Question

Доказать тождество : (с,(a-2c),(2a+b))=(a,b,c) векторы

Answer 1

$(overline{a}, overline{b}, overline{c}) = begin{vmatrix} a_x & a_y & a_z \ b_x & b_y & b_z \ c_x & c_y & c_z end{vmatrix} = a_x begin{vmatrix} b_y & b_z \ c_y & c_zend{vmatrix} - a_y begin{vmatrix} b_x & b_z \ c_x & c_zend{vmatrix} + a_z begin{vmatrix} b_x & b_y \ c_x & c_yend{vmatrix} =\= a_x(b_yc_z - b_zc_y) - a_y(b_xc_z-b_zc_x) + a_z(b_xc_y - b_yc_x)$

$(overline{c}, overline{a - 2c}, overline{2a + b}) = begin{vmatrix} c_x & c_y & c_z \ a_x - 2c_x &a_y - 2c_y & a_z - 2c_z \ 2a_x + b_x & 2a_y + b_y & 2a_z + b_z end{vmatrix} =\= begin{vmatrix} c_x & c_y & c_z \ a_x&a_y&a_z\ 2a_x + b_x & 2a_y + b_y & 2a_z + b_z end{vmatrix} = begin{vmatrix} c_x & c_y & c_z \ a_x&a_y&a_z\ b_x & b_y & b_z end{vmatrix} = -begin{vmatrix}a_x&a_y& a_z \ c_x & c_y & c_z\ b_x & b_y & b_z end{vmatrix} =$

$-(-begin{vmatrix}a_x&a_y&a_z \ b_x & b_y & b_z\c_x & c_y & c_z end{vmatrix}) = begin{vmatrix}a_x&a_y& a_z \ b_x & b_y & b_z\c_x & c_y & c_z end{vmatrix} =\= a_x(b_yc_z - b_zc_y) - a_y(b_xc_z-b_zc_x) + a_z(b_xc_y - b_yc_x)$

Очевидно, тождество верно