A plane is parallel to two lines 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 A, B, C, then the volume of the tetrahedron $\mathrm{OABC}$ is cu. units.

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Aicharya · Accepted Answer

Let the equation of the plane through $(1,1,1)$ be $a(x-1)+b(y-1)+c(z-1)=0$ Since it is parallel to the straight lines having dr's $(1,0,-1)$ and $(-1,1,0)$, therefore $\begin{aligned} & \mathrm{a}-\mathrm{c}=0 \text { and }-\mathrm{a}+\mathrm{b}=0 \ & \Rightarrow \mathrm{a}=\mathrm{b}=\mathrm{c} \end{aligned}$ Therefore, equation of plane is $\mathrm{x}-1+\mathrm{y}-1+\mathrm{z}-1=0 \Rightarrow \frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{3}+\frac{\mathrm{z}}{3}=1$ Its inetrcepts on coordinate axes are $\mathrm{A}(3,0,0), \mathrm{B}(0, 3,0)$ and $\mathrm{C}(0,0, 3)$. Hence, the volume of tetrahedron $\mathrm{OABC}$. $=\frac{1}{6}\left[\begin{array}{lll}a & b & c\end{array}\right]=\frac{1}{6} \left|\begin{array}{lll}3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3\end{array}\right|=\frac{27}{6}=\frac{9}{2}$

A plane is parallel to two lines direction ratios are $(1,0,-1)$ and $(-1,1,0)$ and it contains the point…

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