Mastering Vector Operations: Solving IB SL Math Vector Questions with Step-by-Step Solutions
Example 1: Consider two vectors, u = (3, 1) and v = (-2, 4).
a) Find the magnitude of vector u.
b) Find the unit vector in the direction of vector v.
c) Find the dot product of vectors u and v.
d) Find the angle between vectors u and v.
Let's solve it step by step:
a) The magnitude of a vector v = (v₁, v₂) is given by ∥v∥ = √(v₁² + v₂²).
For vector u, the magnitude is:
∥u∥ = √(3² + 1²) = √(9 + 1) = √10.
Therefore, the magnitude of vector u is √10.
b) The unit vector in the direction of a vector v = (v₁, v₂) is given by u = (v₁/∥v∥, v₂/∥v∥).
For vector **v**, the unit vector is:
u = (-2/∥v∥, 4/∥v∥) = (-2/√(4 + 16), 4/√(4 + 16)) = (-2/√20, 4/√20).
Therefore, the unit vector in the direction of vector v is (-2/√20, 4/√20).
c) The dot product of two vectors u = (u₁, u₂) and v = (v₁, v₂) is given by u · v = u₁v₁ + u₂v₂.
For vectors u and v, the dot product is:
u· v = (3)(-2) + (1)(4) = -6 + 4 = -2.
Therefore, the dot product of vectors u and v is -2.
d) The angle between two vectors u and v can be found using the formula:
cos θ = (u· v) / (∥u∥ ∥v∥).
Plugging in the values we found earlier:
cos θ = -2 / (√10 √20).
To find the angle θ, we can take the inverse cosine (arccos) of the value obtained above:
θ = arccos(-2 / (√10 √20)).
Using a calculator, we find that θ ≈ 128.49 degrees.
Therefore, the angle between vectors u and v is approximately 128.49 degrees.
Example 2: Consider three vectors, a = (2, -1), b = (-3, 4), and c = (1, 3).
a) Calculate the vector sum a + b.
b) Find the scalar product of b and c.
c) Determine the vector d such that c = a + d.
Let's solve it step by step:
a) The vector sum of two vectors a = (a₁, a₂) and b = (b₁, b₂) is given by a + b = (a₁ + b₁, a₂ + b₂).
For vectors a and b, the vector sum is:
a + b = (2 + (-3), -1 + 4) = (-1, 3).
Therefore, the vector sum of a and b is (-1, 3).
b) The scalar product (also known as the dot product) of two vectors a = (a₁, a₂) and b = (b₁, b₂) is given by a · b = a₁b₁ + a₂b₂.
For vectors b and c, the scalar product is:
b · c = (-3)(1) + (4)(3) = -3 + 12 = 9.
Therefore, the scalar product of b and c is 9.
c) To determine the vector d, we can rearrange the equation c = a + d to solve for d.
c = a + d can be rewritten as d = c - a.
Substituting the given values:
d = (1, 3) - (2, -1) = (1 - 2, 3 - (-1)) = (-1, 4).
Therefore, the vector d is (-1, 4).
Example 3: Consider two vectors, u = (4, -2, 1) and v = (-1, 3, 5).
a) Calculate the cross product of vectors u and v.
b) Find the angle between vectors u and v.
c) Determine whether the vectors u and v are orthogonal (perpendicular) to each other.
Let's solve it step by step:
a) The cross product of two vectors u = (u₁, u₂, u₃) and v = (v₁, v₂, v₃) is given by u × v = (u₂v₃ - u₃v₂, u₃v₁ - u₁v₃, u₁v₂ - u₂v₁).
For vectors u and v, the cross product is:
u × v = ( (-2)(5) - (1)(3), (1)(-1) - (4)(5), (4)(3) - (-2)(-1) ) = ( -16, -21, 14 ).
Therefore, the cross product of vectors u and v is (-16, -21, 14).
b) The angle between two vectors u and v can be found using the formula:
cos θ = (u · v) / (∥u∥ ∥v∥).
The dot product of u and v is:
u · v = (4)(-1) + (-2)(3) + (1)(5) = -4 - 6 + 5 = -5.
The magnitude of u is:
∥u∥ = √(4² + (-2)² + 1²) = √(16 + 4 + 1) = √21.
The magnitude of v is:
∥v∥ = √((-1)² + 3² + 5²) = √(1 + 9 + 25) = √35.
Using these values, we can find the angle θ:
θ = arccos(-5 / (√21 √35)).
Using a calculator, we find the value of θ to be approximately 109.47 degrees.
Therefore, the angle between vectors u and v is approximately 109.47 degrees.
c) Two vectors are orthogonal (perpendicular) to each other if their dot product is zero. Let's calculate the dot product of u and v:
u · v = (4)(-1) + (-2)(3) + (1)(5) = -4 - 6 + 5 = -5.
Since the dot product is not zero (-5 ≠ 0), the vectors u and v are not orthogonal.
