Find the angle between the vector A=3i-j+2k
and the negative y axis
To find the angle between vector A and the negative y-axis, we need to determine the components of vector A that are parallel to the negative y-axis.
The negative y-axis can be represented by the vector B = -j.
To find the parallel component of vector A, we need to take the dot product of vector A and vector B, and then divide it by the magnitude of vector B.
A • B = (3i - j + 2k) • (-j)
= 0 - (-1) + 0
= 1
|B| = |-j| = √((-1)^2 + 0^2 + 0^2) = √1 = 1
Therefore, the angle θ between vector A and the negative y-axis can be found using the equation:
cosθ = (A • B) / (|A| |B|)
cosθ = 1 / (3^2 + (-1)^2 + 2^2) = 1 / 14
Taking the inverse cosine of both sides, we get:
θ = arccos(1 / 14)
Using a calculator, we find:
θ ≈ 80.5 degrees
Therefore, the angle between vector A = 3i - j + 2k and the negative y-axis is approximately 80.5 degrees.
To find the angle between a vector and a coordinate axis, we can use the dot product formula.
The dot product between two vectors A and B is given by:
A · B = |A| |B| cos(theta)
Where |A| represents the magnitude of vector A, |B| represents the magnitude of vector B, and theta is the angle between the two vectors.
In this case, we want to find the angle between vector A and the negative y-axis, which can be represented by the vector B=0i-1j+0k.
So, the dot product of vector A and B is:
A · B = (3)(0) + (-1)(-1) + (2)(0) = 1
The magnitudes of vectors A and B are:
|A| = sqrt(3^2 + (-1)^2 + 2^2) = sqrt(14)
|B| = 1
Now, we can use the dot product formula to find the angle theta:
1 = sqrt(14)(1) cos(theta)
cos(theta) = 1/sqrt(14)
Taking the inverse cosine of both sides:
theta = arccos(1/sqrt(14))
Therefore, the angle between vector A and the negative y-axis is approximately 51.32 degrees.