Find the area of a triangle with the given vertices:
(7,3,4)(1,0,6)(4,5,-2)
form 2 vectors
u = (6,3,-2) and v = (3,5,-8)
angle Ø between them , u dot v = |u| |v|cosØ
18 + 15 + 16 = √49√98cosØ
Ø = 45°
area = (1/2)(7)(√98)sin 45
= (7/2)(7√2)(1/√2)
= 24.5
thanks!
To find the area of a triangle with given vertices, you can use the formula:
Area = 1/2 * |(x1(y2-y3) + x2(y3-y1) + x3(y1-y2))|
Let's calculate the area using this formula:
Given Vertices: (7,3,4), (1,0,6), (4,5,-2)
x1 = 7, y1 = 3, z1 = 4
x2 = 1, y2 = 0, z2 = 6
x3 = 4, y3 = 5, z3 = -2
Calculating the determinant part (x1(y2-y3) + x2(y3-y1) + x3(y1-y2)):
(x1(y2 - y3)) = 7*(0 - 5) = 7*(-5) = -35
(x2(y3 - y1)) = 1*(5 - 3) = 1*2 = 2
(x3(y1 - y2)) = 4*(3 - 0) = 4*3 = 12
Now, calculating the absolute value:
|(-35 + 2 + 12)| = |(-35 + 2 + 12)| = |(-35 + 14)| = |-21| = 21
Finally, calculating the area:
Area = 1/2 * |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))|
= 1/2 * 21
= 21/2
= 10.5
Therefore, the area of the triangle with the given vertices is 10.5 square units.
To find the area of a triangle with given vertices, you can use the formula for the area of a triangle in 3D space. The formula is:
Area = 1/2 * |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|
where (x1, y1), (x2, y2), and (x3, y3) are the coordinates of the vertices of the triangle.
In this case, the coordinates of the given vertices are:
(7, 3, 4), (1, 0, 6), and (4, 5, -2).
Let's substitute the values into the formula:
x1 = 7, y1 = 3
x2 = 1, y2 = 0
x3 = 4, y3 = 5
Area = 1/2 * |(7(0 - 5) + 1(5 - 3) + 4(3 - 0))/2|
Now, we can simplify the expression inside the absolute value and calculate the area.
Area = 1/2 * |(-35 + 2 + 12)/2|
Area = 1/2 * |-21/2|
Area = 21/4
Therefore, the area of the triangle with the given vertices (7,3,4), (1,0,6), and (4,5,-2) is 21/4 or 5.25 square units.