7.5 Vector product of two vectors

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 \(\mathbf{a}\) and \(\mathbf{b}\) is a vector \(\mathbf{c}\) such that

• The magnitude of \(\mathrm{c}=c=a b \sin \theta\) where a and b are magnitudes of \(\mathbf{a}\) and \(\mathbf{b}\) and \(\theta\) is the angle between the two vectors.
• \(\mathbf{c}\) is perpendicular to the plane containing \(\mathbf{a}\) and \(\mathbf{b}\).
• If we take a right-handed screw with its head lying in the plane of \(\mathbf{a}\) and \(\mathbf{b}\) and the screw perpendicular to this plane, and if we turn the head in the direction from \(\mathbf{a}\) to \(\mathbf{b}\), then the tip of the screw advances in the direction of \(\mathbf{c}\). 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 \(\mathbf{a}\) and \(\mathbf{b}\) and if the fingers are curled up in the direction from \(\mathbf{a}\) to \(\mathbf{b}\), then the stretched thumb points in the direction of \(\mathbf{c}\), as shown in Fig. 7.15b.

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

It should be remembered that there are two angles between any two vectors \(\mathbf{a}\) and \(\mathbf{b}\). In Fig. 7.15 (a) or (b) they correspond to \(\theta\) (as shown) and \(\left(360^{\circ}-\theta\right)\). While applying either of the above rules, the rotation should be taken through the smaller angle \(\left(<180^{\circ}\right)\) between a and \(\mathbf{b}\). It is \(\theta\) here.

Because of the cross \((x)\) 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, \(\mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{a}\) The vector product, however, is not commutative, i.e. \(\mathbf{a} \times \mathbf{b} \neq \mathbf{b} \times \mathbf{a}\)
  The magnitude of both \(\mathbf{a} \times \mathbf{b}\) and \(\mathbf{b} \times \mathbf{a}\) is the same \((a b \sin \theta)\); also, both of them are perpendicular to the plane of \(\mathbf{a}\) and \(\mathbf{b}\). But the rotation of the right-handed screw in case of \(\mathbf{a} \times \mathbf{b}\) is from \(\mathbf{a}\) to \(\mathbf{b}\), whereas in case of \(\mathbf{b} \times \mathbf{a}\) it is from \(\mathbf{b}\) to \(\mathbf{a}\). This means the two vectors are in opposite directions. We have
  \(
  \mathbf{a} \times \mathbf{b}=-\mathbf{b} \times \mathbf{a}
  \)
• Another interesting property of a vector product is its behaviour under reflection. Under reflection (1.e. on taking the plane mirror image) we have \(x \rightarrow-x, y \rightarrow-y\) and \(z \rightarrow-z\). As a result all the components of a vector change sign and thus \(a \rightarrow-a, b \rightarrow-b\). What happens to \(\mathbf{a} \times \mathbf{b}\) under reflection?
  \(
  \mathbf{a} \times \mathbf{b} \rightarrow(-\mathbf{a}) \times(-\mathbf{b})=\mathbf{a} \times \mathbf{b}
  \)
  Thus, \(\mathbf{a} \times \mathbf{b}\) does not change sign under reflection.
• Both scalar and vector products are distributive with respect to vector addition. Thus,
  \(
  \begin{gathered}
  \mathbf{a} .(\mathbf{b}+\mathbf{c})=\mathbf{a} \cdot \mathbf{b}+\mathbf{a} . \mathbf{c} \\
  \mathbf{a} \times(\mathbf{b}+\mathbf{c})=\mathbf{a} \times \mathbf{b}+\mathbf{a} \times \mathbf{c}
  \end{gathered}
  \)

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

  This follows since magnitude of \(\mathbf{a} \times \mathbf{a}\) is \(a^2 \sin 0^{\circ}=0\)
  From this follow the results
  (i) \(\hat{\mathbf{i}} \times \hat{\mathbf{i}}=\mathbf{0}, \hat{\mathbf{j}} \times \hat{\mathbf{j}}=\mathbf{0}, \hat{\mathbf{k}} \times \hat{\mathbf{k}}=\mathbf{0}\)
  (ii) \(\hat{\mathbf{i}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}}\)

• Note that the magnitude of \(\mathbf{i} \times \hat{\mathbf{j}}\) is \(\sin 90^{\circ}\) or 1 , since \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) both have unit magnitude and the angle between them is \(90^{\circ}\). Thus, \(\hat{\mathbf{i}} \times \hat{\mathbf{j}}\) is a unit vector. A unit vector perpendicular to the plane of \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) and related to them by the right hand screw rule is \(\hat{\mathbf{k}}\). Hence, the above result. You may verify similarly,
  \(
  \hat{\mathbf{j}} \times \hat{\mathbf{k}}=\hat{\mathbf{i}} \text { and } \hat{\mathbf{k}} \times \hat{\mathbf{i}}=\hat{\mathbf{j}}
  \)
  From the rule for commutation of the cross-product, it follows:
  \(
  \hat{\mathbf{j}} \times \hat{\mathbf{i}}=-\hat{\mathbf{k}}, \quad \hat{\mathbf{k}} \times \hat{\mathbf{j}}=-\hat{\mathbf{i}}, \quad \hat{\mathbf{i}} \times \hat{\mathbf{k}}=-\hat{\mathbf{j}}
  \)
  Note if \(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}\) occur cyclically in the above vector product relation, the vector product is positive. If \(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}\) do not occur in cyclic order, the vector product is negative.
  Now,
  \(
  \begin{aligned}
  & \mathbf{a} \times \mathbf{b}=\left(a_x \hat{\mathbf{i}}+a_y \hat{\mathbf{j}}+a_z \hat{\mathbf{k}}\right) \times\left(b_x \hat{\mathbf{i}}+b_y \hat{\mathbf{j}}+b_z \hat{\mathbf{k}}\right) \\
  & =a_x b_y \hat{\mathbf{k}}-a_x b_z \hat{\mathbf{j}}-a_y b_x \hat{\mathbf{k}}+a_y b_z \hat{\mathbf{i}}+a_z b_x \hat{\mathbf{j}}-a_z b_y \hat{\mathbf{i}} \\
  & =\left(a_y b_z-a_z b_y\right) \mathbf{i}+\left(a_z b_x-a_x b_z\right) \mathbf{j}+\left(a_x b_y-a_y b_x\right) \mathbf{k}
  \end{aligned}
  \)
  We have used the elementary cross products in obtaining the above relation. The expression for \(\mathbf{a} \times \mathbf{b}\) can be put in a determinant form which is easy to remember.
  \(
  \mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}
  \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
  a_x & a_y & a_z \\
  b_x & b_y & b_z
  \end{array}\right|
  \)

Example 7.7: Find the scalar and vector products of two vectors. \(\mathbf{a}=(3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})\) \(\operatorname{and} \mathbf{b}=(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\mathbf{3} \hat{\mathbf{k}})\)

Solution:

\(
\begin{aligned}
\mathbf{a} \cdot \mathbf{b} & =(3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \cdot(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \\
& =-6-4-15 \\
& =-25
\end{aligned}
\)
\(
\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
3 & -4 & 5 \\
-2 & 1 & -3
\end{array}\right|=7 \hat{\mathbf{i}}-\hat{\mathbf{j}}-5 \hat{\mathbf{k}}
\)
Note \(\mathbf{b} \times \mathbf{a}=-7 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}}\)