Question

Three lines are given by

$\overrightarrow{r}=\lambda \hat{i},\lambda \in R$

$\overrightarrow{r}=\mu \left(\hat{i}+\hat{j}\right),\mu \in R$ and

$\overrightarrow{r}=v\left(\hat{i}+\hat{j}+\hat{k}\right),v\in R$

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 $△$ then the value of ${(6△)}^2$ equals __________.

Answer 1

Step 1. Finding point $\mathit{A}$:

$\overrightarrow{r}=\lambda \hat{i}$ and $x+y+z=1$, we get

$\Rightarrow$$\lambda +0+0=1$

$\Rightarrow$ $\lambda =1$

$\Rightarrow$ $A=\left(1,0,0\right)$

Step 2. Finding point $\mathit{B}$:

$\overrightarrow{r}=\mu \left(\hat{i}+\hat{j}\right)$ and $x+y+z=1$, we get

$\Rightarrow$$\mu +\mu +0=1$

$\Rightarrow$ $\mu =\frac{1}{2}$

$\Rightarrow$ $B=\left(\frac{1}{2},\frac{1}{2},0\right)$

Step 3. Finding point $\mathit{C}$:

$\overrightarrow{r}=v\left(\hat{i}+\hat{j}+\hat{k}\right)$ and $x+y+z=1$, we get

$\Rightarrow$$v+v+v=1$

$\Rightarrow$ $v=\frac{1}{3}$

$\Rightarrow$ $C=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$

Step 4. Area of $\mathbf{△}\mathit{A}\mathit{B}\mathit{C}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|$

Now, $\overrightarrow{AB}=-\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}$and $\overrightarrow{AC}=-\frac{2}{3}\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{3}\hat{k}$

$\begin{array}{rcl}△& =& \frac{1}{2}\left|\left(-\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}\right)\times \left(-\frac{2}{3}\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{3}\hat{k}\right)\right|\\ & =& \frac{\sqrt{3}}{12}\end{array}$

$\begin{array}{rcl}\therefore {(6△)}^2& =& {\left(\frac{6\sqrt{3}}{12}\right)}^2\\ & =& \frac{3}{4}\\ & =& 0.75\end{array}$

Hence, the value of ${(6△)}^2$ is $\begin{array}{rcl}& & \mathbf{0}\mathbf{.}\mathbf{75}\end{array}$