A plane is parallel to two lines whose direction ratios are $2,0,-2$ and $-2,2,0$ and it contains the point…
A plane is parallel to two lines whose direction ratios are $2,0,-2$ and $-2,2,0$ and it contains the point $(2,2,2)$. If it cuts co-ordinate axes at $\mathrm{A}, \mathrm{B}, \mathrm{C}$ then the volume of tetrahedron $\mathrm{OABC}$ (in cubic units) is
216
6
36
9
Solution
The equation of plane passing through $(2,2,2)$
is $\mathrm{a}(x-2)+\mathrm{b}(y-2)+\mathrm{c}(\mathrm{z}-2)=0$
Also, $2 \mathrm{a}-2 \mathrm{c}=0$
$-2 a+2 b=0$
$\Rightarrow \mathrm{a}=\mathrm{b}=\mathrm{c}$
$\begin{aligned} & x+y+z-6=0 \\ & \Rightarrow \frac{x}{6}+\frac{y}{6}+\frac{z}{6}=1\end{aligned}$
It cuts the co-ordinate axes at $\mathrm{A}(6,0,0)$, $\mathrm{B}(0,6,0)$ and $\mathrm{C}(0,0,6)$
$\therefore \quad \overline{\mathrm{a}}=6 \hat{\mathrm{i}}, \overline{\mathrm{b}}=6 \hat{\mathrm{j}}, \overline{\mathrm{c}}=6 \hat{\mathrm{k}}$
$\therefore \quad$ Volume of tetrahedron $=\frac{1}{6}\left[\begin{array}{lll}-\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
$\begin{aligned} & =\frac{1}{6}\left|\begin{array}{lll}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right| \\ & =36 \text { cubic units }\end{aligned}$