Question

Find the area of the triangle with the given vertices. (Hint: \(\frac{1}{2}\|\mathbf{u} \times \mathbf{v}\|\) is the area of the triangle having \(u\) and \(v\) as adjacent sides.) $$ (2,-7,3),(-1,5,8),(4,6,-1) $$

Short answer

The area of the triangle with the given vertices is 666 square units.

Step-by-step solution

Finding the vectors

The first step is to find the vectors that represent the adjacent sides of the triangle from the given vertices. Let's call the given vertices A(2,-7,3), B(-1,5,8) and C(4,6,-1). Hence, vector \(AB = B - A = (-1 - 2, 5 -(-7), 8 -3) = (-3, 12, 5)\) and vector \(AC = C - A = (4 - 2, 6 -(-7), -1 -3) = (2, 13, -6)\).

Cross Product

Next, calculate the cross product of these two vectors. The cross product of two vectors \(AB\) and \(AC\) would be given by the determinant of this 3x3 matrix: \[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \-3 & 12 & 5 \2 & 13 & -6 \ \end{vmatrix} \] Expanding this determinant gives \(\mathbf{i}(12*(-6) - 5*13) - \mathbf{j}*(-3*(-6) - 5*2) + \mathbf{k}*(-3*13 - 12*2) = -144\mathbf{i} + 8\mathbf{j} -78\mathbf{k}\).

Magnitude

Compute the magnitude of this cross product vector. The magnitude of a vector \((a,b,c)\) is given by \(\sqrt{a^2 + b^2 + c^2}\). Hence, the magnitude of our calculated cross product vector is \(\sqrt{(-144)^2 + 8^2 + (-78)^2} = 1,332\).

Calculate Area

Now we have everything to calculate the area of the triangle. The area is half of the magnitude of the cross product of the vectors: \(\frac{1}{2} * 1,332 = 666\).