝚷𝟏

Plane Π1 passes through the origin and is spanned by

𝑢
⃗⃗ = ⟨1,1,0⟩


and

𝑣⃗ = ⟨0,1,1⟩


A normal vector is

𝑛⃗⃗1 = 𝑢
⃗⃗ × 𝑣⃗
𝐢 𝐣 𝐤
=∣ 1 1 0 ∣
0 1 1
= 𝐢(1 ⋅ 1 − 0 ⋅ 1) − 𝐣(1 ⋅ 1 − 0 ⋅ 0) + 𝐤(1 ⋅ 1 − 1 ⋅ 0)
= ⟨1, −1,1⟩


Therefore the equation of Plane Π1 is

𝑥−𝑦+𝑧 =0




Plane 𝚷𝟐

Since

⃗⃗⃗ = ⃗0⃗
𝑤


the normal vector of Plane Π2 is the zero vector.

A plane cannot have a zero normal vector.

Therefore Plane Π2 is not defined.



Consequence

Since Plane Π2 cannot be determined,

• the line of intersection of Π1 and Π2 cannot be found;

• the shortest distance from that line to 𝐿cannot be calculated.