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

Answer: 0.75

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
$\Delta =\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|$
$\left(i.e.\overrightarrow{AB}=-\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}and\overrightarrow{AC}=-\frac{2}{3}\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{3}\hat{k}\right)$
$\Rightarrow \Delta =\frac{1}{2}\left|\left(-\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}\right)\times \left(-\frac{2}{3}\hat{i}+\frac{1}{3}\hat{j}+\hat{k}\right)\right|$
$=\frac{\sqrt{3}}{12}\Rightarrow {\left(6\Delta \right)}^2={\left(\frac{6\sqrt{3}}{12}\right)}^2=\frac{3}{4}=0.75$