Question

The locus of the point $P(\bar{r})$ which encloses a triangle ABP of area 1 aq. unit with the fixed points $A(\bar{i})$ and $B(\bar{j})$ is

• A. $x^{2}+y^{2}+z^{2}=4$
• B. $(x+2)^{2}+x^{2}+y^{2}=1$
• C. $(x+y-1)^{2}+2 z^{2}=4$
• D. $(x+y-1)^{2}+y^{2}+z^{2}=1$

Answer 1

Answer: (C)

Let point $P(r)$ is $\hat{x i}+y \hat{j}+z \hat{k}$ and $A(\hat{i})$ and $B(\hat{j})$
$AP =(x-1) \hat{i}+y \hat{j}+\hat{z k}$
$\Rightarrow A B=-\hat{i}+\hat{j}$
Area of $\triangle A B P=\frac{1}{2}|A P \times A B|$
$A P \times A B=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ x-1 & y & z \\ -1 & 1 & 0\end{vmatrix}=-z \hat{i}+z \hat{j}+(x+y-1) \hat{k}$
$|A P \times A B|=\sqrt{2 z^{2}+(x+y-1)^{2}}$
$\therefore 1=\frac{1}{2} \sqrt{2 z^{2}+(x+y-1)^{2}}$
$[\because$ Area of $\Delta A B P=1]$
$=(x+y-1)^{2}+2 z^{2}=4$