Find the area of the triangle whose vertices are A(1,0,1), B(2,-1,3), C(5,2,2).
This is calculus?
Find the side lengths, and then use Heron's formula.
AB = √6
and so on
or, find vectors AB and AC and then the area of the triangle is 1/2 their cross-product.
To find the area of a triangle given its vertices in three-dimensional space, we can use the formula:
Area = 1/2 * |(B - A) x (C - A)|
Here, (B - A) represents the vector from point A to B, and (C - A) represents the vector from point A to C. The cross product of these two vectors gives us a vector that is perpendicular to the plane containing the triangle. Taking the magnitude of this cross product gives us the area of the parallelogram formed by these two vectors. Finally, dividing it by 2 gives us the area of the triangle.
First, let's calculate the vectors (B - A) and (C - A):
(B - A) = (2 - 1, -1 - 0, 3 - 1) = (1, -1, 2)
(C - A) = (5 - 1, 2 - 0, 2 - 1) = (4, 2, 1)
Next, let's calculate the cross product of these two vectors:
(B - A) x (C - A) = [(1 * 1) - (-1 * 2), (-1 * 4) - (1 * 1), (1 * 2) - (-1 * 4)] = (3, -3, 6)
Now, let's find the magnitude of this vector:
|(B - A) x (C - A)| = sqrt((3^2) + (-3^2) + (6^2)) = sqrt(9 + 9 + 36) = sqrt(54)
Finally, let's calculate the area:
Area = 1/2 * sqrt(54) = sqrt(54) / 2 ≈ 3.674/2 ≈ 1.837
Therefore, the area of the triangle with vertices A(1,0,1), B(2,-1,3), and C(5,2,2) is approximately 1.837 square units.