EN3: Introduction to Engineering

Teach Yourself Vectors

Division of Engineering

Brown University

5. Multiplication of vectors

5.1 Multiplication by a scalar.

Let be a vector, and a scalar. Then is a vector. The direction of is parallel to and its magnitude is given by .

Note that you can form a unit vector n which is parallel to a by setting

5.2 Formula for the product of a scalar and a vector in Cartesian Components

Let be a vector and a scalar. Find an expression for the components of the vector Then

and hence

Problems

5.1 Find the components of a unit vector parallel to the vector

5.2 Let , . Find 6a+2b, and 6a-2b.

5.3 Dot Product

(also called the scalar product). Let a and b be two vectors. The dot product of a and b is a scalar denoted by , and is defined by

where and denote the magnitudes of a and b, respectively, and is the angle subtended by a and b, as shown in the figure.

Note that , and . If and then if and only if ; i.e. a and b are perpendicular.

5.4 Formula for the dot product of two vectors in Cartesian Components

Let

where are the Cartesian components of vectors a, b in a basis{i,j,k}.

Calculate in terms of .

This time we have to do some real work. Substitute for a and b and see what happens

This is a mess. But recall that I, j and k are mutually perpendicular, so the angle between them is 90 degrees. Recall also that cos(90) = 0. Finally, recall the definition of the dot product. Therefore .

This leaves

Finally, note that the a vector is always parallel to itself, so the angle between a vector and itself is zero. Recall also that I, j and k are all unit vectors. Therefore

, and so on for all three remaining dot products. So, finally

Problems

5.3 Find the dot products of the vectors listed below

(a) ,

(b) ,

5.4 The vectors a and b shown in the figure below have magnitudes , . Calculate .

5.5 Two vectors a and b are mutually perpendicular. What is their dot product?

5.6 Calculate

5.7 Calculate the angle between each pair of vectors listed in Problem 5.3 – i.e. find the angle between a and b in each case

5.8 For the structure shown, calculate the angle between the vectors (i.e. the vector pointing from O to B) and . (Use vectors – it’s possible to do this by long-winded trigonometry and Pythagoras theorem but that’s not the point)

5.5 Dot Product as a Projection

The quantity is sometimes referred to as the component of a in a direction parallel to b. The figure shows why.

The vector a can be thought of as the sum of two vectors: one (OX) parallel to b and another (XA) perpendicular to b. This process of dividing a into two parts is known as projecting a onto components parallel and perpendicular to b.

Recall that |a| is the length of OA. The length of OX is therefore . But recall that , so that , as stated.

Problems

5.9 Let a and b be two vectors. Project a onto components parallel and perpendicular to b as shown in the picture.

(i) Show that the vector

(ii) Verify that the preceding result satisfies , as it should (why?)

(iii ) Show that the component of a in a direction perpendicular to b is

5.10 Let a = 3i+2j+6k, and b = 12i-5j+2k. Calculate the components of a in directions parallel and perpendicular to b.

5.5 Cross Product (also called the vector product).

Let a and b be two vectors. By definition, the cross product of a and b is a vector, denoted by . The direction of c is perpendicular to both a and b, and is chosen so that (a,b,c) form a right handed triad, as shown. The magnitude of c is given by

Note that and .

Calculating the magnitude of the cross product of two vectors is no sweat, but figuring out the direction is a pain. There are various aide-memoirs to help you do this- choose the one you find least confusing, or make up your own.

Right hand rule

To find the direction of , arrange your right hand so that your thumb is parallel to a, your index finger is parallel to b, and the angle between your thumb and index finger is . Now set your middle finger is perpendicular to both a and b. The direction of is parallel to your middle finger. (This rule only really works if , otherwise you permanently damage your hand. Please don’t do this.)

Right hand screw rule

To find the direction of , arrange your right hand so that your thumb is perpendicular to both a and b, and your fingers curl in the direction of the line joining the tip of vector a to the tip of vector b. The direction of is parallel to your thumb.

Bottle-cap rule.

Obtain a twist-top bottle of your favorite beverage. Draw an arrow on the cap. Arrange the bottle so that, by twisting the cap through the angle , you can rotate the arrow from parallel to a to parallel to b. The direction of is parallel to the direction of motion of the bottle-cap as it is turned. (Full beverage containers are not be permitted in EN3 examinations)

If none of these tricks help you

Extend your middle finger into the air. Shout your favorite expletive. This will not help, but it may make you feel better.

Problem

5.11 Let {i,j,k} be a Cartesian basis. Use the definition of the cross product given above to calculate all possible cross products of the basis vectors –i.e., calculate

You will find that the results are all very simple. For example, , since i is parallel to itself. Hence . Similarly , since i and j are both unit vectors and the angle between them is 90 degrees. The rules governing the direction of a cross product also show that is parallel to k. Therefore . See if you can work out the rest on your own.

5.6 Formula for the cross product of two vectors in Cartesian Components.

Let

where are the Cartesian components of vectors a, b in a basis{i,j,k}.

Calculate in terms of .

More work for the wicked. Substitute for a and b

This is another mess. This time, note that (and similarly for and ), note that the angle between i and itself is zero, and recall that sin(0)=0. Therefore .

The remaining cross products between i,j,k have to be calculated laboriously one at a time, using the definition given in the preceding section. The figure shows the direction of all six possible cross products between the basis vectors (magnitudes are not shown to scale, for clarity). Thus, we conclude that

THESE FORMULAS ARE IMPORTANT!

You need to remember them. There is a nice little trick to help you. Write down the 3 vectors i, j, k in a circle, going clockwise, as shown below.

Now, to find the cross product of any pair of basis vectors, you travel around the circle. Thus, to get , you start at i, move to j and then on to k. If you go around the circle clockwise, the answer is positive, if you go counter-clockwise, it is negative. Thus, , and so on, while , etc.

If we substitute these results into our expression for we determine that

Hence

This is not an easy formula to remember, but it is so important that you must memorize it. The following trick is sometimes used to help remember the formula – if you know how to calculate the determinant of a matrix, then you will note that

It is fairly easy to remember the cover-up rule for computing the determinant of a matrix, so this is a popular trick.

Another way to remember the formula is to notice the pattern in the indices. The indices are written out below to show the pattern more clearly

There are two things to notice about this pattern. First, note that the expression for involves only and , similarly, the expression for involves only and , and the third expression has the same feature. Secondly, notice that the indices always appear both forwards (x,y,z or y,z,x or z,x,y) and backwards (x,z,y or y,x,z or z,y,x) in each expression. The forward terms (x,y,z or y,z,x or z,x,y) are all positive, while the backward terms (x,z,y or y,x,z or z,y,x) are all negative.

Problems

5.12 Find the cross products of the vectors listed below

(a) ,

(b) ,

5.13 The vectors a and b shown in the figure have magnitudes , . Calculate . What is the direction of ?

5.14 For the vector triangle shown in 5.13, show that .