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