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Product of Two Vectors > Vector (Cross) Product

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CBSE: Class 12

Definition: Vector Product (Cross Product)

If \[\vec{a}\] and \[\vec{b}\] are two non-zero vectors 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

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 2

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

Key Points: Vector (Cross) Product

  • The cross product of two vectors is a vector quantity.
  • The cross product uses the sine of the angle between the vectors.
    \[ \vec{a} \times \vec{b} = |\vec{a}||\vec{b}| \sin\theta\,\hat{n} \]
  • The resulting vector is perpendicular to both vectors.
  • The cross product is useful in area and direction problems. 
  • For two non-zero vectors, \[ \vec{a} \times \vec{b} = \vec{0} \] indicates that the vectors are parallel or collinear.
  • Area of Triangle: \[ \frac{1}{2}|\vec{a} \times \vec{b}| \]
  •  Area of Parallelogram: \[ |\vec{a} \times \vec{b}| \]

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