FMAT pure1 vectors incompl

Question 1

What is the vector equation of a plane?

Answer 1

The vector equation of a plane is given by:
(r - a) * n = 0
or
r * n = a * n
Where:
- r is the position vector of any point on the plane,
- a is the position vector of a known point on the plane,
- n is the normal vector to the plane.

Question 2

How is the Cartesian equation of a plane derived?

Answer 2

The vector equation:
r * n = a * n
is rewritten as:
n₁x + n₂y + n₃z = d
Where:
- n₁, n₂, n₃ are the components of the normal vector,
- d = a * n.

Question 3

Example: Find the Cartesian equation of the plane passing through (4, 5, -2) with normal vector [2, -1, 3]

Answer 3

1. Substitute normal vector: 2x - y + 3z = d.
2. Use point (4, 5, -2):
   2(4) - 1(5) + 3(-2) = d
   8 - 5 - 6 = -3.
3. Equation: 2x - y + 3z = -3.

Question 4

How do you verify if a point lies on a plane?

Answer 4

Substitute the coordinates of the point into the plane’s equation. If the equation holds true, the point lies on the plane.

Question 5

Example: Verify if point (2, 4, -1) lies on the plane 2x - y + 3z = -3

Answer 5

1. Substitute: 2(2) - 4 + 3(-1) = -3.
   4 - 4 - 3 = -3.
2. The equation holds, so the point lies on the plane.

Question 6

What is another vector form of the equation of a plane?

Answer 6

If b and c are non-parallel vectors lying within the plane and a is a point on the plane, the equation is:
r = a + λb + μc
Where:
- λ and μ are scalar parameters.

Question 7

Example: Find the vector equation of the plane containing points P(2, 3, -1), Q(1, 0, 4), and R(-1, 1, 1).

Answer 7

1. Find vectors in the plane:
   PQ = [1 - 2, 0 - 3, 4 - (-1)] = [-1, -3, 5],
   PR = [-1 - 2, 1 - 3, 1 - (-1)] = [-3, -2, 2].
2. Equation:
   r = [2, 3, -1] + λ[-1, -3, 5] + μ[-3, -2, 2].

Question 8

How can the vector form of a plane be converted to Cartesian form?

Answer 8

1. Expand r = a + λb + μc into component equations.
2. Eliminate λ and μ to obtain the Cartesian equation.

Question 9

Example: Convert the plane
r = [2, 3, -1] + λ[-1, -3, 5] + μ[-3, -2, 2]
to Cartesian form.

Answer 9

1. Expand:
   x = 2 - λ - 3μ,
   y = 3 - 3λ - 2μ,
   z = -1 + 5λ + 2μ.
2. Eliminate λ and μ:
   Combine equations and solve for x, y, z:
   4x + 13y + 7z = 24.

Question 10

What is the significance of the normal vector in a plane equation?

Answer 10

The normal vector determines the orientation of the plane in space. Parallel planes share the same normal vector.

Question 11

How can you find the equation of a plane given three points?

Answer 11

1. Use the points to find two vectors lying within the plane.
2. Take the cross product of these vectors to find the normal vector.
3. Use the normal vector and one of the points to find the equation of the plane.

Question 12

What is the relationship between parallel planes?

Answer 12

Parallel planes have the same normal vector but different values of d in the Cartesian equation n₁x + n₂y + n₃z = d.

Question 13

What are the three possible relationships between a line and a plane?

Answer 13

1. The line and the plane are parallel, so there are no intersections.
2. The line lies in the plane, so all points on the line are in the plane.
3. The line intersects the plane at a single point.

Question 14

How can you determine if a line is parallel to a plane?

Answer 14

If the direction vector of the line is perpendicular to the normal vector of the plane, the line is parallel to the plane. Test a point on the line to check if it lies in the plane.

Question 15

Example: Find the intersection of the line
r = [2, 3, -4] + λ[-1, 2, 1]
and the plane 2x - 3y + z = 5.

Answer 15

1. Parametrize the line:
   x = 2 - λ, y = 3 + 2λ, z = -4 + λ.
2. Substitute into the plane equation:
   2(2 - λ) - 3(3 + 2λ) + (-4 + λ) = 5.
3. Solve for λ:
   4 - 2λ - 9 - 6λ - 4 + λ = 5.
   -7λ = 14 → λ = -2.
4. Substitute λ = -2 into the line:
   x = 4, y = -1, z = -6.
   Intersection point: (4, -1, -6).

Question 16

How do you find the shortest distance from a point to a plane?

Answer 16

The shortest distance d from a point P(x₁, y₁, z₁) to a plane ax + by + cz + d = 0 is:
d = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²).

Question 17

Example: Find the distance from P(2, 2, -4) to the plane 2x - y + 3z + 6 = 0.

Answer 17

1. Use the distance formula:
   d = |2(2) - 1(2) + 3(-4) + 6| / √(2² + (-1)² + 3²).
2. Simplify:
   d = |4 - 2 - 12 + 6| / √(4 + 1 + 9).
   d = |-4| / √14.
   d = 4 / √14 ≈ 1.07

Question 18

How do you find the angle between a line and a plane?

Answer 18

The angle α between a line and a plane is related to the angle θ between the direction vector d of the line and the normal vector n of the plane:
α = 90° - θ

Question 19

Example: Find the angle between the line
r = [2, -3, 1] + λ[1, 2, -1]
and the plane x - y + z = 3.

Answer 19

1. Direction vector of line: d = [1, 2, -1].
2. Normal vector of plane: n = [1, -1, 1].
3. cos(θ) = (d * n) / (|d| |n|):
   d * n = (1)(1) + (2)(-1) + (-1)(1) = -2.
   |d| = √(1² + 2² + (-1)²) = √6,
   |n| = √(1² + (-1)² + 1²) = √3.
   cos(θ) = -2 / (√6 √3) = -√2 / 3.
4. θ = cos⁻¹(-√2 / 3) ≈ 125.26°.
5. α = 90° - 125.26° = 54.74°.

Question 20

How do you find the line of intersection of two planes?

Answer 20

1. Solve the plane equations as simultaneous equations to express x, y, and z in terms of a parameter λ.
2. Use the parameterized equations to write the line’s vector equation.

Question 21

Example: Find the line of intersection for planes 2x + y - z = 4 and x - y + 2z = 1

Answer 21

1. Solve for z in terms of x and y:
   From the second equation: z = (1 - x + y) / 2.
2. Substitute into the first equation:
   2x + y - (1 - x + y) / 2 = 4.
   Multiply through by 2: 4x + 2y - (1 - x + y) = 8.
   Simplify: 5x + y = 9 → y = 9 - 5x.
3. Parameterize:
   Let x = λ, y = 9 - 5λ, z = (1 - λ + (9 - 5λ)) / 2.
   z = (10 - 6λ) / 2 = 5 - 3λ.
4. Equation of the line:
   r = [0, 9, 5] + λ[1, -5, -3]

Question 22

What is the vector product (cross product)?

Answer 22

The vector product of vectors a and b, written as a × b, is:
a × b = |a||b|sin(θ)n̂
Where:
- θ is the angle between a and b,
- n̂ is a unit vector perpendicular to both a and b.
The result is a vector, not a scalar.

Question 23

How is the vector product calculated in component form?

Answer 23

a × b = det([[i, j, k], [a₁, a₂, a₃], [b₁, b₂, b₃]])
Where:
- a₁, a₂, a₃ are components of a,
- b₁, b₂, b₃ are components of b.

Question 24

Example: Compute a × b for a = [2, 1, -2] and b = [1, 3, 4]

Answer 24

1. Use the determinant formula:
   a × b = [[i, j, k], [2, 1, -2], [1, 3, 4]].
2. Expand:
   = i((1)(4) - (3)(-2)) - j((2)(4) - (1)(-2)) + k((2)(3) - (1)(1)).
   = i(4 + 6) - j(8 + 2) + k(6 - 1).
   = [10, -10, 5]

Question 25

What are the key properties of the vector product?

Answer 25

1. Anti-commutative: a × b = -b × a.
2. Distributive: a × (b + c) = a × b + a × c.
3. Scalar multiples: (ka) × b = k(a × b).
4. a × a = 0 (a vector crossed with itself is zero).

Question 26

Example: Show that a × b = -b × a for a = [2, -3, 1] and b = [1, -1, 2]

Answer 26

1. Compute a × b:
   a × b = [[i, j, k], [2, -3, 1], [1, -1, 2]] = [5, 3, 1].
2. Compute b × a:
   b × a = [[i, j, k], [1, -1, 2], [2, -3, 1]] = [-5, -3, -1].
3. Result: a × b = -b × a.

Question 27

How is the vector product used to find a vector perpendicular to two vectors?

Answer 27

The result of a × b is always perpendicular to both a and b.

Question 28

Example: Find a vector perpendicular to p = [2, -3, -1] and q = [4, 2, -3].

Answer 28

1. Compute p × q:
   p × q = [[i, j, k], [2, -3, -1], [4, 2, -3]].
   = [11, 10, 16].
2. Result: [11, 10, 16] is perpendicular to both p and q.

Question 29

How is the vector product used to find the equation of a plane?

Answer 29

1. Use two vectors in the plane to compute their vector product (a × b).
2. The result is the normal vector to the plane.
3. Use the normal vector and a point in the plane to find the plane’s equation.

Question 30

Example: Find the equation of the plane through A(2, 1, 3), B(0, 1, -4), and C(5, 3, -2).

Answer 30

1. Compute AB and AC:
   AB = [0 - 2, 1 - 1, -4 - 3] = [-2, 0, -7],
   AC = [5 - 2, 3 - 1, -2 - 3] = [3, 2, -5].
2. Compute AB × AC:
   AB × AC = [[i, j, k], [-2, 0, -7], [3, 2, -5]] = [14, -1, -4].
3. Equation of the plane:
   14x - y - 4z = d.
4. Substitute A(2, 1, 3):
   14(2) - 1(1) - 4(3) = d → d = 15.
5. Plane equation: 14x - y - 4z = 15.

Question 31

How is the vector product used to find the line of intersection of two planes?

Answer 31

1. Use the vector product of the normal vectors of the planes to find the direction vector.
2. Solve the plane equations simultaneously to find a point on the line.
3. Combine the direction vector and point to form the line’s equation.

Question 32

Example: Find the line of intersection of planes 2x + y - z = 4 and x - y + 2z = 1.

Answer 32

1. Normal vectors: n₁ = [2, 1, -1], n₂ = [1, -1, 2].
2. Compute n₁ × n₂:
   n₁ × n₂ = [[i, j, k], [2, 1, -1], [1, -1, 2]] = [1, -5, -3].
3. Solve plane equations with x = 0:
   2(0) + y - z = 4, 0 - y + 2z = 1 → y = -3, z = -7.
4. Line equation:
   r = [0, -3, -7] + λ[1, -5, -3]