MAT pure2&3 vectors

Question 1

What are the two forms of representing vectors?

Answer 1

1. Magnitude-direction form: (r, θ), where r is the magnitude and θ is the angle measured anticlockwise from the positive x-axis.
2. Component form: xi + yj, where i and j are unit vectors in the x and y directions.

Question 2

How do you convert a vector from magnitude-direction form to component form?

Answer 2

Given (r, θ), use:
x = r * cos(θ)
y = r * sin(θ)
The vector is x * i + y * j.

Question 3

Example: Convert (10, 70°) to component form.

Answer 3

x = 10 * cos(70°) = 3.42
y = 10 * sin(70°) = 9.40
Vector = 3.42 * i + 9.40 * j

Question 4

How do you calculate the magnitude of a vector given in component form?

Answer 4

For vector a = xi + yj:
|a| = sqrt(x^2 + y^2)

Question 5

Example: Find the magnitude of 5i + 3j.

Answer 5

|a| = sqrt(5^2 + 3^2) = sqrt(25 + 9) = sqrt(34)

Question 6

How do you convert a vector from component form to magnitude-direction form?

Answer 6

1. Magnitude: r = sqrt(x^2 + y^2)
2. Direction: θ = tan^(-1)(y / x)

Question 7

Example: Convert 5i + 3j to magnitude-direction form.

Answer 7

Magnitude = sqrt(5^2 + 3^2) = sqrt(34)
Direction = tan^(-1)(3 / 5) = 31.0°
Vector = (sqrt(34), 31.0°)

Question 8

What happens when you multiply a vector by a scalar?

Answer 8

Each component is multiplied by the scalar:
k * (xi + yj) = (k * x)i + (k * y)j

Question 9

How do you add or subtract vectors?

Answer 9

Add or subtract their respective components:
(x1i + y1j) + (x2i + y2j) = (x1 + x2)i + (y1 + y2)j

Question 10

What is a unit vector and how is it calculated?

Answer 10

A unit vector has a magnitude of 1. For vector a:
Unit vector = a / |a|

Question 11

Example: Find the unit vector in the direction of 2i - 3j.

Answer 11

|a| = sqrt(2^2 + (-3)^2) = sqrt(4 + 9) = sqrt(13)
Unit vector = (2/sqrt(13))i - (3/sqrt(13))j

Question 12

How do you calculate the magnitude of a 3D vector?

Answer 12

For vector a = xi + yj + zk:
|a| = sqrt(x^2 + y^2 + z^2)

Question 13

Example: Find the magnitude of vector 4i - 3j + 2k.

Answer 13

|a| = sqrt(4^2 + (-3)^2 + 2^2) = sqrt(16 + 9 + 4) = sqrt(29)

Question 14

How do you find the vector AB given points A(x1, y1, z1) and B(x2, y2, z2)?

Answer 14

AB = (x2 - x1)i + (y2 - y1)j + (z2 - z1)k

Question 15

Example: Find AB for points A(4, -1, 2) and B(-1, 3, 1).

Answer 15

AB = (-1 - 4)i + (3 - (-1))j + (1 - 2)k
AB = -5i + 4j - k
|AB| = sqrt((-5)^2 + 4^2 + (-1)^2) = sqrt(42)

Question 16

What is required to find the cartesian equation of a line?

Answer 16

• The coordinates of one point on the line.
• The gradient of the line (or the coordinates of a second point).

Question 17

What is required to find the vector equation of a line?

Answer 17

• The position vector of one point on the line.
• The direction vector of the line (or the position vector of a second point).

Question 18

What is the general vector equation of a line?

Answer 18

r = a + λd
Where:
- a is the position vector of a point on the line.
- d is the direction vector.
- λ is a parameter.

Question 19

Example: Find the vector equation of a line through A(4, -1) and parallel to vector (4, 2).

Answer 19

a = (4, -1)
d = (4, 2)
Simplified direction vector: (2, 1)
Vector equation: r = (4, -1) + λ(2, 1)

Question 20

How is the vector equation of a line joining points A and B written?

Answer 20

r = a + λ(b - a)
Where:
- a and b are the position vectors of points A and B.
- λ is a parameter.

Question 21

Example: Find the vector equation of a line joining A(5, -2) and B(1, 3).

Answer 21

a = (5, -2)
b = (1, 3)
Vector equation: r = (5, -2) + λ(-4, 5)

Question 22

What is the vector equation of a line in three dimensions?

Answer 22

r = a + λd
Where:
- a is the position vector of a point on the line.
- d is the direction vector.
- λ is a parameter.

Question 23

Example: Find the vector equation of the line through A(1, -2, -3) and B(0, -2, 1).

Answer 23

a = (1, -2, -3)
b = (0, -2, 1)
d = b - a = (-1, 0, 4)
Vector equation: r = (1, -2, -3) + λ(-1, 0, 4)

Question 24

How do you find the angle between two lines in three dimensions?

Answer 24

Scalar product: (3)(1) + (2)(0) + (-2)(-3) = 9
Magnitude of (3, 2, -2): sqrt(3^2 + 2^2 + (-2)^2) = sqrt(14)
Magnitude of (1, 0, -3): sqrt(1^2 + 0^2 + (-3)^2) = sqrt(10)
cos(θ) = 9 / (sqrt(14) * sqrt(10))
θ ≈ 40.5°

Question 25

How do you find the intersection of two lines in two dimensions?

Answer 25

Equate the vector equations of the two lines and solve for the parameters to find the common point.

Question 26

What are skew lines?

Answer 26

In three dimensions, skew lines are lines that do not intersect and are not parallel.

Question 27

How do you find the distance of a point from a line?

Answer 27

Find the perpendicular distance from the point to the line by identifying the foot of the perpendicular and calculating its length.

Question 28

What is the vector equation of a line in two dimensions?

Answer 28

The vector equation of a line is given by:
r = a + λd
Where:
- r is the position vector of a point on the line,
- a is the position vector of a known point on the line,
- d is the direction vector of the line,
- λ is a scalar parameter.

Question 29

Example: Find the vector equation of the line through A(4, -1) parallel to vector (4, 2).

Answer 29

1. Position vector of A: a = [4, -1].
2. Direction vector: d = [4, 2].
3. Equation:
   r = a + λd
   r = [4, -1] + λ[4, 2].

Question 30

How is the vector equation of a line in three dimensions written?

Answer 30

The general form is:
r = a + λd
Where:
- a is the position vector of a point on the line,
- d is the direction vector,
- λ is a scalar parameter.

Question 31

Example: Find the vector equation of the line joining A(1, 2, -3) and B(0, -2, 1)

Answer 31

1. Position vectors:
   a = [1, 2, -3], b = [0, -2, 1].
2. Direction vector:
   d = b - a = [0 - 1, -2 - 2, 1 + 3] = [-1, -4, 4].
3. Equation:
   r = a + λd
   r = [1, 2, -3] + λ[-1, -4, 4].

Question 32

How do you find the angle between two lines?

Answer 32

The angle θ between two lines is found using their direction vectors a and b:
cos(θ) = (a * b) / (|a| |b|)
Where:
- a * b is the scalar product,
- |a| and |b| are the magnitudes of the vectors.

Question 33

Example: Find the angle between lines with direction vectors [3, 1, -2] and [1, 0, -3]

Answer 33

1. Scalar product:
   a * b = (3)(1) + (1)(0) + (-2)(-3) = 3 + 0 + 6 = 9.
2. Magnitudes:
   |a| = √(3² + 1² + (-2)²) = √14,
   |b| = √(1² + 0² + (-3)²) = √10.
3. cos(θ) = 9 / (√14 * √10) = 9 / √140.
   θ ≈ 40.5°.

Question 34

What is the condition for two lines to intersect?

Answer 34

Two lines intersect if there exists a common position vector r that satisfies both equations. Solve simultaneously:
r₁ = a₁ + λd₁
r₂ = a₂ + μd₂

Question 35

Example: Find the intersection point of the lines:
r₁ = [5, 0] + λ[4, 1]
r₂ = [-2, 1] + μ[1, 3]

Answer 35

1. Solve:
   [5 + 4λ, λ] = [-2 + μ, 1 + 3μ].
   Equating components:
   5 + 4λ = -2 + μ,
   λ = 1 + 3μ.
2. Solve simultaneously:
   λ = 2, μ = 3.
3. Substitute into r₁:
   r = [5 + 4(2), 0 + 2] = [13, 2].
   Intersection point: (13, 2)

Question 36

What is the distance of a point from a line in 3D?

Answer 36

The shortest distance is the perpendicular distance:
d = |(p - a) × d| / |d|
Where:
- p is the position vector of the point,
- a is a point on the line,
- d is the direction vector of the line,
- × represents the vector cross product.

Question 37

Example: Find the distance of point P(2, 1, -2) from the line r = [1, 1, 2] + λ[2, -1, 3].

Answer 37

1. Direction vector: d = [2, -1, 3].
2. Vector from point to line: v = p - a = [2 - 1, 1 - 1, -2 - 2] = [1, 0, -4].
3. Cross product:
   v × d = [[i, j, k], [1, 0, -4], [2, -1, 3]] = [4, -11, -1].
4. Magnitude:
   |v × d| = √(4² + (-11)² + (-1)²) = √138.
   |d| = √(2² + (-1)² + 3²) = √14.
5. Distance:
   d = |v × d| / |d| = √138 / √14 ≈ 3.14