Let the height of a triangle be l, where a triangle has a base of 5 units. Points are given as A(1, -1, 2), B(-2, 1, 0) (direction cosine of the line), and C(3, 0, 4). Find the area of the triangle (in square units, rounded to 3 decimal places).
In a right angle \(\triangle ABC\), \(\angle A = 90°\) and sides \(a, b, c\) are, respectively, 5 cm, 4 cm and 3 cm. If a force \(\vec{F}\) has moments 0, 9 and 16 in N cm units, respectively, about vertices \(A\), \(B\) and \(C\), then magnitude of \(\vec{F}\) is
For a triangle ABC, let $\vec{p}=\overrightarrow{BC}$, $\vec{q}=\overrightarrow{CA}$ and $\vec{r}=\overrightarrow{BA}$. If $|\vec{p}|=2\sqrt{3}$, $|\vec{q}|=2$ and $\cos\theta=\dfrac{1}{\sqrt{3}}$, where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$, then $|\vec{p}\times(\vec{q}-3\vec{r})|^2+3|\vec{r}|^2$ is equal to: