Question

Three lines are given by
$\vec{r}=\lambda\,\hat{i},\,\lambda\,\in\,\mathbb{R}$
$\vec{r}=\mu\left(\hat{i}+\hat{j}\right), \mu\,\in\,\mathbb{R}$ and
$\vec{r}=v\left(\hat{i}+\hat{j}+\hat{k}\right), v\,\in\,\mathbb{R}.$
Let the lines cut the plane $x + y + z = 1$ at the points A, B and C respectively. If the area of the triangle ABC is A then the value of ($6\Delta)^2$ equals___

Answer 1

Finding point A
$\vec{r}=\lambda\hat{i}$ and $x + y + z = 1,$ we get
$\lambda+0+0=1 \Rightarrow \lambda=1 \Rightarrow A=\left(1, 0 , 0\right)$
Similarly for point B
$\vec{r}=\mu\left(\hat{i}+\hat{j}\right)$ and $x+y+z=1$
We get $\mu+\mu+0=1 \Rightarrow \mu=\frac{1}{2} \Rightarrow B\left(\frac{1}{2}, \frac{1}{2},0\right)$
and for point C $\vec{r}=v\left(\hat{i}+\hat{j}+\hat{k}\right) \Rightarrow v=\frac{1}{3} \Rightarrow C\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$
Area of triangle ABC
$\Delta=\frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|$
$\left(i.e. \overrightarrow{AB}=-\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}\,and \,\overrightarrow{AC}=-\frac{2}{3}\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{3}\hat{k}\right)$
$\Rightarrow \Delta=\frac{1}{2}\left|\left(-\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}\right)\times\left(-\frac{2}{3}\hat{i}+\frac{1}{3}\hat{j}+\hat{k}\right)\right|$
$=\frac{\sqrt{3}}{12} \Rightarrow \left(6\Delta\right)^{2}=\left(\frac{6\sqrt{3}}{12}\right)^{2}=\frac{3}{4}=0.75$