Contents
Physics Topics such as mechanics, thermodynamics, and electromagnetism are fundamental to many other scientific fields.
What are the Properties of Vector or Cross Product?
The product of two vectors can be a vector. For example,
angular velocity × position vector = linear velocity
[For details see the chapter Circular Motion]
Here, both angular velocity and position vector are vector quantities and their product, linear velocity, is also a vector quantity Such type of product of two vectors is called a vector product or cross product and is represented by putting a cross (×) sign between the two vectors.
Definition: The vector product or cross product of two vectors \(\vec{a}\) and \(\vec{b}\) is defined as \(\vec{a}\) × \(\vec{b}\) = ab sinθ \(\hat{n}\), where θ is the angle from \(\vec{a}\) to \(\vec{b}\), and \(\hat{n}\) is a unit vector. The direction of \(\hat{n}\) is the direction of advance of a right-handed screw when it is rotated from \(\vec{b}\) to \(\vec{b}\). [Fig.].
Here, \(\hat{n}\) ⊥ \(\vec{a}\) and \(\hat{n}\) ⊥ \(\vec{b}\). So the product \(\vec{a}\) × \(\vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\). The unit vector \(\hat{n}\) has a magnitude, |\(\hat{n}\)| = 1. So the magnitude of the vector product is
\(|\vec{a} \times \vec{b}|\) = ab sinθ.
Some properties of vector product:
i) \(\vec{A} \times \vec{A}\) = \(\overrightarrow{\mathbf{0}}\) i.e., vector product of a vector with itself is a null vector.
The angle between the same vector taken twice is zero.
∴ The magnitude of \(\vec{A} \times \vec{A}\) = |\(\vec{A} \times \vec{A}\)| = AAsin0° = 0.
Hence, \(\vec{A} \times \vec{A}\) = 0
ii) \(\vec{A} \times \vec{B}\) = \(-\vec{B} \times \vec{A}\) i.e., vector product is not commutative.
If the angle from \(\vec{A}\) to \(\vec{B}\) is θ, the angle from \(\vec{B}\) to \(\vec{A}\) is -θ. As sin (-θ) = – sinθ, we get, \(\vec{A}\) × \(\vec{B}\) = –\(\vec{B}\) × \(\vec{A}\).
iii) Vector product of two mutually perpendicular vectors:
If \(\vec{A}\) ⊥ \(\vec{B}\) then, the magnitude of \(\vec{A}\) × \(\vec{B}\) = –\(\vec{B}\) × \(\vec{A}\) = ABsin90° = AB, and the direction of \(\vec{A}\) × \(\vec{B}\) is perpendicular to both of \(\vec{A}\) and \(\vec{B}\) [Fig.]. Hence, \(\vec{A}\), \(\vec{B}\) and (\(\vec{A}\) × \(\vec{B}\)) —all these three vectors are mutually perpendicular, i.e., if \(\vec{A}\) and \(\vec{B}\) are along x and y axes respectively, then their product \(\vec{A}\) × \(\vec{B}\) will be along z -axis.
iv) Vector products of mutually perpendicular unit vectors:
\(\hat{i}\), \(\hat{j}\), \(\hat{k}\) are the unit vectors along positive x, y and z axes respectively. The magnitude of each is 1.
As \(\vec{A}\) × \(\vec{A}\) = 0, we get, \(\hat{i}\) × \(\hat{i}\) = \(\hat{j}\) × \(\hat{j}\) = \(\hat{k}\) × \(\hat{k}\) = o
Again, the magnitude of
\(\hat{i}\) × \(\hat{j}\) = |\(\hat{i}\) × \(\hat{j}\)| = (1)(1)sin90° = 1
According to the right-handed corkscrew rule, the direction of \(\hat{i}\) × \(\hat{j}\) is along the z-axis [Fig.]. Again, the unit vector along z-axis is \(\hat{k}\). Hence, \(\hat{i}\) × \(\hat{j}\) = \(\hat{k}\)
Similarly, \(\hat{i}\) × \(\hat{j}\) = \(\hat{k}\), \(\hat{j}\) × \(\hat{k}\) = \(\hat{i}\), \(\hat{k}\) × \(\hat{i}\) = \(\hat{j}\)
Again, since \(\vec{A}\) × \(\vec{B}\) = –\(\vec{B}\) × \(\vec{A}\), \(\hat{j}\) × \(\hat{j}\) = –\(\hat{k}\), \(\hat{k}\) × \(\hat{j}\) = –\(\hat{i}\), \(\hat{i}\) × \(\hat{k}\) = –\(\hat{j}\)
Here, the cyclic order of \(\hat{i}\), \(\hat{j}\) \(\hat{k}\) is important. If this cyclic order is maintained, the product of the first two vectors leads to the third with a positive sign, whereas if the cyclic order is not maintained, the product becomes the third vector with a negative sign [Fig.].
v) Vector product in terms of positional coordinates:
Out of the 9 terms of this product, the terms containing \(\hat{i}\) × \(\hat{i}\), \(\hat{j}\) × \(\hat{j}\) and \(\hat{k}\) × \(\hat{k}\) are zero. Writing the remaining 6 terms using the relations like \(\hat{i}\) × \(\hat{j}\) = \(\hat{k}\),
we get,
\(\vec{A} \times \vec{B}\) = \(\hat{i}\)(AyBz – AzBy) + \(\hat{j}\)(AzBx – AxBz) + \(\hat{k}\)(AxBy – AyBx)
It is convenient to write this product in a determinant form:
\(\vec{A} \times \vec{B}\) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
A_x & A_y & A_z \\
B_x & B_y & B_z
\end{array}\right|\)
Area of triangle: If two vectors \(\vec{a}\) and \(\vec{b}\) represents the two sides of a triangle then the half of the magnitude of their cross product will give the area of the triangle.
In Δ ABC [Fig.], \(\overrightarrow{C B}\) = \(\vec{a}\), \(\overrightarrow{C A}\) = \(\vec{b}\) and the angle between them = C.
Height of the triangle, AD = ACᐧ\(\frac{A D}{A C}\) = bsin C
∴ Area of ΔABC = \(\frac{1}{2}\) × (base) × (height)
= \(\frac{1}{2}\) × (BC) × (AD) = \(\frac{1}{2}\)absinC
= \(\frac{1}{2}|\vec{a} \times \vec{b}|\)
Area of Parallelogram: ABCD is a parallelogram [Fig.]. Let us denote the two adjacent sides BC and BA by two vectors, \(\vec{a}\) = \(\overrightarrow{B C}\) and \(\vec{b}\) = \(\overrightarrow{B A}\); θ = angle between the vectors \(\vec{a}\) and \(\vec{b}\).
Now, we draw AD ⊥ BC. From the figure, AD = ABᐧ\(\frac{A D}{A B}\) = b sinθ. So, the area of the parallelogram,
S = base × height = (BC)(AD) = ab sinθ = |\(\vec{a}\) × \(\vec{b}\)|
This means that the magnitude of the cross product of two vectors is geometrically represented by the area of a parallelogram whose adjacent sides stand for the two given vectors.
Surface area as a vetor: The preceding relation S = |\(\vec{a}\) × \(\vec{b}\)| hints at the possibility of writing S as a vector product,
|\(\vec{S}\) = |\(\vec{a}\) × \(\vec{b}\)|
But it would mean that the area S is a vector quantity Now, we have to check whether it can be true. For this, let us consider a plane mirror in a room. To describe its effect precisely, we have to specify not only its surface area but also its orientation, i.e., how it is placed in the room. A statement like ‘a mirror of area 600 cm2 ‘ is not sufficient; it should be stated like ‘a mirror of area 600 cm2 facing north’. This confirms that a plane surface should indeed be treated as a vector quantity; its magnitude equals the area of the surface, and its direction is perpendicular to the surface. A curved surface does not have a definite direction, and cannot be treated as a vector. However, an infinitesimally small area ds on this surface is effectively plane, and may be treated as a vector \(\overrightarrow{d s}\). This concept is widely used in different branches of physics.
In this context, the relation \(\vec{S}\) = \(\vec{a} \times \vec{b}\) is valid for a parallelogram. Consequently, the cross product of two vectors is geometrically represented by the area vector of a parallelogram whose adjacent sides stand for the two given vectors.
Numerical Examples
Example 1.
\(\overrightarrow{\boldsymbol{A}}\) = 2\(\hat{i}\) + 3\(\hat{j}\) + 4\(\hat{k}\) are two vectors. Find \(\overrightarrow{\boldsymbol{A}}\) × \(\overrightarrow{\boldsymbol{B}}\) [HS(XI) ’06]
Solution:
Example 2.
Using vector method in a triangle, prove that,
(i) \(\frac{a}{\sin A}\) = \(\frac{b}{\sin B}\) = \(\frac{c}{\sin C}\)
(ii) cosA = \(\frac{b^2+c^2-a^2}{2 b c}\)
Solution:
According to Fig., \(\vec{a}\) + \(\vec{c}\) = \(\vec{b}\)
i) \(\vec{a}\) × \(\vec{b}\) = \(\vec{a}\) × (\(\vec{a}\) + \(\vec{c}\)) = \(\vec{a}\) × \(\vec{c}\)
or, ab sinC = ac sinθ
(where θ is the angle between \(\vec{a}\) and \(\vec{c}\) )
or, bsinC = csin (180° – B)
or, b sin C = c sin B or, \(\frac{b}{\sin B}\) = \(\frac{c}{\sin C}\)
Proceeding in the same way, we get
\(\frac{a}{\sin A}\) = \(\frac{b}{\sin B}\) = \(\frac{c}{\sin C}\)
ii) b2 = \(\vec{b} \cdot \vec{b}\) = \((\vec{a}+\vec{c}) \cdot(\vec{a}+\vec{c})\) = \(\vec{a} \cdot \vec{a}\) + \(\vec{c} \cdot \vec{c}\) + 2\(\vec{a} \cdot \vec{c}\)
= a2 + c2 + 2ac cosθ
= a2 + c2 + 2ac cos(180° – B)
= a2 + c2 = 2 ac cosB
Similarly, a2 = b2 + c2 – 2bc cos A
or, cosA = \(\frac{b^2+c^2-a^2}{2 b c}\)
Example 3.
If the two diagonals of a parallelogram are given by \(\vec{R}_1\) = 3\(\hat{i}\) – 2\(\hat{j}\) + 7\(\hat{k}\) and \(\vec{R}_2\) = 5\(\hat{i}\) + 6\(\hat{j}\) – 3\(\hat{k}\) , find out the area of this parallelogram.
Solution:
Let, the two adjacent sides of the parallelogram be \(\vec{A}\) and \(\vec{B}\).Then,
Example 4.
Find the angle between force \(\overrightarrow{\boldsymbol{F}}\) = (3\(\hat{i}\) + 4\(\hat{j}\) – 5\(\hat{k}\)) and displacement \(\overrightarrow{\boldsymbol{d}}\) = (5\(\hat{i}\) + 4\(\hat{j}\) – 3\(\hat{k}\)). Also find the projection of \(\vec{F}\) on \(\vec{d}\).
Solution:
Let θ be the angle between the vectors \(\vec{F}\) and \(\vec{d}\).
Then by the definition of scalar product of two vectors, we have
Example 5.
Prove that \(|\vec{P} \times \vec{Q}|^2\) = \(|\vec{P}|^2|\vec{Q}|^2\) – \(|\overrightarrow{\boldsymbol{P}} \cdot \overrightarrow{\boldsymbol{Q}}|^2\)
Solution:
Example 6.
The resultant of two vectors \(\vec{A}\) and \(\vec{B}\) acting through a point O is \(\vec{R}\). A certain straight line intersects the lines representing the vectors \(\vec{A}\), \(\vec{B}\) and \(\vec{R}\) at points P, Q and S, respectively. Prove that \(\frac{A}{O P}\) + \(\frac{B}{O Q}\) = \(\frac{R}{O S}\)
Solution:
Let, the unit vectors along \(\vec{A}\), \(\vec{B}\) and \(\vec{R}\) be \(\hat{a}\), \(\hat{b}\) and \(\hat{c}\) respectively. Then