Question

If the line $\frac{x - 1}{2}=\frac{y - 2}{3}=\frac{z - 4}{4}$ intersect the $xy$ and $yz$ plane at points $A$ and $B$ respectively. If the volume of the tetrahedron $OABC$ is $V$ cubic units (where, $O$ is the origin) and point $C$ is $\left(1,0 , 4\right)$ , then the value of $102\,V$ is equal to

Answer 1

On the $xy$ plane, $z=0$
$\Rightarrow \frac{x - 1}{2}=\frac{y - 2}{3}=-1\Rightarrow x=-1,y=-1$
$\Rightarrow $ Coordinates of point $A$ are $\left(- 1 , - 1,0\right)$
On the $yz$ plane, $x=0$
$\Rightarrow \frac{- 1}{2}=\frac{y - 2}{3}=\frac{z - 4}{4}\Rightarrow y=\frac{1}{2},z=2$
$\Rightarrow $ Coordinates of point $B$ are $\left(0 , \frac{1}{2} , 2\right)$
$\overset{ \rightarrow }{O A}=-\hat{i}-\hat{j}$
$\overset{ \rightarrow }{O B}=\frac{1}{2}\hat{j}+2\hat{k}$
$\overset{ \rightarrow }{O C}=\hat{i}+4\hat{k}$
The volume of the tetrahedron $OABC$ is $V=\frac{1}{6}\left[\overset{ \rightarrow }{O A} \,\,\,\overset{ \rightarrow }{O B} \,\,\,\overset{ \rightarrow }{O C}\right]$
$=\frac{1}{6}\begin{vmatrix} -1 & -1 & 0 \\ 0 & \frac{1}{2} & 2 \\ 1 & 0 & 4 \end{vmatrix}$
$=\left|\frac{1}{6} \left[- 1 \left(2\right) + 1 \left(- 2\right)\right]\right|=\frac{4}{6}=\frac{2}{3}$ cubic units
Hence, $102\,V=68$