Calculating the Resultant Vector and its Angle

When working with vectors in physics and engineering, it's often necessary to find the magnitude of the resultant vector and the angle between vectors. This article explains how to calculate these using basic vector principles, such as the Law of Cosines and the Law of Sines.

Introduction

Consider two vectors, (vec{A}) and (vec{B}), with magnitudes of 10 units and 6 units respectively. The angle between these vectors is 30 degrees. We will calculate the magnitude of the resultant vector (vec{R}) and the angle between (vec{R}) and each of the constituent vectors.

Calculating the Magnitude of the Resultant Vector

To find the magnitude of the resultant vector (vec{R}), we can use the Law of Cosines. The formula is:

[ R sqrt{A^2 B^2 - 2AB cos theta} ]

Given:

(A 10) (magnitude of vector (vec{A})) (B 6) (magnitude of vector (vec{B})) (theta 30^circ) (angle between vectors (vec{A}) and (vec{B}))

First, calculate (cos 30^circ):

[ cos 30^circ frac{sqrt{3}}{2} approx 0.866 ]

Substitute the values into the formula:

[ R sqrt{10^2 6^2 - 2 times 10 times 6 times 0.866} ]

Calculate each term:

(10^2 100) (6^2 36) (2 times 10 times 6 times 0.866 103.92)

Now, sum these values:

[ R sqrt{100 36 - 103.92} sqrt{239.92} approx 15.49 text{ units} ]

So, the magnitude of the resultant vector (vec{R}) is approximately 15.49 units.

Calculating the Angle Between the Resultant Vector and Each Constituent Vector

To find the angle between the resultant vector (vec{R}) and each constituent vector, we can use the Law of Sines. The formula is:

[ frac{R}{sin phi} frac{B}{sin theta} ]

Since (sin 30^circ 0.5), we have:

[ frac{15.49}{sin phi} frac{6}{0.5} 12 ]

Now, substituting (R):

[ frac{15.49}{sin phi} 12 ]

Solving for (sin phi):

[ sin phi frac{15.49}{12} approx 1.2917 ]

Since (sin phi) cannot exceed 1, this indicates that the angle calculated this way is not valid. Therefore, we calculate the angle (alpha) between (vec{R}) and (vec{B}) using the Law of Sines:

[ frac{R}{sin alpha} frac{A}{sin theta} ]

Given that: (sin theta 0.5), we have:

[ frac{15.49}{sin alpha} frac{10}{0.5} 20 ]

Thus:

[ sin alpha frac{15.49}{20} approx 0.7745 ]

Now, calculate (alpha):

[ alpha approx sin^{-1}0.7745 approx 50.4^circ ]

To find the angle between (vec{R}) and (vec{A}), we use the previously calculated angle (theta) and subtract (alpha):

(180^circ - 30^circ - 50.4^circ approx 99.6^circ)

Final Results:

Magnitude of the resultant vector (vec{R}): approximately 15.49 units Angle (phi) between (vec{R}) and (vec{A}): approximately 50.4 degrees Angle (alpha) between (vec{R}) and (vec{B}): approximately 39.6 degrees

These calculations show the resultant vector's magnitude and the angles involved between the vectors.