Find angles between a vector and b vector if a vector + b vector + c vector = 0, a vector magnitude = 3, b vector magnitude = 5 and c vector magnitude = 7
clearly, a,b,c form a triangle.
The law of cosines says that the angle θ is given by
3^2 + 5^2 - 2*3*5 cosθ = 7^2
More likely, you want pi-θ
To find the angles between the vectors, we can use the dot product formula:
a · b = |a| |b| cosθ
Given that a + b + c = 0, we can rewrite it as:
c = -a - b
Now, let's find the angle between a and b:
a · b = |a| |b| cosθ
a · b = (-a - b) · b
Using the properties of the dot product:
a · b = -a · b - b · b
Simplifying:
a · b + a · b = 0
2(a · b) = 0
Since the dot product is the sum of the products of the corresponding components of the vectors, if the dot product is zero, it means the vectors are orthogonal and the angle between them is 90 degrees. Therefore, the angle between vector a and vector b is 90 degrees.
Now, let's find the angle between vector b and vector c:
c = -a - b
|c| = |-a - b|
|c| = sqrt((-a - b) · (-a - b))
|c| = sqrt(a · a + b · b + 2(a · b))
Using the dot product formula:
|c| = sqrt(|a|^2 + |b|^2 + 2(a · b))
Substituting the given magnitudes:
7 = sqrt(3^2 + 5^2 + 2(a · b))
49 = 34 + 2(a · b)
(a · b) = 49 - 34
(a · b) = 15
Now, let's find the angle between vector b and vector c:
b · c = |b| |c| cosθ
Using the given magnitudes:
(b · c) = 5 * 7 * cosθ
(b · c) = 35 * cosθ
From the equation derived above:
15 = 35 * cosθ
cosθ = 15 / 35
cosθ ≈ 0.4286
Now we can find the angle:
θ = arccos(0.4286)
θ ≈ 64.65 degrees
Therefore, the angles between vector a and vector b is 90 degrees, and the angle between vector b and vector c is approximately 64.65 degrees.
To find the angles between vectors, we can use the dot product formula:
a · b = |a| |b| cos(θ)
Given that a · b + b · c + c · a = 0, we can use this information to find the angles between the vectors.
Let's start by calculating the dot product for the given vectors:
a · b = |a| |b| cos(θ_ab) => 3 * 5 * cos(θ_ab)
b · c = |b| |c| cos(θ_bc) => 5 * 7 * cos(θ_bc)
c · a = |c| |a| cos(θ_ca) => 7 * 3 * cos(θ_ca)
Given that a · b + b · c + c · a = 0, we have:
3 * 5 * cos(θ_ab) + 5 * 7 * cos(θ_bc) + 7 * 3 * cos(θ_ca) = 0
Now, we can solve this equation to find the angles between the vectors.