Question

Three lines are given by
$\overrightarrow{r}=\lambda \hat{i},\lambda \in R$
$\overrightarrow{r}=\mu (\hat{i}+\hat{j}),\mu \in R$
$\overrightarrow{r}=\nu (\hat{i}+\hat{j}+\hat{k})\nu \in R.$
Let the lines cut the plane $x+y+z=1$ at the points $A,B\text{and}C$ respectively. If the area of the triangle $ABC$ is $△,$ then the value of$(12△{)}^2$ equals to

Answer 1

First Line: $\overrightarrow{r}=\lambda \hat{i},\Rightarrow x_1=\lambda ,y_1=0,z_1=0$
Second Line: $\overrightarrow{r}=\mu (\hat{i}+\hat{j}),\Rightarrow x_2=\mu ,y_2=\mu ,z_2=0$
Third Line: $\overrightarrow{r}=\nu (\hat{i}+\hat{j}+\hat{k}),\Rightarrow x_3=\nu ,y_3=\nu ,z_3=\nu$
Given that lines cut the plane $x+y+z=1$
So, first line cuts the plane at point $\lambda +0+0=1\Rightarrow \lambda =1$ so point $A=(1,0,0)$
Similarly second line cuts the plane at point $\mu +\mu +0=1\Rightarrow 2\mu =1$ so point $B=(\frac{1}{2},\frac{1}{2},0)$
and third line cuts the plane at point $\nu +\nu +\nu =1\Rightarrow 3\nu =1$ so point $C=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$
$\therefore$ Area of triangle, $△=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$ where,
$\overrightarrow{AB}=[(\frac{1}{2}-1)\hat{i}+\frac{1}{2}\hat{j}+0\hat{k})=-\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}]$
$\overrightarrow{AC}=[(\frac{1}{3}-1)\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{3}\hat{k})=-\frac{2}{3}\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{3}\hat{k}]$
So, $△=\frac{1}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -\frac{1}{2}& \frac{1}{2}& 0\\ -\frac{2}{3}& \frac{1}{3}& \frac{1}{3}\end{array}\right|$
$=\frac{1}{2}|\frac{1}{6}\hat{i}-(-\frac{1}{6})\hat{j}+(-\frac{1}{6}+\frac{2}{6})\hat{k}|$
$=\frac{1}{2}|\frac{1}{6}\hat{i}+\frac{1}{6}\hat{j}+\frac{1}{6})\hat{k}|$
$=\frac{1}{2}\sqrt{(\frac{1}{6}{)}^2+(\frac{1}{6}{)}^2+(\frac{1}{6}{)}^2}=\frac{\sqrt{3}}{12}$
$\therefore (12△{)}^2={(12\times \frac{\sqrt{3}}{12})}^2=3$