Question

Решите срочно на 30 баллов ​

Answer 1

Объяснение:

1.

$|\^a|=13\ \ \ \ |\^b|=19 \ \ \ \ \ |\^a-\^b|=22\ \ \ \ |\^a+\^b|=?\\|\^a-\^b|=\sqrt{|\^a|^2+|\^b|^2-2*|\^a|*|\^b|*cos\alpha } \\\sqrt{13^2+19^2-2*13*19*cos\alpha } =22\\(\sqrt{169+361-2*13*19*cos\alpha })^2 =22^2\\530-494*cos\alpha =484\\494*cos\alpha =46.\\|\^a+\^b|=\sqrt{|\^a|^2+|\^b|^2+2*|\^a|*|\^b|*cos\alpha}=\sqrt{530+46}=\sqrt{576}=24.$

Ответ: 24.

2.

$A(-1;-2;4)\ \ \ \ B(-4;-2;0)\ \ \ \ C(3;-2;1).\\AB=(-4-(-1);-2-(-2);0-4)=(-3;0;-4).\\|AB|=\sqrt{(-3)^2+0^2+(-4)^2}=\sqrt{9+0+16} =\sqrt{25}=5.\\BC=(3-(-4);-2-(-2);1-0)=(7;0;1). \\|BC|=\sqrt{7^2+0^2+1^2} =\sqrt{49+0+1}=\sqrt{50} =\sqrt{25*2} =5\sqrt{2}.\\AC=(3-(-1); -2-(-2);1-4)=(4;0;-3).\\|AC|=\sqrt{4^2+0^2+(-3)^2}=\sqrt{16+0+9} =\sqrt{25} =5.\\cos\^A=\frac{|(-3)4+0*0+(-4)*(-3)|}{5*5} =\frac{|-12+0+12|}{25}=\frac{0}{25} =0.\\\^A=\frac{\pi }{2} .$

$cos\^B=\frac{|(-3)*7+0*0+(-4)*1|}{5*5\sqrt{2} } =\frac{|-21+0-4|}{25\sqrt{2} } =\frac{|-25|}{25\sqrt{2} } =\frac{1}{\sqrt{2} } =\frac{\sqrt{2} }{2}.\\\^B=\frac{\pi }{4}.\\cos\^C= \frac{|7*4+0*0+(-3)*1|}{5*5\sqrt{2} }=\frac{|28+0-3|}{25\sqrt{2} } =\frac{|25|}{25\sqrt{2} }=\frac{1}{\sqrt{2} }=\frac{\sqrt{2} }{2}.\\\^C=\frac{\pi }{4}.$

Ответ: |AB|=5. |BC|=5√2, |AC|=5, ∠A=π/2, ∠B=∠C=π/4.