Answer:
a. The area of the triangle is [tex]Area=\frac{1}{2}\sqrt{209}\approx 7.228[/tex]
b. Using Heron's formula the area is [tex]Area=\sqrt{52.25} \approx 7.228[/tex]
Step-by-step explanation:
a. The magnitude of the product u × v is by definition the area of the parallelogram spanned by u and v when placed tail-to-tail. Hence we can use the vector product to compute the area of a triangle formed by three points A, B and C in space.
It follows that the area of the triangle is
[tex]Area=\frac{1}{2} |\vec{AB}\times \vec{AC}|[/tex]
Let A = (2, 0, 0), B = (0, 3, 0), and C = (1, 1, 4).
We have [tex]\vec{AB}=(0,3,0)-(2,0,0)=(-2,3,0)[/tex] and [tex]\vec{AC}=(1,1,4)-(2,0,0)=(-1,1,4)[/tex]
The cross product of two vectors [tex]a=<a_1,a_2,a_3>[/tex] and [tex]b=<b_1,b_2,b_3>[/tex] is given by
[tex]\vec{a}\times \vec{b}=<a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1>[/tex]
Applying the above formula we get
[tex]\vec{AB}\times \vec{AC}=(3\cdot(4)-(1)\cdot(0))-(-2\cdot(4)-(-1)\cdot(0))+(-2\cdot(1)-(-1)\cdot(3))\\\vec{AB}\times \vec{AC}=(12,8,1)[/tex]
The magnitude of a vector is given by
[tex]\|\vc{a}\| = \sqrt{a_1^2+a_2^2+a_3^2}.[/tex]
Thus,
[tex]|\vec{AB}\times \vec{AC}|=\sqrt{12^2+8^2+1^2}\\ |\vec{AB}\times \vec{AC}|=\sqrt{209}[/tex]
The area of the triangle is
[tex]Area=\frac{1}{2}\sqrt{209}\approx 7.228[/tex]
b. The Heron's formula is given by
[tex]A = \sqrt{s(s-a)(s-b)(s-c)}[/tex]
where [tex]s=\frac{a+b+c}{2}[/tex] and a, b, and c are the sides of the triangle.
To find the sides of the triangle given the vertices you must,
[tex]a=|BC|=|B-C|\\a^2=(B_x-C_x)^2+(B_y-C_y)^2+(B_z-C_z)^2\\a=\sqrt{(B_x-C_x)^2+(B_y-C_y)^2+(B_z-C_z)^2} \\a=\sqrt{(0-1)^2+(3-1)^2+(0-4)^2} \\a=\sqrt{21}=4.58[/tex]
[tex]b=|AC|=|A-C|\\b^2=(A_x-C_x)^2+(A_y-C_y)^2+(A_z-C_z)^2\\b=\sqrt{(A_x-C_x)^2+(A_y-C_y)^2+(A_z-C_z)^2} \\b=\sqrt{(2-1)^2+(0-1)^2+(0-4)^2} \\b=\sqrt{18}=4.24[/tex]
[tex]c=|AB|=|A-B|\\c^2=(A_x-B_x)^2+(A_y-B_y)^2+(A_z-B_z)^2\\c=\sqrt{(A_x-B_x)^2+(A_y-B_y)^2+(A_z-B_z)^2} \\c=\sqrt{(2-0)^2+(0-3)^2+(0-0)^2} \\c=\sqrt{13}=3.61[/tex]
[tex]s=\frac{4.58+4.24+3.61}{2} =6.22[/tex]
[tex]A = \sqrt{s(s-a)(s-b)(s-c)}\\\\A=\sqrt{6.22(6.22-4.58)(6.22-4.24)(6.22-3.61)}\\\\A=\sqrt{52.25} \approx 7.228[/tex]
Answer:
The magnitude of the product u × v is by definition the area of the parallelogram spanned by u and v when placed tail-to-tail. Hence we can use the vector product to compute the area of a triangle formed by three points A, B and C in space.
It follows that the area of the triangle is
Let A = (2, 0, 0), B = (0, 3, 0), and C = (1, 1, 4).
We have and
The cross product of two vectors and is given by
Applying the above formula we get
The magnitude of a vector is given by
Thus,
The area of the triangle is
b. The Heron's formula is given by
where and a, b, and c are the sides of the triangle.
To find the sides of the triangle given the vertices you must,
Compute side a, b, and c from coordinates using the Pythagorean theorem
Find the value of s
Apply the Heron's formula
Step-by-step explanation:
