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        please explain me obout cross product of vectos n detail
7 years ago

105 Points
										Dear aditya,
The cross product of two vectors a and b is denoted by a × b. In physics, sometimes the notation a∧b is used,   though this is avoided in mathematics to avoid confusion with the exterior product.
The cross product a × b is defined as a vector c that is perpendicular to both a and b, 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
$\mathbf{a} \times \mathbf{b} = a b \sin \theta \ \mathbf{\hat{n}}$
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 $\scriptstyle\mathbf{\hat{n}}$ 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 $\scriptstyle\mathbf{\hat{n}}$ 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 $\scriptstyle\mathbf{\hat{n}}$ 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 = jj × i = −k           k × j = −i           i × k = −ji × i = j × j = k × k = 0.
The cross product can be calculated by distributive cross-multiplication:
a × b = (a1i + a2j + a3k) × (b1i + b2j + b3k) a × b = a1i × (b1i + b2j + b3k) + a2j × (b1i + b2j + b3k) + a3k × (b1i + b2j + b3k) a × b = (a1i × b1i) + (a1i × b2j) + (a1i × b3k) + (a2j × b1i) + (a2j × b2j) + (a2j × b3k) + (a3k × b1i) + (a3k × b2j) + (a3k × b3k).

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7 years ago
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