Question

A plane is parallel to lines whose direction ratios are $(1,0,-1)$ $(-1,1,0)$ and it contain the point $(1,1,1)$. If it cuts coordinate axes at $A,B,C$, then the volume of the tetrahedron $OABC$ is

• A. $\frac{9}{5}$ cu units
• B. $\frac{9}{4}$ cu units
• C. $\frac{9}{2}$ cu units
• D. None of these

Answer 1

Answer: (C)

$\frac{9}{2}$ cu units

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

$a-c=0$ and $-a+b=0$

$\Rightarrow a=b=c$

Therefore, equation of plane is $x-1+y-1+z-1=0$ or $\frac{x}{3}+\frac{y}{3}+\frac{z}{3}=1.$

Its intercepts on coordinate axes are $A(3,0,0),B(0,3,0)$ and $C(0,0,3).$ Hence, the voluem of tetrahedrron $OABC.$

$=\frac{1}{6}\left[abc\right]$

$=\frac{1}{6}\left|\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right|=\frac{27}{6}=\frac{9}{2}$ cu. units.