MAT1503 Assignment FOUR 2025

MAT1503 ASSIGNMENT 4 2025

Question 1

(1.1)

Normal vector for −𝑥 + 3𝑦 − 2𝑧 = 6 is 〈−1, 3, −2〉

Parallel vectors have the same normal vectors.

Equation of plane:

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

Where 〈𝑎, 𝑏, 𝑐 〉 is a normal vector and ( 𝑥0 , 𝑦0 , 𝑧0 ) is a point on the plane

−1(𝑥 − 0) + 3(𝑦 − 0) − 2(𝑧 − 0) = 0

−𝑥 + 3𝑦 − 2𝑧 = 0

(1.2)

The distance between a point (𝑢, 𝑣, 𝑤 ) and a plane 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑑 = 0 is:

|𝑎𝑢 + 𝑏𝑣 + 𝑐𝑤 + 𝑑 |
√𝑎2 + 𝑏2 + 𝑐 2

where | . | is the absolute value of.

3𝑥 − 𝑦 + 4𝑧 = −2

⇒ 3𝑥 − 𝑦 + 4𝑧 + 2 = 0

Point (−1, −2, 0)

Distance

|3 (−1) − 1( −2) + 4(0) + 2|
=
√(3) 2 + (−1)2 + (4) 2

| −3 + 2 + 0 + 2|
=
√9 + 1 + 16

, |1|
=
√26

1
=
√26


Question 2

(2.1)

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 ‖ = 0 then 𝜃 = 90° we have orthogonal/perpendicular vectors.
⃗ ‖‖𝑣⃗ ‖
𝑢

𝑢
⃗ ⋅𝑣⃗
(b) If ‖ 𝑢⃗‖‖𝑣⃗ ‖ = 1 then 𝜃 = 0° the vectors are positive multiples of each other, going the
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 < ‖ < 0 then 90° < 𝜃 < 180° (We have an obtuse angle)
⃗ ‖‖𝑣⃗‖
𝑢



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


⃗ = 〈−1, 1, 0, −1〉, 𝑣 = 〈1, −1, 3, −2〉
𝑢

Let 𝜃 be the angle between the two above vectors.

𝑢
⃗ ⋅𝑣
cos 𝜃 =
‖𝑢
⃗ ‖ ‖𝑣 ‖

−1 × 1 + 1 × −1 + 0 × 3 − 1 × −2
cos 𝜃 =
√(−1) 2 + (1)2 + 02 + (−1) 2 √12 + ( −1)2 + 32 + (−2)2

, −1 − 1 + 0 + 2
cos 𝜃 =
√1 + 1 + 0 + 1√1 + 1 + 9 + 4


0
cos 𝜃 =
√3√15

cos 𝜃 = 0

𝜃 = 90°

The above two vectors are perpendicular.

(2.2)

3
𝑟 = 〈0, −1, −2, 〉
4

‖𝑟‖


3 2
= √ 02 + (−1)2 + (−2) 2 + ( )
4


9
= √0 + 1 + 4 +
16


9
= √5 +
16


5 9
=√ +
1 16


80 9
=√ +
16 16

89
=√
16

,Direction cosines:

0
cos 𝜃1 =
√89
16

−1
cos 𝜃2 =
√89
16

−2
cos 𝜃3 =
√89
16

3
(4)
cos 𝜃4 =
√89
16

cos 𝜃1 = 0


16
cos 𝜃2 = −√
89


16
cos 𝜃3 = −2√
89


3 16
cos 𝜃4 = ( ) √
4 89

cos 𝜃1 = 0

4
cos 𝜃2 = −
√89

4
cos 𝜃3 = −2
√89

3 4
cos 𝜃4 = ( )
4 √89

,cos 𝜃1 = 0

4
cos 𝜃2 = −
√89

8
cos 𝜃3 = −
√89

3
cos 𝜃4 =
√89

Direction angles

𝜃1 = 90°

4
𝜃2 = 180° − cos −1 ( )
√89

8
𝜃3 = 180° − cos −1 ( )
√89

3
𝜃4 = cos −1 ( )
√89

(2.3)

1
𝑟(𝑡) = 〈𝑡, − , 𝑡 2 − 2〉
𝑡

𝑟(𝑡) = 〈𝑡, −𝑡 −1 , 𝑡 2 − 2〉

𝑑
𝑟(𝑡) = 〈1, −1 × −1𝑡 −1−1 , 2𝑡 − 0〉
𝑑𝑡

𝑑
𝑟(𝑡) = 〈1, 𝑡 −2 , 2𝑡〉
𝑑𝑡

When 𝑡 = 1

𝑑
𝑟(1) = 〈1, (1)−2 , 2(1)〉
𝑑𝑡

𝑑
𝑟(1) = 〈1, 1,2〉
𝑑𝑡

, 𝑑 𝑑 𝑑
(𝑉(𝑡) ⋅ 𝑟(𝑡) ) = (𝑉(𝑡) ) ⋅ 𝑟(𝑡) + 𝑉(𝑡) ⋅ (𝑟(𝑡))
𝑑𝑡 𝑑𝑡 𝑑𝑡


𝑑 𝑑
(𝑉(𝑡) ⋅ 𝑟(𝑡) ) = 𝑉 ′ (𝑡) ⋅ 𝑟(𝑡) + 𝑉(𝑡) ⋅ (𝑟(𝑡) )
𝑑𝑡 𝑑𝑡

When 𝑡 = 1

𝑑 𝑑
(𝑉(𝑡) ⋅ 𝑟(𝑡) ) = 𝑉 ′ (1) ⋅ 𝑟(1) + 𝑉(1) ⋅ (𝑟(1))
𝑑𝑡 𝑑𝑡

1
𝑟(𝑡) = 〈𝑡, − , 𝑡 2 − 2〉
𝑡

1
𝑟(1) = 〈1, − , (1) 2 − 2〉
1

𝑟(1) = 〈1, −1, −1〉

𝑑
(𝑉(𝑡) ⋅ 𝑟(𝑡) ) = 〈1, −2,2〉 ⋅ 〈1, −1, −1〉 + 〈−1, 1, −2〉 ⋅ 〈1, 1,2〉
𝑑𝑡

𝑑
(𝑉(𝑡) ⋅ 𝑟(𝑡) ) = 1 × 1 − 2 × −1 + 2 × −1 + ( −1 × 1 + 1 × 1 − 2 × 1)
𝑑𝑡

𝑑
(𝑉(𝑡) ⋅ 𝑟(𝑡) ) = 1 + 2 − 2 + (−1 + 1 − 2)
𝑑𝑡

𝑑
(𝑉(𝑡) ⋅ 𝑟(𝑡) ) = 1 + (−2)
𝑑𝑡

𝑑
(𝑉(𝑡) ⋅ 𝑟(𝑡) ) = 1 − 2
𝑑𝑡

𝑑
(𝑉(𝑡) ⋅ 𝑟(𝑡) ) = −1
𝑑𝑡