Find a vector with a magnitude of 1.00 that bisects the angle between the vectors 5.00i + 11.0j and 2.00i − 1.00j. Give your answer in rectangular coordinates. Please explain how you arrived at the solution.
Thanks!
To find a vector that bisects the angle between two given vectors, we can follow these steps:
Step 1: Normalize the given vectors.
Step 2: Calculate the sum of the normalized vectors.
Step 3: Normalize the sum vector.
Step 4: Multiply the normalized sum vector by the desired magnitude.
Let's go through each step:
Step 1: Normalize the given vectors.
To normalize a vector, divide each of its components by its magnitude.
The magnitude of the first vector, 5.00i + 11.0j, can be found using the Pythagorean theorem:
|v1| = sqrt((5.00)^2 + (11.0)^2) = sqrt(25.00 + 121.00) = sqrt(146.00) ≈ 12.08
Thus, the normalized vector v1' is calculated by dividing each component by its magnitude:
v1' = (5.00i + 11.0j) / 12.08 ≈ (0.41i + 0.91j)
Similarly, the magnitude of the second vector, 2.00i − 1.00j, can be found:
|v2| = sqrt((2.00)^2 + (-1.00)^2) = sqrt(4.00 + 1.00) = sqrt(5.00) ≈ 2.24
The normalized vector v2' is:
v2' = (2.00i − 1.00j) / 2.24 ≈ (0.89i − 0.45j)
Step 2: Calculate the sum of the normalized vectors.
Add the normalized vectors v1' and v2' together:
v_sum = v1' + v2' ≈ (0.41i + 0.91j) + (0.89i − 0.45j) ≈ (1.30i + 0.46j)
Step 3: Normalize the sum vector.
Calculate the magnitude of the sum vector:
|v_sum| = sqrt((1.30)^2 + (0.46)^2) = sqrt(1.69 + 0.21) = sqrt(1.90) ≈ 1.38
Divide the sum vector by its magnitude to normalize it:
v_sum' = (1.30i + 0.46j) / 1.38 ≈ (0.94i + 0.33j)
Step 4: Multiply the normalized sum vector by the desired magnitude.
We want the vector to have a magnitude of 1.00, so we need to scale the normalized sum vector, v_sum', accordingly:
Final answer = 1.00 * v_sum' = 1.00 * (0.94i + 0.33j) ≈ (0.94i + 0.33j)
Therefore, a vector that bisects the angle between the vectors 5.00i + 11.0j and 2.00i − 1.00j and has a magnitude of 1.00 in rectangular coordinates is approximately (0.94i + 0.33j).