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

Answer 1

Answer: (C)

Let point $P(r)$ is $\widehat{xi}+y\hat{j}+z\hat{k}$ and $A(\hat{i})$ and $B(\hat{j})$
$AP=(x-1)\hat{i}+y\hat{j}+\widehat{zk}$
$\Rightarrow AB=-\hat{i}+\hat{j}$
Area of $△ABP=\frac{1}{2}|AP\times AB|$
$AP\times AB=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ x-1& y& z\\ -1& 1& 0\end{array}\right|=-z\hat{i}+z\hat{j}+(x+y-1)\hat{k}$
$|AP\times AB|=\sqrt{2z^2+(x+y-1{)}^2}$
$\therefore 1=\frac{1}{2}\sqrt{2z^2+(x+y-1{)}^2}$
$[\because$ Area of $\Delta ABP=1]$
$=(x+y-1{)}^2+2z^2=4$