Question

The two displacement vectors A ⇀ = 5.4 i ^ - 1.4 j ^ + 0.8 k ^ (m) and B ⇀ with length (1.5 m) makes an angle of (40 ° ) to the east of the north. The angle (in degrees) between these two vectors is (Round your answer to two digits after decimal point):

Answer 1

The angle between the two vectors is approximately 57.51 degrees, rounded to two decimal places.

To find the angle between two vectors, you can use the dot product formula:

A dot B = |A| * |B| * cos(theta)

where:

A dot B is the dot product of vectors A and B,

|A| and |B| are the magnitudes of vectors A and B, respectively,

theta is the angle between the two vectors.

First, let's find the magnitude of vector B:

|B| = 1.5 m

Now, find the dot product of A and B:

A dot B = (5.4 m) * cos(40 degrees)

Calculate this value.

Then, use the dot product formula to find cos(theta):

cos(theta) = (A dot B) / (|A| * |B|)

Finally, find the angle theta using the inverse cosine function:

theta = arccos((A dot B) / (|A| * |B|))

Make sure to convert the angle from radians to degrees. Round your answer to two digits after the decimal point.

Now, let's calculate it:

A dot B = (5.4 m) * cos(40 degrees)

A dot B ≈ (5.4 m) * (0.766)

|B| = 1.5 m

cos(theta) = ((5.4 m) * (0.766)) / ((5.4 m) * (1.5 m))

Now, find theta:

theta = arccos(0.514)

theta ≈ 1.003 radians

Convert theta to degrees:

theta ≈ 1.003 radians * (180 degrees / pi) ≈ 57.51 degrees

So, the angle between the two vectors is approximately 57.51 degrees. Rounded to two digits after the decimal point, the final answer is 57.51 degrees.