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imes mathbf a mathbf a times mathbf b times mathbf a left mathbf a right 2 mathbf b mathbf a cdot mathbf b mathbf a rearrange to solve for b to give b c a a 2 a b a 2 a displaystyle mathbf b frac mathbf c times mathbf a left mathbf a right 2 frac mathbf a cdot mathbf b left mathbf a right 2 mathbf a the coefficient of the last term can be simplified to just the arbitrary constant t to yield the result shown above lagrange s identity edit the relation a b 2 det a a a b a b b b a 2 b 2 a b 2 displaystyle left mathbf a times mathbf b right 2 det begin bmatrix mathbf a cdot mathbf a mathbf a cdot mathbf b mathbf a cdot mathbf b mathbf b cdot mathbf b end bmatrix left mathbf a right 2 left mathbf b right 2 mathbf a cdot mathbf b 2 can be compared with another relation involving the right hand side namely lagrange s identity expressed as 14 1 i j n a i b j a j b i 2 a 2 b 2 a b 2 displaystyle sum _ 1 leq i j leq n left a_ i b_ j a_ j b_ i right 2 left mathbf a right 2 left mathbf b right 2 mathbf a cdot b 2 where a and b may be n dimensional vectors this also shows that the riemannian volume form for surfaces is exactly the surface element from vector calculus in the case where n 3 combining these two equations results in the expression for the magnitude of the cross product in terms of its components 15 a b 2 1 i j 3 a i b j a j b i 2 a 1 b 2 b 1 a 2 2 a 2 b 3 a 3 b 2 2 a 3 b 1 a 1 b 3 2 displaystyle begin aligned mathbf a times mathbf b 2 sum _ 1 leq i j leq 3 a_ i b_ j a_ j b_ i 2 a_ 1 b_ 2 b_ 1 a_ 2 2 a_ 2 b_ 3 a_ 3 b_ 2 2 a_ 3 b_ 1 a_ 1 b_ 3 2 end aligned the same result is found directly using the components of the cross product found from a b det i j k a 1 a 2 a 3 b 1 b 2 b 3 displaystyle mathbf a times mathbf b det begin bmatrix hat mathbf i hat mathbf j hat mathbf k a_ 1 a_ 2 a_ 3 b_ 1 b_ 2 b_ 3 end bmatrix in r 3 lagrange s equation is a special case of the multiplicativity vw v w of the norm in the quaternion algebra it is a special case of another formula also sometimes called lagrange s identity which is the three dimensional case of the binet cauchy identity 16 17 a b c d a c b d a d b c displaystyle mathbf a times mathbf b cdot mathbf c times mathbf d mathbf a cdot mathbf c mathbf b cdot mathbf d mathbf a cdot mathbf d mathbf b cdot mathbf c if a c and b d this simplifies to the formula above alternative ways to compute edit conversion to matrix multiplication edit the vector cross product also can be expressed as the product of a skew symmetric matrix and a vector 16 a b a b 0 a 3 a 2 a 3 0 a 1 a 2 a 1 0 b 1 b 2 b 3 a b b t a 0 b 3 b 2 b 3 0 b 1 b 2 b 1 0 a 1 a 2 a 3 displaystyle begin aligned mathbf a times mathbf b mathbf a _ times mathbf b begin bmatrix 0 a_ 3 a_ 2 a_ 3 0 a_ 1 a_ 2 a_ 1 0 end bmatrix begin bmatrix b_ 1 b_ 2 b_ 3 end bmatrix mathbf a times mathbf b mathbf b _ times mathrm t mathbf a begin bmatrix 0 b_ 3 b_ 2 b_ 3 0 b_ 1 b_ 2 b_ 1 0 end bmatrix begin bmatrix a_ 1 a_ 2 a_ 3 end bmatrix end aligned where the superscript t refers to the transpose operation and a is defined by a d e f 0 a 3 a 2 a 3 0 a 1 a 2 a 1 0 displaystyle mathbf a _ times stackrel rm def begin bmatrix 0 a_ 3 a_ 2 a_ 3 0 a_ 1 a_ 2 a_ 1 0 end bmatrix the columns a i of the skew symmetric matrix for a vector a can be also obtained by calculating the cross product with unit vectors that is a i a e i i 1 2 3 displaystyle mathbf a _ times i mathbf a times mathbf hat e _ i i in 1 2 3 or a i 1 3 a e i e i displaystyle mathbf a _ times sum _ i 1 3 left mathbf a times mathbf hat e _ i right otimes mathbf hat e _ i where displaystyle otimes is the outer product operator also if a is itself expressed as a cross product a c d displaystyle mathbf a mathbf c times mathbf d then a d c t c d t displaystyle mathbf a _ times mathbf d mathbf c mathrm t mathbf c mathbf d mathrm t proof by substitution evaluation of the cross product gives a c d c 2 d 3 c 3 d 2 c 3 d 1 c 1 d 3 c 1 d 2 c 2 d 1 displaystyle mathbf a mathbf c times mathbf d begin pmatrix c_ 2 d_ 3 c_ 3 d_ 2 c_ 3 d_ 1 c_ 1 d_ 3 c_ 1 d_ 2 c_ 2 d_ 1 end pmatrix hence the left hand side equals a 0 c 2 d 1 c 1 d 2 c 3 d 1 c 1 d 3 c 1 d 2 c 2 d 1 0 c 3 d 2 c 2 d 3 c 1 d 3 c 3 d 1 c 2 d 3 c 3 d 2 0 displaystyle mathbf a _ times begin bmatrix 0 c_ 2 d_ 1 c_ 1 d_ 2 c_ 3 d_ 1 c_ 1 d_ 3 c_ 1 d_ 2 c_ 2 d_ 1 0 c_ 3 d_ 2 c_ 2 d_ 3 c_ 1 d_ 3 c_ 3 d_ 1 c_ 2 d_ 3 c_ 3 d_ 2 0 end bmatrix now for the right hand side c d t c 1 d 1 c 1 d 2 c 1 d 3 c 2 d 1 c 2 d 2 c 2 d 3 c 3 d 1 c 3 d 2 c 3 d 3 displaystyle mathbf c mathbf d mathrm t begin bmatrix c_ 1 d_ 1 c_ 1 d_ 2 c_ 1 d_ 3 c_ 2 d_ 1 c_ 2 d_ 2 c_ 2 d_ 3 c_ 3 d_ 1 c_ 3 d_ 2 c_ 3 d_ 3 end bmatrix and its transpose is d c t c 1 d 1 c 2 d 1 c 3 d 1 c 1 d 2 c 2 d 2 c 3 d 2 c 1 d 3 c 2 d 3 c 3 d 3 displaystyle mathbf d mathbf c mathrm t begin bmatrix c_ 1 d_ 1 c_ 2 d_ 1 c_ 3 d_ 1 c_ 1 d_ 2 c_ 2 d_ 2 c_ 3 d_ 2 c_ 1 d_ 3 c_ 2 d_ 3 c_ 3 d_ 3 end bmatrix evaluation of the right hand side gives d c t c d t 0 c 2 d 1 c 1 d 2 c 3 d 1 c 1 d 3 c 1 d 2 c 2 d 1 0 c 3 d 2 c 2 d 3 c 1 d 3 c 3 d 1 c 2 d 3 c 3 d 2 0 displaystyle mathbf d mathbf c mathrm t mathbf c mathbf d mathrm t begin bmatrix 0 c_ 2 d_ 1 c_ 1 d_ 2 c_ 3 d_ 1 c_ 1 d_ 3 c_ 1 d_ 2 c_ 2 d_ 1 0 c_ 3 d_ 2 c_ 2 d_ 3 c_ 1 d_ 3 c_ 3 d_ 1 c_ 2 d_ 3 c_ 3 d_ 2 0 end bmatrix comparison shows that the left hand side equals the right hand side this result can be generalized to higher dimensions using geometric algebra in particular in any dimension bivectors can be identified with skew symmetric matrices so the product between a skew symmetric matrix and vector is equivalent to the grade 1 part of the product of a bivector and vector 18 in three dimensions bivectors are dual to vectors so the product is equivalent to the cross product with the bivector instead of its vector dual in higher dimensions the product can still be calculated but bivectors have more degrees of freedom and are not equivalent to vectors 18 this notation is also often much easier to work with for example in epipolar geometry from the general properties of the cross product follows immediately that a a 0 displaystyle mathbf a _ times mathbf a mathbf 0 and a t a 0 displaystyle mathbf a mathrm t mathbf a _ times mathbf 0 and from fact that a is skew symmetric it follows that b t a b 0 displaystyle mathbf b mathrm t mathbf a _ times mathbf b 0 the above mentioned triple product expansion bac cab rule can be easily proven using this notation as mentioned above the lie algebra r 3 with cross product is isomorphic to the lie algebra so 3 whose elements can be identified with the 3 3 skew symmetric matrices the map a a provides an isomorphism between r 3 and so 3 under this map the cross product of 3 vectors corresponds to the commutator of 3x3 skew symmetric matrices matrix conversion for cross product with canonical base vectors denoting with e i r 3 1 displaystyle mathbf e _ i in mathbf r 3 times 1 the i displaystyle i th canonical base vector the cross product of a generic vector v r 3 1 displaystyle mathbf v in mathbf r 3 times 1 with e i displaystyle mathbf e _ i is given by v e i c i v displaystyle mathbf v times mathbf e _ i mathbf c _ i mathbf v where c 1 0 0 0 0 0 1 0 1 0 c 2 0 0 1 0 0 0 1 0 0 c 3 0 1 0 1 0 0 0 0 0 displaystyle mathbf c _ 1 begin bmatrix 0 0 0 0 0 1 0 1 0 end bmatrix quad mathbf c _ 2 begin bmatrix 0 0 1 0 0 0 1 0 0 end bmatrix quad mathbf c _ 3 begin bmatrix 0 1 0 1 0 0 0 0 0 end bmatrix these matrices share the following properties c i t c i displaystyle mathbf c _ i textrm t mathbf c _ i skew symmetric both trace and determinant are zero rank c i 2 displaystyle text rank mathbf c _ i 2 c i c i t p e i displaystyle mathbf c _ i mathbf c _ i textrm t mathbf p _ mathbf e _ i perp see below the orthogonal projection matrix of a vector v 0 displaystyle mathbf v neq mathbf 0 is given by p v v v t v 1 v t displaystyle mathbf p _ mathbf v mathbf v left mathbf v textrm t mathbf v right 1 mathbf v t the projection matrix onto the orthogonal complement is given by p v i p v displaystyle mathbf p _ mathbf v perp mathbf i mathbf p _ mathbf v where i displaystyle mathbf i is the identity matrix for the special case of v e i displaystyle mathbf v mathbf e _ i it can be verified that p e 1 0 0 0 0 1 0 0 0 1 p e 2 1 0 0 0 0 0 0 0 1 p e 3 1 0 0 0 1 0 0 0 0 displaystyle mathbf p _ mathbf e _ 1 perp begin bmatrix 0 0 0 0 1 0 0 0 1 end bmatrix quad mathbf p _ mathbf e _ 2 perp begin bmatrix 1 0 0 0 0 0 0 0 1 end bmatrix quad mathbf p _ mathbf e _ 3 perp begin bmatrix 1 0 0 0 1 0 0 0 0 end bmatrix for other properties of orthogonal projection matrices see projection linear algebra index notation for tensors edit the cross product can alternatively be defined in terms of the levi civita tensor e ijk and a dot product η mi which are useful in converting vector notation for tensor applications c a b c m i 1 3 j 1 3 k 1 3 η m i e i j k a j b k displaystyle mathbf c mathbf a times b leftrightarrow c m sum _ i 1 3 sum _ j 1 3 sum _ k 1 3 eta mi e_ ijk a j b k where the indices i j k displaystyle i j k correspond to vector components this characterization of the cross product is often expressed more compactly using the einstein summation convention as c a b c m η m i e i j k a j b k displaystyle mathbf c mathbf a times b leftrightarrow c m eta mi e_ ijk a j b k in which repeated indices are summed over the values 1 to 3 in a positively oriented orthonormal basis η mi δ mi the kronecker delta and e i j k ε i j k displaystyle e_ ijk varepsilon _ ijk the levi civita symbol in that case this representation is another form of the skew symmetric representation of the cross product ε i j k a j a displaystyle varepsilon _ ijk a j mathbf a _ times in classical mechanics representing the cross product by using the levi civita symbol can cause mechanical symmetries to be obvious when physical systems are isotropic an example consider a particle in a hooke s law potential in three space free to oscillate in three dimensions none of these dimensions are special in any sense so symmetries lie in the cross product represented angular momentum which are made clear by the above mentioned levi civita representation citation needed mnemonic edit mnemonic to calculate a cross product in vector form xyzzy mnemonic redirects here for other uses see xyzzy the word xyzzy can be used to remember the definition of the cross product if a b c displaystyle mathbf a mathbf b times mathbf c where a a x a y a z b b x b y b z c c x c y c z displaystyle mathbf a begin bmatrix a_ x a_ y a_ z end bmatrix mathbf b begin bmatrix b_ x b_ y b_ z end bmatrix mathbf c begin bmatrix c_ x c_ y c_ z end bmatrix then a x b y c z b z c y displaystyle a_ x b_ y c_ z b_ z c_ y a y b z c x b x c z displaystyle a_ y b_ z c_ x b_ x c_ z a z b x c y b y c x displaystyle a_ z b_ x c_ y b_ y c_ x the second and third equations can be obtained from the first by simply vertically rotating the subscripts x y z x the problem of course is how to remember the first equation and two options are available for this purpose either to remember the relevant two diagonals of sarrus s scheme those containing i or to remember the xyzzy sequence since the first diagonal in sarrus s scheme is just the main diagonal of the above mentioned 3 3 matrix the first three letters of the word xyzzy can be very easily remembered cross visualization edit similarly to the mnemonic device above a cross or x can be visualized between the two vectors in the equation this may be helpful for remembering the correct cross product formula if a b c displaystyle mathbf a mathbf b times mathbf c then a b x b y b z c x c y c z displaystyle mathbf a begin bmatrix b_ x b_ y b_ z end bmatrix times begin bmatrix c_ x c_ y c_ z end bmatrix if we want to obtain the formula for a x displaystyle a_ x we simply drop the b x displaystyle b_ x and c x displaystyle c_ x from the formula and take the next two components down a x b y b z c y c z displaystyle a_ x begin bmatrix b_ y b_ z end bmatrix times begin bmatrix c_ y c_ z end bmatrix when doing this for a y displaystyle a_ y the next two elements down should wrap around the matrix so that after the z component comes the x component for clarity when performing this operation for a y displaystyle a_ y the next two components should be z and x in that order while for a z displaystyle a_ z the next two components should be taken as x and y a y b z b x c z c x a z b x b y c x c y displaystyle a_ y begin bmatrix b_ z b_ x end bmatrix times begin bmatrix c_ z c_ x end bmatrix a_ z begin bmatrix b_ x b_ y end bmatrix times begin bmatrix c_ x c_ y end bmatrix for a x displaystyle a_ x then if we visualize the cross operator as pointing from an element on the left to an element on the right we can take the first element on the left and simply multiply by the element that the cross points to in the right hand matrix we then subtract the next element down on the left multiplied by the element that the cross points to here as well this results in our a x displaystyle a_ x formula a x b y c z b z c y displaystyle a_ x b_ y c_ z b_ z c_ y we can do this in the same way for a y displaystyle a_ y and a z displaystyle a_ z to construct their associated formulas applications edit the cross product has applications in various contexts for example it is used in computational geometry physics and engineering a non exhaustive list of examples follows computational geometry edit the cross product appears in the calculation of the distance of two skew lines lines not in the same plane from each other in three dimensional space the cross product can be used to calculate the normal for a triangle or polygon an operation frequently performed in computer graphics for example the winding of a polygon clockwise or anticlockwise about a point within the polygon can be calculated by triangulating the polygon like spoking a wheel and summing the angles between the spokes using the cross product to keep track of the sign of each angle in computational geometry of the plane the cross product is used to determine the sign of the acute angle defined by three points p 1 x 1 y 1 p 2 x 2 y 2 displaystyle p_ 1 x_ 1 y_ 1 p_ 2 x_ 2 y_ 2 and p 3 x 3 y 3 displaystyle p_ 3 x_ 3 y_ 3 it corresponds to the direction upward or downward of the cross product of the two coplanar vectors defined by the two pairs of points p 1 p 2 displaystyle p_ 1 p_ 2 and p 1 p 3 displaystyle p_ 1 p_ 3 the sign of the acute angle is the sign of the expression p x 2 x 1 y 3 y 1 y 2 y 1 x 3 x 1 displaystyle p x_ 2 x_ 1 y_ 3 y_ 1 y_ 2 y_ 1 x_ 3 x_ 1 which is the signed length of the cross product of the two vectors to use the cross product simply extend the 2d vectors p 1 p 2 p 3 displaystyle p_ 1 p_ 2 p_ 3 to co planar 3d vectors by setting z k 0 displaystyle z_ k 0 for each of them in the right handed coordinate system if the result is 0 the points are collinear if it is positive the three points constitute a positive angle of rotation around p 1 displaystyle p_ ...