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Introducing vectors for engineering applications
Introducing vectors for engineering applications

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4.4 Finding the angle between two vectors

The scalar product of two vectors has an important application in calculating the angle between two vectors. If we start with the definition of the scalar product in terms of the magnitudes and directions of the vectors, and rearrange it, then we get the following result.

Angle between two vectors

The angle theta between any two non-zero vectors bold a and bold b is given by

cosine of theta equals a dot operator b divided by absolute value of a times absolute value of b full stop

We can use this result to find the angle between two vectors in component form.

Example 8 Calculating the angle between two vectors in component form

Alice and Bob have attached ropes to a face of the block of ice and are pulling it in different directions, see Figure 34. Vector bold a describes the force applied by Alice, and in component form is given by a equals 100 times i plus 50 times j . Vector bold b describes the force applied by Bob, and in component form is given by b equals 90 times i minus 70 times j .

What is the angle between these vectors, to one decimal place?

Described image
Figure 34 Alice and Bob pulling a block of ice

Solution

First let’s use the components of bold a and bold b to find a dot operator b , absolute value of a and absolute value of b . We have

equation sequence part 1 a dot operator b equals part 2 left parenthesis 100 times i plus 50 times j right parenthesis dot operator left parenthesis 90 times i minus 70 times j right parenthesis equals part 3 left parenthesis 100 multiplication 90 right parenthesis plus left parenthesis 50 multiplication left parenthesis negative 70 right parenthesis right parenthesis times equals 5500 comma
equation sequence part 1 absolute value of a equals part 2 Square root of 100 squared plus 50 squared equals part 3 Square root of 12 500 equals part 4 111.80 horizontal ellipsis comma
equation sequence part 1 absolute value of b equals part 2 Square root of 90 squared plus left parenthesis negative 70 right parenthesis squared equals part 3 Square root of 8100 plus 4900 equals part 4 Square root of 13 000 equals part 5 114.017 horizontal ellipsis full stop

Using these we can calculate cosine of theta :

equation sequence part 1 cosine of theta equals part 2 a dot operator b divided by absolute value of a times absolute value of b equals part 3 5500 divided by Square root of 12 500 multiplication Square root of 13 000 equals part 4 0.431 horizontal ellipsis full stop

So

equation sequence part 1 theta equals part 2 cosine super negative one of 5500 divided by Square root of 12 500 multiplication Square root of 13 000 equals part 3 64.440 times ellipsis super ring operator full stop

Therefore the angle between the vectors is 64.4° (to 1 d.p.).

Activity 22

Find, to the nearest degree, the angle between the vectors

a equals vector element 1 two element 2 two and b equals vector element 1 one element 2 three full stop