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
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
