Ответ:

a=2CB+AC–3BA(–15; 5; –11)

Пошаговое объяснение:

А(1; 3; 2)

В(–2; 4; –1)

С(1; 3; –2)

НАЙТИ: а=2СВ+АС–3ВА

РЕШЕНИЕ:

СВ=(Вх–Сх; Ву–Су; Вz–Cz)=(–2–1; 4–3; –1–(–2)=(–3; 1; –1+2)=(–3; 1; 1)

CB(–3; 1; 1)

2CB=(2×CBx; 2×CBy; 2×CBz)=(2×(–3); 2×1; 2×1)=(–6; 2; 2)

2CB(–6; 2; 2)

AC(1–1; 3–3; –2–2)=(0; 0; –4)

ВА(1–(–2); 3–4; 2–(–1))=(1+2; –1; 2+1)=

=(3; –1; 3)

3ВА(3×3; 3×(–1); 3×3)=(9; –3; 9)

2СВ(–6; 2; 2); АС(0; 0; –4); 3ВА(9; –3; 9)

2СВ+АС–3ВА=

=(2СВх+АСх–3ВАх; 2СВу+АСу–3ВАу; 2СВz+ACz–3BAz)=(–6+0–9; 2+0–(–3); 2–4–9)=(–15; 2+3; 2–13)=(–15; 5; –11)