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Product of Two Vectors

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Estimated time: 14 minutes
CBSE: Class 12

Introduction

Vectors can be combined in more than one way, and each type of product gives different information. The two most important products are the scalar (dot) product and the vector (cross) product, which are used to find angles, projections, area, and direction-related results.

CBSE: Class 12
Maharashtra State Board: Class 12

Definition: Scalar Product (Dot Product)

If \[\vec{a}\] and \[\vec{b}\] are two vectors and \[\theta\] is the angle between them, then their scalar product is given by:

\[\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\]
 
Angle Between Two Vectors: 
\[\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}\]
CBSE: Class 12

Properties of Dot Product

  • Commutative: \[\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\]
  • Distributive: \[\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}\]

  • \[\vec{a} \cdot \vec{a} = |\vec{a}|^2\]

  • If \[\vec{a} \cdot \vec{b} = 0\], the vectors are perpendicular if both are non-zero.

  • \[\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\]
  • \[\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1\]

    \[\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0\]
CBSE: Class 12

Definition: Projection of One Vector on Another

Projection is the part of one vector in the direction of another vector.

Scalar projection of \[\vec{a}\] on \[\vec{b}\]

\[\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\]

Vector projection of \[\vec{a}\] on \[\vec{b}\]

\[\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b}\]
CBSE: Class 12

Definition: Vector Product (Cross Product)

If \[\vec{a}\] and \[\vec{b}\] are two vectors with angle \[\theta\] between them, then their vector product is:

\[\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta \hat{n}\]

where \[\hat{n}\] is a unit vector perpendicular to both \[\vec{a}\] and \[\vec{b}\], in the direction given by the right-hand rule.

Cross Product Angle: \[\sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|}\]

CBSE: Class 12

Properties of Cross Product

  • Not commutative: \[\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})\].

  • Distributive over addition.

  • \[\vec{a} \times \vec{a} = \vec{0}\].

  • The result is perpendicular to both vectors.

  • If vectors are parallel, then \[\sin \theta = 0\], so \[\vec{a} \times \vec{b} = \vec{0}\].

  • If vectors are perpendicular, then \[|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\].

  • Unit Vector Cross Product Relations: 
    \[\hat{i} \times \hat{j} = \hat{k}\]
    \[\hat{j} \times \hat{k} = \hat{i}\]
    \[\hat{k} \times \hat{i} = \hat{j}\]

    and

    \[\hat{j} \times \hat{i} = -\hat{k}\]
    \[\hat{k} \times \hat{j} = -\hat{i}\]
    \[\hat{i} \times \hat{k} = -\hat{j}\]
  • Determinant form: 

    \[\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}\]
CBSE: Class 12

Example 1

For any two vectors \[\vec{a}\] and \[\vec{b}\], we always have \[|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|\] (triangle inequality).

                Fig 10.21

Solution: The inequality holds trivially in case either \[\vec{a} = \vec{0}\] or \[\vec{b} = \vec{0}\] (How?). So, let \[|\vec{a}| \neq 0 \neq |\vec{b}|\]. Then,

\[|\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b})^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b})\]

\[= \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}\]

\[= |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2\] (scalar product is commutative)

\[\leq |\vec{a}|^2 + 2|\vec{a} \cdot \vec{b}| + |\vec{b}|^2\] (since \[x \leq |x| \forall x \in \mathbf{R}\])

\[\leq |\vec{a}|^2 + 2|\vec{a}| |\vec{b}| + |\vec{b}|^2\] (from Example 19)

\[= (|\vec{a}| + |\vec{b}|)^2\]

Hence \[|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|\]

CBSE: Class 12

Example 2

Find \[|\vec{a} \times \vec{b}|\], if \[\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}\] and \[\vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k}\]

Solution: We have

\[\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix}\]
\[= \hat{i}(-2 - 15) - (-4 - 9)\hat{j} + (10 - 3)\hat{k} = -17\hat{i} + 13\hat{j} + 7\hat{k}\]

Hence \[|\vec{a} \times \vec{b}| = \sqrt{(-17)^2 + (13)^2 + (7)^2} = \sqrt{507}\]

CBSE: Class 12

Example 3

Find the area of a parallelogram whose adjacent sides are given by the vectors \[\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}\] and \[\vec{b} = \hat{i} - \hat{j} + \hat{k}\].

Solution: The area of a parallelogram with \[\vec{a}\] and \[\vec{b}\] as its adjacent sides is given by \[|\vec{a} \times \vec{b}|\].

Now

\[\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} = 5\hat{i} + \hat{j} - 4\hat{k}\]

Therefore \[|\vec{a} \times \vec{b}| = \sqrt{25 + 1 + 16} = \sqrt{42}\]

and hence, the required area is \[\sqrt{42}\].

CBSE: Class 12
Maharashtra State Board: Class 12

Key Points: Product of Vector in Algebra

  • Dot product result is a scalar.

  • Cross product result is a vector.

  • Dot product uses cosine; cross product uses sine.

  • Dot product helps in angle and projection questions.

  • Cross product helps in area and direction questions.

  • \[\vec{a} \cdot \vec{b} = 0\] indicates perpendicular non-zero vectors.

  • \[\vec{a} \times \vec{b} = \vec{0}\] indicates parallel vectors.

  • Applications of Cross Product: 

    Area of Triangle:

    \[\frac{1}{2}|\vec{a} \times \vec{b}|\]

    Area of Parallelogram:

    \[|\vec{a} \times \vec{b}|\]

Shaalaa.com | Vector Algebra part 16 (Concepts: Scalar , Vector Product)

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