Question

Find the area of the triangle determined by the points \(\mathrm{P}_{1}(2,2,0), \mathrm{P}_{2}(-1,0,1)\) and \(\mathrm{P}_{3}(0,4,3)\) by using the cross-product.

Short answer

The area of the triangle determined by points \(\mathrm{P}_{1}(2,2,0)\), \(\mathrm{P}_{2}(-1,0,1)\), and \(\mathrm{P}_{3}(0,4,3)\) is \(\frac{\sqrt{116}}{2}\).

Step-by-step solution

Find the Vectors

We are given points \(\mathrm{P}_{1}(2,2,0)\), \(\mathrm{P}_{2}(-1,0,1)\), and \(\mathrm{P}_{3}(0,4,3)\). To find the vectors formed by these points, subtract the coordinates of each point as follows: $$\vec{v} = \mathrm{P}_{2} - \mathrm{P}_{1} = (-1-2, 0-2, 1-0) = (-3, -2, 1)$$ $$\vec{w} = \mathrm{P}_{3} - \mathrm{P}_{1} = (0-2, 4-2, 3-0) = (-2, 2, 3)$$

Compute the Cross-Product

Now we will compute the cross-product of the vectors \(\vec{v}\) and \(\vec{w}\) as follows: $$\vec{v} \times \vec{w} = \begin{bmatrix} i & j & k \\ -3 & -2 & 1 \\ -2 & 2 & 3 \end{bmatrix} = i(-2 \cdot 3 - 1 \cdot 2) - j(-3 \cdot 3 - 1 \cdot 2) + k(-3 \cdot 2 - (-2) \cdot 2)$$ $$\vec{v} \times \vec{w} = (-6,-8,4)$$

Calculate the Magnitude

Next, we will compute the magnitude of the cross-product: $$\Vert \vec{v} \times \vec{w} \Vert = \sqrt{(-6)^{2} + (-8)^{2} + (4)^{2}} = \sqrt{36 + 64 + 16} = \sqrt{116}$$

Compute the Area of the Triangle

Finally, the area of the triangle determined by the given points is half the magnitude of the cross-product: Area \(= \frac{1}{2} \Vert \vec{v} \times \vec{w} \Vert = \frac{1}{2} \sqrt{116} = \frac{\sqrt{116}}{2}\) The area of the triangle determined by points \(\mathrm{P}_{1}(2,2,0)\), \(\mathrm{P}_{2}(-1,0,1)\), and \(\mathrm{P}_{3}(0,4,3)\) is \(\frac{\sqrt{116}}{2}\).