Question

Помогите пожалуйста с математикой, вектора

Answer 1

Ответ:

1) (13; -11; -33)

2)

$\sqrt{74} \: \: \sqrt{53} \: \: \sqrt{14}$

3) $\cos(AB, BC) = \frac{-57}{\sqrt{3922} }$

$\cos(BC, CD) = \frac{-5}{\sqrt{742} }$

4) 79

Пошаговое объяснение:

А(-3;1;4) В(1; -2;-3) С(2;2;3) D(5;3;1)

1) АВ-3ВС+4СD=(4;-3;-7) - (3;12;18)+ (12;4;-8)= (13; -11; -33)

AB= (1+3; -2-1; -3-4)=(4; -3; -7)

BC=(2-1; 2+2; 3+3)=( 1; 4; 6) 3BC=(3;12;18)

CD=(5-2; 3-2; 1-3)= (3; 1; -2) 4CD= (12;4;-8)

2) AB(4;-3;-7)

$|AB| = \sqrt{16 + 9 + 49} = \sqrt{74}$

BC(1;4;6)

$|BC| = \sqrt{1 + 16 + 36} = \sqrt{53}$

CD(3;1;-2)

$|CD| = \sqrt{9 + 1 + 4} = \sqrt{14}$

3) AB и BC

$\cos(AB, BC) = \frac{AB \times BC}{ |AB| \times |BC| } </p><p>AB×BC=4×1-3×4-7×6= 4-12-49= -57</p><p>[tex] |AB| \times |BC| = \sqrt{72} \times \sqrt{53} = \sqrt{3922}$

$\cos(AB, BC) = \frac{-57}{\sqrt{3922} }$

BC и СD

BC×CD= 1×3+4×1-6×2=3+4-12=-5

$|BC| \times |CD| = \sqrt{53} \times \sqrt{14} = \sqrt{742}$

$\cos(BC, CD) = \frac{-5}{\sqrt{742} }$

4) (AB+CD)AD= AB×AD+CD×AD=47+32= 79

AD=(5+3;3-1;1-4)= (8;2;-3)

AB×AD=4×8-3×2+7×3= 32-6+21= 47

CD×AD= 3×8+1×2+2×3= 24+2+6=32