MAT1503 Assignment THREE 2025

MAT1503 ASSIGNMENT 3 2025

Question 1

The angle, 𝜃 between two vectors is such that 0° ≤ 𝜃 ≤ 180°.
For vectors 𝑢
⃗ , 𝑣 be vectors, then

𝑢⃗ ⋅𝑣
cos(𝜃 ) =
‖𝑢
⃗ ‖‖𝑣 ‖

𝑢
⃗ ⋅𝑣⃗
(a) If ‖ 𝑢⃗‖‖𝑣⃗ ‖ = 0 then 𝜃 = 90° we have orthogonal/perpendicular vectors.

𝑢
⃗ ⋅𝑣⃗
(b) If ‖ = 1 then 𝜃 = 0° the vectors are positive multiples of each other, going the
⃗ ‖‖𝑣⃗ ‖
𝑢
same direction.

𝑢
⃗ ⋅𝑣⃗
(c) If ‖ 𝑢⃗‖‖𝑣⃗ ‖ = −1 then 𝜃 = 180° the vectors are negative multiples of each other, going
opposite directions.

𝑢
⃗ ⋅𝑣⃗
(d) −1 < ‖𝑢⃗‖‖𝑣⃗‖ < 0 then 90° < 𝜃 < 180° (We have an obtuse angle)

𝑢
⃗ ⋅𝑣⃗
(e) 0 < ‖𝑢⃗‖‖𝑣⃗‖ < 1 then 0° < 𝜃 < 90° (We have an acute angle)


(1.1)

⃗ = 〈1, −3, −1〉, 𝑎 = 〈2, 1, −1〉
𝑢

𝑢⃗ ⋅𝑎
cos(𝜃 ) =
‖𝑢
⃗ ‖‖𝑎‖

〈1, −3, −1〉 ⋅ 〈2, 1, −1〉
cos(𝜃 ) =
√12 + ( −3)2 + (−1)2 √22 + 12 + (−1) 2

1 × 2 + (−3) × 1 + ( −1) × −1
cos(𝜃 ) =
√1 + 9 + 1√4 + 1 + 1

2−3+1
cos(𝜃 ) =
√11√6

, 0
cos(𝜃 ) =
√66

cos(𝜃 ) = 0

𝜃 = 90°

The vectors are orthogonal, so there is neither an acute nor an obtuse angle between them.

(1.2)

𝑣 = 〈1, 0, −2〉, 𝑏⃗ = 〈2, − 1, −1〉

𝑣 ⋅ 𝑏⃗
cos(𝜃 ) =
‖𝑣 ‖‖𝑏⃗‖

〈1, 0, −2〉 ⋅ 〈2, − 1, −1〉
cos(𝜃 ) =
√12 + 02 + ( −2)2 √22 + (−1)2 + (−1) 2

1 × 2 + 0 × −1 + (−2) × (−1)
cos(𝜃 ) =
√1 + 0 + 4√4 + 1 + 1

2−0+2
cos(𝜃 ) =
√5√6

4
cos(𝜃 ) =
√5 × 6

4
cos(𝜃 ) =
√30

0 < 16 < 36

√0 < √16 < √36

0 < 4 < √36

0 4 √36
< <
√36 √36 √36

4
0< <1
√30

, 𝑢⃗ ⋅𝑣
0< <1
‖𝑢
⃗ ‖‖𝑣‖

So, the angle between 𝑣 and 𝑏⃗ is acute.

(1.3)

𝑣 = 〈1, 0, −2〉, 𝑏⃗ = 〈2, − 1, −1〉
𝑣 ⋅ 𝑏⃗
proj𝑏⃗ 𝑣 = ( ⃗
2) 𝑏
‖𝑏⃗‖

〈1, 0, −2〉 ⋅ 〈2, − 1, −1〉
proj𝑏⃗ 𝑣 = ( ) 〈2, − 1, −1〉
22 + (−1)2 + ( −1)2

2−0+2
proj𝑏⃗ 𝑣 = ( ) 〈2, − 1, −1〉
4+1+1

4
proj𝑏⃗ 𝑣 = 〈2, − 1, −1〉
6

2
proj𝑏⃗ 𝑣 = 〈2, − 1, −1〉
3

2 2 2
proj𝑏⃗ 𝑣 = 〈2 × , − 1 × , −1 × 〉
3 3 3

4 2 2
proj𝑏⃗ 𝑣 = 〈 , − , − 〉
3 3 3