Vector Product Of Two Vectors

We are already familiar with vectors and their use in physics. In chapter 6 (Work, Energy, Power) we defined the scalar product of two vectors. An important physical quantity, work, is defined as a scalar product of two vector quantities, force and displacement.

We shall now define another product of two vectors. This product is a vector. Two important quantities in the study of rotational motion, namely, moment of a force and angular momentum, are defined as vector products.

Definition of Vector Product

A vector product of two vectors a and b is a vector c such that

(i) magnitude of c = c = ab sinθ where a and b are magnitudes of a and b and θ is the angle between the two vectors. 

(ii) c is perpendicular to the plane containing a and b. 

(iii) if we take a right handed screw with its head lying in the plane of a and b and the screw perpendicular to this plane, and if we turn the head in the direction from a to b, then the tip of the screw advances in the direction of c. This right handed screw rule is illustrated in Fig. 7.15a.

Alternately, if one curls up the fingers of right hand around a line perpendicular to the plane of the vectors a and b and if the fingers are curled up in the direction from a to b, then the stretched thumb points in the direction of c, as shown in Fig. 7.15b.

Fig.1

Fig.1: (a) Rule of the right handed screw for defining the direction of the vector product of two vectors. (b) Rule of the right hand for defining the direction of the vector product.

A simpler version of the right hand rule is the following : Open up your right hand palm and curl the fingers pointing from a to b. Your stretched thumb points in the direction of c.

It should be remembered that there are two angles between any two vectors a and b. In Fig. 7.15 (a) or (b) they correspond to θ (as shown) and (360$^0$ – θ). While applying either of the above rules, the rotation should be taken through the smaller angle (<180$^0$) between a and b. It is θ here.

Because of the cross (×) used to denote the vector product, it is also referred to as cross product.

• Note that scalar product of two vectors is commutative as said earlier, a.b = b.a

The vector product, however, is not commutative, i.e. a × b ≠ b × a

The magnitude of both a × b and b × a is the same (ab sinθ); also, both of them are perpendicular to the plane of a and b. But the rotation of the right-handed screw in case of a × b is from a to b, whereas in case of b × a it is from b to a. This means the two vectors are in opposite directions. We have

a × b = − b × a

• Another interesting property of a vector product is its behaviour under reflection. Under reflection (i.e. on taking the plane mirror image) we have x → −x, y → −y and z → −z. As a result all the components of a vector change sign and thus a→ −a, b → −b. What happens to a × b under reflection?

a × b → (−a) × (−b) = a × b

Thus, a × b does not change sign under reflection.

• Both scalar and vector products are distributive with respect to vector addition. Thus,

                        a.(b + c) = a.b + a.c

                        a × (b + c) = a × b + a × c

• We may write c = a × b in the component form. For this we first need to obtain some elementary cross products:
• a × a = 0 (0 is a null vector, i.e. a vector with zero magnitude)

This follows since magnitude of a × a is a$^2$ sin 0° = 0.

From this follow the results

(i) $\mathbf{\hat{i}} \times \mathbf{\hat{i}}=0$; $\mathbf{\hat{j}} \times \mathbf{\hat{j}}=0$ and $\mathbf{\hat{k}} \times \mathbf{\hat{k}}=0$

(ii) $\mathbf{\hat{i}} \times \mathbf{\hat{j}}=\mathbf{\hat{k}}$

Note that the magnitude of $\mathbf{\hat{i}} \times \mathbf{\hat{j}}$ is sin 90$^0$ or 1, since $\mathbf{\hat{i}}$ and $\mathbf{\hat{j}}$ both have unit magnitude and the angle between them is 90$^0$. Thus, $\mathbf{\hat{i}} \times \mathbf{\hat{j}}$ is a unit vector. A unit vector perpendicular to the plane of $\mathbf{\hat{i}}$ and $\mathbf{\hat{j}}$ and related to them by the right hand screw rule is $\mathbf{\hat{k}}$. Hence, the above result. You may verify similarly,

$\mathbf{\hat{j}} \times \mathbf{\hat{k}}=\mathbf{\hat{i}}$ and $\mathbf{\hat{k}} \times \mathbf{\hat{i}}=\mathbf{\hat{j}}$

From the rule for commutation of the cross product, it follows:

$\mathbf{\hat{j}} \times \mathbf{\hat{i}}=-\mathbf{\hat{k}}$; $\mathbf{\hat{k}} \times \mathbf{\hat{j}}=-\mathbf{\hat{k}}$ and $\mathbf{\hat{i}} \times \mathbf{\hat{k}}=-\mathbf{\hat{j}}$

Note if $\mathbf{\hat{i}}$, $\mathbf{\hat{j}}$, $\mathbf{\hat{k}}$ occur cyclically in the above vector product relation, the vector product is positive. If $\mathbf{\hat{i}}$, $\mathbf{\hat{j}}$, $\mathbf{\hat{k}}$ do not occur in cyclic order, the vector product is negative.

Now,

$\mathbf{a} \times \mathbf{b}=(a_x\hat{\mathbf{i}}+a_y\hat{\mathbf{j}}+a_z\hat{\mathbf{k}}) \times (b_x\hat{\mathbf{i}}+b_y\hat{\mathbf{j}}+b_z\hat{\mathbf{k}})$

$=a_xb_y\hat{\mathbf{k}}-a_xb_z\hat{\mathbf{j}}-a_yb_x\hat{\mathbf{k}}+a_yb_z\hat{\mathbf{i}}+a_xb_z\hat{\mathbf{j}}-a_zb_y\hat{\mathbf{i}}$

$=(a_yb_z-a_zb_y)\hat{\mathbf{i}}+(a_zb_x-a_xb_z)\hat{\mathbf{j}}+(a_xb_y-a_yb_x)\hat{\mathbf{k}}$

We have used the elementary cross products in obtaining the above relation. The expression for a × b can be put in a determinant form which is easy to remember.

$\mathbf{a}\times \mathbf{b}=\begin{vmatrix}\hat{\mathbf{i}} &\hat{\mathbf{j}}  &\hat{\mathbf{k}}\\ a_x& a_y & a_z\\b_x& b_y & b_z\end{vmatrix}$

Example 1

Find the scalar and vector products of two vectors. a = (3$\mathbf{\hat{i}}$ – 4$\mathbf{\hat{j}}$ + 5$\mathbf{\hat{k}}$) and b = (–2$\mathbf{\hat{i}}$ + $\mathbf{\hat{j}}$ – 3$\mathbf{\hat{k}}$)

Answer

a.b = (3$\mathbf{\hat{i}}$ – 4$\mathbf{\hat{j}}$ + 5$\mathbf{\hat{k}}$).(–2$\mathbf{\hat{i}}$ + $\mathbf{\hat{j}}$ – 3$\mathbf{\hat{k}}$)

= –6 – 4 – 15 = –25

$\mathbf{a}\times \mathbf{b}=\begin{vmatrix}\hat{\mathbf{i}} &\hat{\mathbf{j}}  &\hat{\mathbf{k}}\\ 3& -4 & 5\\-2& 1 & -3\end{vmatrix}=7\mathbf{\hat{i}}-\mathbf{\hat{j}}-5\mathbf{\hat{k}}$

Note $\mathbf{b} \times \mathbf{a}=-7\mathbf{\hat{i}}+\mathbf{\hat{j}}+5\mathbf{\hat{k}}$