Find the angle between the vector A=31-j+2k
and the negative y axis
To find the angle between vector A and the negative y-axis, we can use the dot product of the two vectors and the formula for the angle between two vectors:
Let the negative y-axis be represented by the vector B = 0-i-0k.
The dot product of vectors A and B is given by:
A · B = (31)(0) + (-1)(-1) + (2)(0) = 1
The magnitude of vector A is given by:
|A| = sqrt((31)^2 + (-1)^2 + (2)^2) = sqrt(972) = 31
The magnitude of vector B is given by:
|B| = sqrt((0)^2 + (-1)^2 + (0)^2) = 1
Using the formula cosθ = A · B / (|A| |B|), we can find the angle θ:
cosθ = 1 / (31 * 1) = 1 / 31
θ = arccos(1 / 31) ≈ 85.47 degrees
Therefore, the angle between vector A and the negative y-axis is approximately 85.47 degrees.
To find the angle between a vector and the negative y-axis, we can use the dot product and trigonometry.
First, let's find the dot product between vector A and the negative y-axis vector (0, -1, 0).
A · (-y-axis) = (31, -1, 2) · (0, -1, 0) = (0 * 31) + (-1 * -1) + (2 * 0) = 1
Next, we need to calculate the magnitudes of both vectors.
|A| = √(31² + (-1)² + 2²) = √(961 + 1 + 4) = √966 ≈ 31.08
|y-axis| = √(0² + (-1)² + 0²) = √(0 + 1 + 0) = √1 = 1
Now, we can calculate the cosine of the angle between the vectors using the dot product formula:
cos(angle) = A · (-y-axis) / (|A| * |y-axis|)
= 1 / (31.08 * 1)
= 1 / 31.08
≈ 0.032
Finally, we can find the angle by taking the inverse cosine (arccos) of the cosine value:
angle = arccos(0.032)
≈ 89.88 degrees or approximately 90 degrees
Therefore, the angle between vector A and the negative y-axis is approximately 90 degrees.