Find the length of Vector A+B given the length of vector A, the length of vector B and the angle between them

Question

Suppose that the length of vector A, [[A]] = 4, the length of vector B, [[B]] =√3, the angle between them is Pi/6. Find the length of A+B, [[A+B]]

Answer 1

There are two ways of solving this. One decomposes the vectors into their horizontal and vertical components, and then uses the pythagorean theorem to find the length of the sum. The other uses the law of cosines. I'll present both methods here.

Decomposing the vectors into components:

The problem never said which directions the vectors were facing, only that they were separated by a 30 degree angle. (The problem actually specified pi/6 radians, but we know that pi/6 radians is 30 degrees.) The good news is that we can assume any direction for the first vector and always get the same answer.

I'll assume that the first vector is perfectly horizontal and that the second vector points 30 degrees upward from the horizontal.

We want to find the horizontal and vertical components of each of the two vectors, then add those components to get the horizontal and vertical components of the sum, (called the resultant), and finally, calculate its length using the pythagorean theorem.

The first vector is easy, because it's horizontal. The problem says it has length 4. So we get

A_horizontal = 4.

A_vertical = 0.

Next, we do the same for vector B. Because it points diagonally, we have to use trigonometry to find its horizontal and vertical components. Because it has length √(3) and points at a 30 degree angle, we get the following equations:

B_horizontal = √3 cos (30) = 1.5.

B_vertical = √3 sin (30) = √3 /2.

Now we add the vectors to get the components of the resultant:

(A+B)_horizontal = A_horizontal + B_Horizontal = 4 + 1.5 = 5.5

(A+B)_vertical = A_vertical + B_vertical = 0 + √3 /2 = √3 /2.

Finally, we use the Pythagorean theorem to find the total length of the resultant vector A+B.

||A+B|| = √( ((A+B)_horizontal)² + ((A+B)_vertical)² ) = √ (5.5² + (√3 /2)²) = √(31)

Law of cosines:

At first glance, you might think "Why would the law of cosines be useful here. The law of cosines is used for triangles. This isn't a triangle, it's the sum of two vectors." Recall how to add two vectors. If you draw one vector, then another vector that starts where the first one ends, then the sum, or resultant, of the two vectors is the vector that begins at the start of the first one and travels in a straight line, ending at the end of the second vector. These three vectors, (A, B, and the resultant) form a triangle.

Notice that we have the lengths of the two vectors and the angle between them. There is one other detail that it's easy to mess up. The two vectors meet at a pi/6 radian (or 30 degree) angle when they start at the same point. If vector B is moved to start where vector A ends instead, they will actually form a 150 degree angle. (I would draw a picture to demonstrate, but this format doesn't permit it. I may add a video later.)

Now that we have the lengths of the two vectors and the angle between them, we can use the law of cosines:

||A+B||² = ||A||² + ||B||² - 2||A|| ||B|| cos(150)

||A+B||² = 4² + √3² - 2 x 4 x √3 x -√3 / 2

||A+B||² = 16 + 3 - (-12)

||A+B||² = 31

||A+B|| = √31