A plane is parallel to two lines whose direction ratios are $1,0,-1$ and $-1,1,0$ and it contains the point…
A plane is parallel to two lines whose direction ratios are $1,0,-1$ and $-1,1,0$ and it contains the point $(1,1,1)$. If it cuts the co-ordinate axes at $\mathrm{A}, \mathrm{B}, \mathrm{C}$, then the volume of the tetrahedron $\mathrm{OABC}$ (in cubic units) is
$\frac{9}{4}$
$\frac{9}{2}$
$9$
$27$
Solution
Equation of the plane passing through $(1,1,1)$ is given as
$\mathrm{a}(x-1)+\mathrm{b}(y-1)+\mathrm{c}(\mathrm{z}-1)=0$
As the plane is parallel to the lines having direction ratios $1,0,-1$ and $-1,1,0$, we get
$\begin{aligned}
& a-c=0 \text { and }-a+b=0 \\
& \Rightarrow a=b=c
\end{aligned}$
$\therefore \quad$ From (i) and (ii), we get
$\begin{aligned}
& x-1+y-1+z-1=0 \\
& \therefore \quad x+y+z=3 \quad \Rightarrow \frac{x}{3}+\frac{y}{3}+\frac{z}{3}=1
\end{aligned}$
$\therefore \quad$ Co-ordinates of A, B, C are $(3,0,0),(0,3,0)$ and $(0,0,3)$ respectively.
$\therefore \quad$ Volume of tetrahedron $\mathrm{OABC}$
$\begin{aligned}
& =\frac{1}{6}\left|\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right| \\
& =\frac{1}{6} \times 27 \\
& =\frac{9}{2} \text { cu. units }
\end{aligned}$