Question

Find the area of the triangle whose vertices area A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2).

Answer 1

Answer:

Area of triangle

$=\frac{3\sqrt{3} }{2}\:\:units$

Step-by-step explanation:

Direction ratio's of AB(2-1,3-2,1-3)

(1,1,-2)

Direction ratio's of AC(3-1,1-2,2-3)

(2,-1,-1)

$cos\:A=\frac{2(1)+(-1)1+(-1)(-2)}{\sqrt{1+1+4} \sqrt{4+1+1} } \\\\cos\:A=\frac{3}{\sqrt{6}\sqrt{6} } \\\\cos\:A=\frac{1}{2} \\\\A=\frac{\pi}{3}$

Distance AB=

$\sqrt{(2-1)^{2} +(3-2)^{2} +(1-3)^{2} } \\\\=\sqrt{1+1+4} \\\\=\sqrt{6} \\\\AC=\sqrt{(3-1)^{2}+(1-2)^{2} +(2-3)^{2}} \\\\=\sqrt{4+1+1} \\\\=\sqrt{6}$

Area of triangle ABC

$\frac{1}{2} |AB||AC|\:\:sin A\\\\=\frac{1}{2} \sqrt{6}\sqrt{6}\:sin\:\frac{\pi }{3}\\\\=\frac{1}{2}(6)\frac{\sqrt{3}}{2}\\\\=\frac{3\sqrt{3}}{2}\:\:units$

Answer 2

Answer:

Step-by-step explanation:

See attachment

It is not m it is units