Sketch the triangle with vertices at the origin O, P=(5,4,0), and Q=(2,1,4), and compute its area using cross products.

Where, [tex] \sf \vec{OP} [/tex] and [tex] \sf \vec{OQ} [/tex] are the vectors from [tex] \sf O [/tex] to [tex] \sf P [/tex] and [tex] \sf O [/tex] to [tex] \sf Q [/tex].

Given:

[tex] \sf P = (5, 4, 0) [/tex] and [tex] \sf Q = (2, 1, 4) [/tex]

Vectors:

[tex] \sf \vec{OP} = (5, 4, 0) [/tex] and [tex] \sf \vec{OQ} = (2, 1, 4) [/tex].

The cross product [tex] \sf \vec{OP} \times \vec{OQ} [/tex] is computed as follows:

[tex]\vec{OP} \times \vec{OQ} =\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 4 & 0 \\ 2 & 1 & 4 \end{vmatrix}[/tex]

The cross product of two vectors [tex] \sf \vec{a} = (a_1, a_2, a_3) [/tex] and [tex] \sf \vec{b} = (b_1, b_2, b_3) [/tex] in three-dimensional space is given by the formula:

[tex] \vec{a} \times \vec{b} = \left( a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \right)[/tex]

In this case, [tex] \sf \vec{OP} = (5, 4, 0) [/tex] and [tex] \sf \vec{OQ} = (2, 1, 4) [/tex], so:

[tex] \sf a_1 = 5 [/tex], [tex] \sf a_2 = 4 [/tex], [tex] \sf a_3 = 0 [/tex]
[tex] \sf b_1 = 2 [/tex], [tex] \sf b_2 = 1 [/tex], [tex] \sf b_3 = 4 [/tex]

Applying the formula:

For the [tex] \sf \mathbf{i} [/tex] component (x-component): [tex] \sf a_2b_3 - a_3b_2 = 4 \cdot 4 - 0 \cdot 1 = 16 - 0 = 16 [/tex]
For the [tex] \sf \mathbf{j} [/tex] component (y-component): [tex] \sf a_3b_1 - a_1b_3 = 0 \cdot 2 - 5 \cdot 4 = 0 - 20 = -20 [/tex] (Notice that the sign here indicates the direction of the component along the [tex] \sf \mathbf{j} [/tex] axis, meaning it's in the opposite direction of the [tex] \sf \mathbf{j} [/tex] axis.)
For the [tex] \sf \mathbf{k} [/tex] component (z-component): [tex] \sf a_1b_2 - a_2b_1 = 5 \cdot 1 - 4 \cdot 2 = 5 - 8 = -3 [/tex]

[tex]= (16 - 0) \mathbf{i} - (20 - 0) \mathbf{j} + (5 - 8) \mathbf{k}[/tex]

[tex]= (16, -20, -3)[/tex]

The magnitude of this cross-product is:

[tex]\| \vec{OP} \times \vec{OQ} \| [/tex]

[tex]= \sqrt{16^2 + (-20)^2 + (-3)^2} [/tex]

[tex]= \sqrt{256 + 400 + 9} [/tex]

[tex]= \sqrt{665}[/tex]

Thus, the area of the triangle is:

[tex]\textbf{Area} = \frac{1}{2} \sqrt{665}[/tex]