Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

To find point 'A', consider the line $\vec{r}=\lambda \hat{i}$ and the plane $x+y+z=1$.

The general point on the line is $(\lambda ,0,0)$.

Substituting this into the plane equation:

$\lambda +0+0=1$

$\lambda =1$

Thus, point A is $(1,0,0)$ ...(i)

For point 'B', consider the line $\vec{r}=\mu (\hat{i}+\hat{j})$.

The general point on this line is $(\mu ,\mu ,0)$.

Substituting this into the plane equation:

$\mu +\mu +0=1$

$2\mu =1$

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

Thus, point B is $\left(\frac{1}{2},\frac{1}{2},0\right)$ ...(ii)

Similarly, for point 'C', consider the line $\vec{r}=\nu (\hat{i}+\hat{j}+\hat{k})$.

The general point on this line is $(\nu ,\nu ,\nu )$.

Substituting this into the plane equation:

$\nu +\nu +\nu =1$

$3\nu =1$

$\nu =\frac{1}{3}$

Thus, point C is $\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$ ...(iii)

Now, we find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:

$\overrightarrow{AB}=B-A=\left(\frac{1}{2}-1\right)\hat{i}+\left(\frac{1}{2}-0\right)\hat{j}+(0-0)\hat{k}=-\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}$

$\overrightarrow{AC}=C-A=\left(\frac{1}{3}-1\right)\hat{i}+\left(\frac{1}{3}-0\right)\hat{j}+\left(\frac{1}{3}-0\right)\hat{k}=-\frac{2}{3}\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{3}\hat{k}$

The area of triangle $△ABC$ is given by:

$△ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$

First, calculate the cross product $\overrightarrow{AB}\times \overrightarrow{AC}$:

$$\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -\frac{1}{2}& \frac{1}{2}& 0\\ -\frac{2}{3}& \frac{1}{3}& \frac{1}{3}\end{array}\right|$$

$$=\hat{i}\left(\frac{1}{2}\cdot \frac{1}{3}-0\cdot \frac{1}{3}\right)-\hat{j}\left(-\frac{1}{2}\cdot \frac{1}{3}-0\cdot -\frac{2}{3}\right)+\hat{k}\left(-\frac{1}{2}\cdot \frac{1}{3}-\frac{1}{2}\cdot -\frac{2}{3}\right)$$

$$=\hat{i}\left(\frac{1}{6}\right)-\hat{j}\left(-\frac{1}{6}\right)+\hat{k}\left(-\frac{1}{6}+\frac{1}{3}\right)$$

$$=\frac{1}{6}\hat{i}+\frac{1}{6}\hat{j}+\frac{1}{6}\hat{k}$$

Now, find the magnitude of the cross product:

$$|\overrightarrow{AB}\times \overrightarrow{AC}|=\sqrt{{\left(\frac{1}{6}\right)}^2+{\left(\frac{1}{6}\right)}^2+{\left(\frac{1}{6}\right)}^2}=\sqrt{\frac{1}{36}+\frac{1}{36}+\frac{1}{36}}=\sqrt{\frac{3}{36}}=\sqrt{\frac{1}{12}}=\frac{1}{2\sqrt{3}}$$

So, the area of $△ABC$ is:

$$△=\frac{1}{2}\cdot \frac{1}{2\sqrt{3}}=\frac{1}{4\sqrt{3}}=\frac{\sqrt{3}}{12}$$

Finally, calculate $(6△{)}^2$:

$$(12△{)}^2=3$$