Question

A plane is passing through $P(h,k,l)$ which is drawn at right angle to $OP$, where $O$ is the origin. If the plane meets the coordinates axis at $A,B,C$ respectively. Let $OP=p;(p>0)$, then which of the following is/are correct.

• A. When $h=k=l=x$, then area of the $△ABC$ is $\frac{9\sqrt{3}{x}^2}{2}$
• B. When $h+k+l=0$, then $hk+kl+hl<0$
• C. Coordinates of $B=(0,\frac{p^2}{k},0)$
• D. When the plane passes through $(1,1,1)\text{and}h+l=0$, then $p\in (0,1)$

Answer 1

Answer: (A), (B), (C), (D)

When the plane passes through $(1,1,1)\text{and}h+l=0$, then $p\in (0,1)$

Using distance formula,
$OP=\sqrt{h^2+k^2+l^2}=p\cdots \left(1\right)$

When $h+k+l=0$,
$h^2+k^2+l^2>0\Rightarrow (h+k+l{)}^2-2(hk+kl+hl)>0\Rightarrow hk+kl+hl<0$

Dr's of $OP$,
$\frac{h}{\sqrt{h^2+k^2+l^2}},\frac{k}{\sqrt{h^2+k^2+l^2}},\frac{l}{\sqrt{h^2+k^2+l^2}}=\frac{h}{p},\frac{k}{p},\frac{l}{p}$
Let the equation of the plane is,
$\frac{h}{p}x+\frac{k}{p}y+\frac{l}{p}z=d$
It contains $(h,k,l)$,
$\frac{h^2+k^2+l^2}{p}=d\Rightarrow d=p$
Therefore the equation of the plane is,
$hx+ky+lz=p^2$

When the plane passing through $(1,1,1)$ and $h+l=0$
$h+k+l=p^2\Rightarrow k=p^2$
From equation $\left(1\right)$,
$\sqrt{h^2+p^4+l^2}=p\Rightarrow h^2+p^4+l^2=p^2\Rightarrow h^2+l^2=p^2-p^4=p^2(1-p^2)$
We know that,
$h^2+l^2>0\Rightarrow p^2(1-p^2)>0\Rightarrow p\in (0,1)$

Equation of the plane is,
$hx+ky+lz=p^2$
So coordinate of
$A=(\frac{p^2}{h},0,0)B=(0,\frac{p^2}{k},0)C=(0,0,\frac{p^2}{l})$

Now,
$\overrightarrow{AB}=-\frac{p^2}{h}\hat{i}+\frac{p^2}{k}\hat{j}\overrightarrow{BC}=-\frac{p^2}{k}\hat{j}+\frac{p^2}{l}\hat{k}$
So the area of the $△ABC$
$\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{BC}|=\frac{1}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -\frac{p^2}{h}& \frac{p^2}{k}& 0\\ 0& -\frac{p^2}{k}& \frac{p^2}{l}\end{array}\right|=\frac{p^4}{2}|\frac{\hat{i}}{kl}+\frac{\hat{j}}{hl}+\frac{\hat{k}}{kh}|=\frac{p^5}{2hkl}$

When $h=k=l=x$, then area of the
$△ABC=\frac{9\sqrt{3}{x}^2}{2}$