Using vectors, find the area of the triangle ABC with vertices A(1, 2…

The correct answer is Option (1) → $\frac{1}{2} \sqrt{274}$ ##

Here, $AB = (2 - 1)\hat{\mathbf{i}} + (-1 - 2)\hat{\mathbf{j}} + (4 - 3)\hat{\mathbf{k}}$

$[∵AB = (x_2 - x_1)\hat{\mathbf{i}} + (y_2 - y_1)\hat{\mathbf{j}} + (z_2 - z_1)\hat{\mathbf{k}}]$

$= \hat{\mathbf{i}} - 3\hat{\mathbf{j}} + \hat{\mathbf{k}}$

and $AC = (4 - 1)\hat{\mathbf{i}} + (5 - 2)\hat{\mathbf{j}} + (-1 - 3)\hat{\mathbf{k}}$

$= 3\hat{\mathbf{i}} + 3\hat{\mathbf{j}} - 4\hat{\mathbf{k}}$

$∴AB \times AC = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix}$

$= \hat{\mathbf{i}}(12 - 3) - \hat{\mathbf{j}}(-4 - 3) + \hat{\mathbf{k}}(3 + 9)$

$= 9\hat{\mathbf{i}} + 7\hat{\mathbf{j}} + 12\hat{\mathbf{k}}$

and $|AB \times AC| = \sqrt{9^2 + 7^2 + 12^2}$

$= \sqrt{81 + 49 + 144}$

$= \sqrt{274}$

$∴ \text{Area of } \triangle ABC = \frac{1}{2} |AB \times AC|$

$= \frac{1}{2}\sqrt{274} \text{ sq units}$