Vector Algebra Question 18
Question 18 - 31 January - Shift 2
The foot of perpendicular from the origin O to a plane P which meets the co-ordinate axes at the points A, B, C is $(2, a, 4), a \in N$. If the volume of the tetrahedron $OABC$ is 144 unit $^{3}$, then which of the following points is NOT on P?
(1) $(2,2,4)$
(2) $(0,4,4)$
(3) $(3,0,4)$
(4) $(0,6,3)$
Show Answer
Answer: (3)
Solution:
Formula: Volume of tetrahedron, Equation of a plane
Equation of Plane:
$ \begin{aligned} & (2 \hat{i}+a + 4 \hat{k}) \cdot[(x-2) \hat{i}+(y-a) \hat{j}+(z-4) \hat{k}]=0 \\ & \Rightarrow 2 x+ay+4 z=20+a^{2} \\ & \Rightarrow A \equiv(\frac{20+a^{2}}{2}, 0,0) \\ & B \equiv(0, \frac{20+a^{2}}{a}, 0) \\ & C \equiv(0,0, \frac{20+a^{2}}{4}) \\ & \Rightarrow \text{ Volume of tetrahedron } \\ & =\frac{1}{6}[\overrightarrow{{}a} \overrightarrow{{}c}] \\ & =\frac{1}{6} \overrightarrow{{}a} \cdot(\overrightarrow{{}b} \times \overrightarrow{{}c}) \\ & \Rightarrow \frac{1}{6}(\frac{20+a^{2}}{2}) \cdot(\frac{20+a^{2}}{a}) \cdot(\frac{20+a^{2}}{4})=144 \\ & \Rightarrow(20+a^{2})^{3}=144 \times 48 \times a \\ & \Rightarrow m=2 \\ & \Rightarrow \text{ Equation of plane is } 2 x+2 y+4 z=24 \end{aligned} $
Or $x+y+2 z=12$
$\Rightarrow(3,0,4) \quad$ Not lies on the Plane P.
$x+y+2 z=12$
