if a vector+ b vector+ c vector =0, & a vector=3, b=5, c=9, find angle between a & b vector
They must form a triangle with sides 3,5, and 9
but that is an impossible triangle since
3+5 < 9
To find the angle between two vectors a and b, we need to use the dot product formula:
a · b = |a| * |b| * cos(θ)
where a · b is the dot product of vectors a and b, |a| and |b| are the magnitudes of vectors a and b respectively, and θ is the angle between them.
In this case, we are given that the vector sum of a + b + c is equal to zero, and the values of a, b, and c are 3, 5, and 9 respectively.
Since a + b + c = 0, we can rearrange the equation to solve for c:
c = -a - b
Substituting the given values, we get:
c = -3i - 5j - 9k
To find the angle between a and b, we first need to calculate the magnitude of each vector. The magnitude of a is given as 3, and the magnitude of b is 5.
Now let's calculate the dot product of a and b:
a · b = |a| * |b| * cos(θ)
3 * 5 * cos(θ) = a · b
15 * cos(θ) = a · b
To find the dot product, we need the components of a and b vectors. Assuming both vectors are in 3D space, let's take their components as follows:
a = 3i + 0j + 0k (assuming a is along the x-axis)
b = 5cos(θ)i + 5sin(θ)j + 0k (assuming b is in the xy-plane at an angle θ with the x-axis)
Now, we can calculate the dot product of a and b:
a · b = (3i + 0j + 0k) · (5cos(θ)i + 5sin(θ)j + 0k)
a · b = 15cos(θ)
Now, we can substitute the dot product back into the equation:
15cos(θ) = a · b
15cos(θ) = 15 * cos(θ)
cos(θ) = (a · b) / (|a| * |b|)
cos(θ) = 15 / (3 * 5)
cos(θ) = 1
Therefore, θ = 0 degrees.
Hence, the angle between vector a and b is 0 degrees.