Question

Find the area of a triangle with vertices: (2,0,0),(0,3,0),

and (1,1,4) using the dot product only. Verify youranswer is
correct by using Heron's formula.

Answer 1

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

Explanation:

Answer 2

a. The area of the triangle is
`Area=(1)/(2)√(209)\approx 7.228`

b. Using Heron's formula the area is
`Area=√(52.25) \approx 7.228`

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

`Area=(1)/(2) |\vec{AB}* \vec{AC}|`

Let A = (2, 0, 0), B = (0, 3, 0), and C = (1, 1, 4).

We have
`\vec{AB}=(0,3,0)-(2,0,0)=(-2,3,0)` and
`\vec{AC}=(1,1,4)-(2,0,0)=(-1,1,4)`

The cross product of two vectors
`a=<a_1,a_2,a_3>` and
`b=<b_1,b_2,b_3>` is given by

`\vec{a}* \vec{b}=<a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1>`

Applying the above formula we get

`\vec{AB}* \vec{AC}=(3\cdot(4)-(1)\cdot(0))-(-2\cdot(4)-(-1)\cdot(0))+(-2\cdot(1)-(-1)\cdot(3))\\\vec{AB}* \vec{AC}=(12,8,1)`

The magnitude of a vector is given by

`\|\vc{a}\| = √(a_1^2+a_2^2+a_3^2).`

Thus,

`|\vec{AB}* \vec{AC}|=√(12^2+8^2+1^2)\\ |\vec{AB}* \vec{AC}|=√(209)`

The area of the triangle is

`Area=(1)/(2)√(209)\approx 7.228`

b. The Heron's formula is given by

`A = √(s(s-a)(s-b)(s-c))`

where
`s=(a+b+c)/(2)` 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

`a=|BC|=|B-C|\\a^2=(B_x-C_x)^2+(B_y-C_y)^2+(B_z-C_z)^2\\a=√((B_x-C_x)^2+(B_y-C_y)^2+(B_z-C_z)^2) \\a=√((0-1)^2+(3-1)^2+(0-4)^2) \\a=√(21)=4.58`

`b=|AC|=|A-C|\\b^2=(A_x-C_x)^2+(A_y-C_y)^2+(A_z-C_z)^2\\b=√((A_x-C_x)^2+(A_y-C_y)^2+(A_z-C_z)^2) \\b=√((2-1)^2+(0-1)^2+(0-4)^2) \\b=√(18)=4.24`

`c=|AB|=|A-B|\\c^2=(A_x-B_x)^2+(A_y-B_y)^2+(A_z-B_z)^2\\c=√((A_x-B_x)^2+(A_y-B_y)^2+(A_z-B_z)^2) \\c=√((2-0)^2+(0-3)^2+(0-0)^2) \\c=√(13)=3.61`

• Find the value of s

`s=(4.58+4.24+3.61)/(2) =6.22`

• Apply the Heron's formula

`A = √(s(s-a)(s-b)(s-c))\\\\A=√(6.22(6.22-4.58)(6.22-4.24)(6.22-3.61))\\\\A=√(52.25) \approx 7.228`