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The study of Physics Topics involves the exploration of matter, energy, and the forces that govern the universe.
What is Dot Product (Scalar Product)? What are the Properties of a Scalar Dot Product?
Product of a Scalar and a Vector
A scalar has magnitude but no direction. On the other hand, a vector has both magnitude and direction. Therefore, when a scalar is multiplied with a vector, the product has both magnitude and direction. Thus, the product is a vector.
i) When a vector is multiplied by a scalar, the product is also a vector. If the scalar is positive then the direction of the vector remains unaltered but, if the scalar is negative, the direction of the vector becomes opposite to that of the original vector.
Example: 10 eastwards × 5 = 50 eastwards
but, 10 eastwards × (-5) = (-50) eastwards
= 50 westwards
ii) Product of a scalar quantity with a vector quantity produces, in general, some other vector quantity
Example:
iii) If the vector is a unit vector, then the product’s magnitude is the same as the scalar’s. For example, if \(\vec{n}\) is a unit vector due east, then \(\vec{n}\) × 5 = 5 due east or, \(\vec{n}\) × 5 m ᐧ s-1 speed = 5 m ᐧ s-1 velocity eastwards.
Product of Two Vectors
To vectors, when multiplied, may produce either a scalar or a vector. Accordingly they are called scalar or dot product and vector or cross product.
Scalar or Dot Product
Two vectors may give a scalar product like:
force ᐧ displacement = work
[For details see the chapter Work and Energy]
Here, although force and displacement are both vector quantities, their product, work, is a scalar quantity Such type of product of two vectors is called a scalar product or dot product. A dot (ᐧ) symbol is used between the two vectors to represent such a product mathematically.
Definition: The scalar product of two vectors \(\vec{A}\) and \(\vec{B}\) is defined as \(\vec{A}\) ᐧ \(\vec{B}\) = AB cosθ, where θ is the angle between \(\vec{A}\) and \(\vec{B}\).
Some properties of scalar products:
i) \(\vec{A}\).\(\vec{A}\) = A2, i.e., the scalar product of a vector with itself is the square of its magnitude.
If the same vector is taken twice, then the angle between them = 0°.
∴ \(\vec{A} \cdot \vec{A}\) = AAcosθ = A2
Hence, A = \(\sqrt{\vec{A} \cdot \vec{A}}\)
This rule is often used for determining the magnitude of any vector.
ii) \(\vec{A} \cdot \vec{B}\) = \(\vec{B} \cdot \vec{A}\), ie., scalar products are commutative.
If the angle from \(\vec{A}\) to \(\vec{B}\) is θ, the angle from \(\vec{B}\) to \(\vec{A}\) is -θ.
∴ \(\vec{A} \cdot \vec{B}\) = ABcosθ
and \(\vec{B} \cdot \vec{A}\) = BAcos(-θ) = ABcosθ
Hence, \(\vec{A} \cdot \vec{B}\) = \(\vec{B} \cdot \vec{A}\)
iii) Scalar product of two orthogonal vectors, ie., the vectors which are mutually at right angles, is zero.
If \(\vec{A}\) ⊥ \(\vec{B}\), then the angle between them, θ = 90°. Hence,
cosθ = 0°
∴ \(\vec{A} \cdot \vec{B}\) = ABcosθ = 0
Alternatively, if \(\vec{A} \cdot \vec{B}\) = o for two non-zero vectors \(\vec{A}\) and \(\vec{B}\), then \(\vec{A}\) ⊥ \(\vec{B}\)
iv) If θ is the angle between two vectors \(\vec{A}\) and \(\vec{B}\) then
\(\vec{A} \cdot \vec{B}\) = ABcosθ or, \(\frac{\vec{A} \cdot \vec{B}}{A B}\) = cosθ
or, θ = cos-1\(\left(\frac{\vec{A} \cdot \vec{B}}{A B}\right)\)
v) Scalar products of mutually perpendicular unit vectors: This rule is often used to determine the angle between two vectors, \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) are three unit vectors along the directions of positive x, y, z axes respectively. Each has a value of 1 and they are mutually perpendicular. If we take the same vector twice, the angle between them will be 00; for example, the angle between \(\hat{i}\) and \(\hat{i}\) is 0°. If we take two different vectors, then the angle between them will be 90°; for example, the angle between \(\hat{i}\) and \(\hat{j}\) is 90°. Therefore,
\(\hat{i}\)ᐧ\(\hat{i}\) = (1)(1)cos0° = 1
\(\hat{i}\)ᐧ\(\hat{j}\) = (1)(1)cos90° = 0
So, the general rule for scalar product of these unit vectors is
\(\hat{i}\)ᐧ\(\hat{i}\) = 1, \(\hat{j}\)ᐧ\(\hat{j}\) = 1, \(\hat{k}\)ᐧ\(\hat{k}\) = 1
and \(\hat{i}\)ᐧ\(\hat{j}\) = \(\hat{j}\)ᐧ\(\hat{i}\) = \(\hat{j}\)ᐧ\(\hat{k}\) = \(\hat{k}\)ᐧ\(\hat{j}\) = \(\hat{k}\)ᐧ\(\hat{i}\) = \(\hat{i}\)ᐧ\(\hat{k}\) = 0
vi) Scalar product of two vectors using the position coordinates:
Following the rules of simple multiplication, and omitting the 6 terms containing \(\hat{i}\)ᐧ\(\hat{j}\), \(\hat{i}\)ᐧ\(\hat{k}\), etc. that are equal to zero,
\(\vec{A} \cdot \vec{B}\) = AxBx + AyBy + AzBz
vii) Dot product of a vector with \(\hat{i}\), \(\hat{j}\), \(\hat{k}\),:
Thus, the dot product of any vector with \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) denotes the magnitudes of the x, y, z components respectively of the vector itself. This property stands for any unit vector \(\hat{n}\)
\(\hat{n} \cdot \vec{A}\) = magnitude of the component of \(\vec{A}\) along the direction of the unit vector \(\hat{n}\). i.e., the projection of \(\vec{A}\) along the direction of \(\hat{n}\).
Numerical Examples
Example 1.
Find the magnitude of the vector 3\(\hat{i}\) – 4\(\hat{j}\) + 12\(\hat{k}\).
Solution:
Example 2.
Find the angle between the vectors \(\hat{i}\) + \(\hat{j}\) and \(\hat{i}\) – \(\hat{k}\).
Solution:
Let us consider, \(\vec{A}\) = (\(\hat{i}\) + \(\hat{j}\)) and \(\vec{B}\) = (\(\hat{i}\) – \(\hat{k}\))
∴ \(\vec{A} \cdot \vec{A}\) = \((\hat{i}+\hat{j}) \cdot(\hat{i}+\hat{j})\) = 12 + 12 = 2 = A2
∴ A = \(\sqrt{2}\) units
Example 3.
Find the angle between the vectors \(\overrightarrow{\boldsymbol{A}}\) = 2\(\hat{i}\) + 3\(\hat{j}\) and \(\overrightarrow{\boldsymbol{B}}\) = -3\(\hat{i}\) + 2\(\hat{j}\). [HS(XI)’06]
Solution:
Let the angle between the two vectors \(\vec{A}\) and \(\vec{B}\) be θ.
Now, \(\vec{A} \cdot \vec{B}\) = (2\(\hat{i}\) + 3\(\hat{j}\)) ᐧ (-3\(\hat{i}\) + 2\(\hat{j}\)) = -6 + 6 = 0
As this dot product is zero, \(\vec{A}\) and \(\vec{B}\) are perpendicular to
each other. So, the angle between them = 90°.
Example 4.
Prove that the diagonals of a rhombus are perpendicular to each other.
Solution:
In Fig., ABCD is a rhombus. \(\overrightarrow{A B}\) = \(\overrightarrow{D C}\) = \(\vec{a}\) and \(\overrightarrow{A D}\) = \(\overrightarrow{B C}\) = \(\vec{b}\); the values of \(\vec{a}\) and \(\vec{b}\) are the same for a rhombus.
As the dot product is 0, the angle between the vectors = 90°, i.e., \(\overrightarrow{A C}\) and \(\overrightarrow{B D}\) are perpendicular to each other.
Example 5.
\(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are three unit vectors. Show that, \(|\vec{a}-\vec{b}|^2\) + \(|\vec{b}-\vec{c}|^2\) + \(|\vec{c}-\vec{a}|^2\) ≤ 9
Solution:
Example 6.
Find the projection of the vector \(\overrightarrow{\boldsymbol{P}}\) = 2\(\hat{i}\) – 3\(\hat{j}\) + 6\(\hat{k}\) on the vector \(\overrightarrow{\boldsymbol{Q}}\) = \(\hat{i}\) + 2\(\hat{j}\) + 2\(\hat{k}\).
Solution:
The projection of \(\overrightarrow{\boldsymbol{P}}\) on \(\overrightarrow{\boldsymbol{Q}}\) is
Example 7.
A particle is moving in a curvilinear path defined by the equations x = 2t2, y = t2 – 4t and z = 3t – 5. Find out the magnitudes of the components of velocity and acceleration along (\(\hat{i}\) – 3\(\hat{j}\) + 2\(\hat{k}\)) at time t = 1.
Solution:
Example 8.
Prove that a right angled triangle can be formed using the vectors \(\vec{A}\) = \(\hat{i}\) – 3\(\hat{j}\) + 5\(\hat{k}\), \(\vec{B}\) = 2\(\hat{i}\) + \(\hat{j}\) – 4\(\hat{k}\) and \(\vec{C}\) = 3\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\).
Solution:
∴ In the triangle formed by \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\), the angle between \(\vec{B}\) and \(\vec{C}\) is 90°. Thus, it is a right-angled triangle.
[The dot product \(\vec{B}\) ᐧ \(\vec{C}\) has been calculated, because \(\vec{A}\) forms the largest side of the triangle. An angle of 90° is always opposite to the largest side. In this example,