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` please explain me obout cross product of vectos n detail`

6 years ago

Dear aditya,

The cross product of two vectors

aandbis denoted bya×b. In physics, sometimes the notationa∧bis used,^{ }though this is avoided in mathematics to avoid confusion with the exterior product.The cross product

a×bis defined as a vectorcthat is perpendicular to bothaandb, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.The cross product is defined by the formula

^{}^{}

where *θ* is the measure of the smaller angle between **a** and **b** (0° ≤ *θ* ≤ 180°), *a* and *b* are the magnitudes of vectors **a** and **b**, and is a unit vector perpendicular to the plane containing **a** and **b** in the direction given by the right-hand rule as illustrated. If the vectors **a** and **b** are parallel (i.e., the angle *θ* between them is either 0° or 180°), by the above formula, the cross product of **a** and **b** is the zero vector **0**.

The direction of the vector is given by the right-hand rule, where one simply points the forefinger of the right hand in the direction of **a** and the middle finger in the direction of **b**. Then, the vector is coming out of the thumb (see the picture on the right). Using this rule implies that the cross-product is anti-commutative, i.e., **b** × **a** = -(**a** × **b**). By pointing the forefinger toward **b** first, and then pointing the middle finger toward **a**, the thumb will be forced in the opposite direction, reversing the sign of the product vector.

**i**×**j**=**k****j**×**k**=**i****k**×**i**=**j****j**×**i**= −**k****k**×**j**= −**i****i**×**k**= −**j****i**×**i**=**j**×**j**=**k**×**k**=**0**.

The cross product can be calculated by distributive cross-multiplication:

**a**×**b**= (*a*_{1}**i**+*a*_{2}**j**+*a*_{3}**k**) × (*b*_{1}**i**+*b*_{2}**j**+*b*_{3}**k**)

**a**×**b**=*a*_{1}**i**× (*b*_{1}**i**+*b*_{2}**j**+*b*_{3}**k**) +*a*_{2}**j**× (*b*_{1}**i**+*b*_{2}**j**+*b*_{3}**k**) +*a*_{3}**k**× (*b*_{1}**i**+*b*_{2}**j**+*b*_{3}**k**)

**a**×**b**= (*a*_{1}**i**×*b*_{1}**i**) + (*a*_{1}**i**×*b*_{2}**j**) + (*a*_{1}**i**×*b*_{3}**k**) + (*a*_{2}**j**×*b*_{1}**i**) + (*a*_{2}**j**×*b*_{2}**j**) + (*a*_{2}**j**×*b*_{3}**k**) + (*a*_{3}**k**×*b*_{1}**i**) + (*a*_{3}**k**×*b*_{2}**j**) + (*a*_{3}**k**×*b*_{3}**k**).-
GOODLUCK

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6 years ago

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