Question

If $A_{xy},A_{yz},A_{zx}$ be the area of projections of an area $A$ on the $xy,yz$ abd $zx$ planes respectively, then considering $\overrightarrow{A}$, a vector quantity whose direction in normal to surface $A$ and its component along the coordinate axes be $A_x,A_y$ and $A_z$ such that $\overrightarrow{A}=A_{x}i+A_{y}j+A_{z}k$, therefore we can say $\left|A_x\right|=A_{zy},\left|A_y\right|=A_{xz},\left|A_z\right|=A_{xy}$ and hence the area $\left|\overrightarrow{A}\right|=\sqrt{{A_x}^2+{A_y}^2+{A_z}^2}\Rightarrow A^2={A_{xy}}^2+{A_{yz}}^2+{A_{zx}}^2$

Through a point $P(h,k,l)$ a plane is drawn at right angle to $OP$ to meet the coordinate axes in $A,B,C$ and $OP=p,$ then the area of $△ABC$ is

• A. $\frac{p^5}{3hkl}$
• B. $\frac{p^5}{hkl}$
• C. $\frac{p^5}{2hkl}$
• D. None of these.

Answer 1

The correct option is C $\frac{p^5}{2hkl}$

$OP=p=\sqrt{h^2+k^2+l^2}$

Direction cosines of $OP$ are $(\frac{h}{\sqrt{h^2+k^2+l^2}},\frac{k}{\sqrt{h^2+k^2+l^2}},\frac{l}{\sqrt{h^2+k^2+l^2}})$

$\because OP$ is normal to plane; equation will be

$\left(\frac{h}{\sqrt{h^2+k^2+l^2}}\right)x+\left(\frac{k}{\sqrt{h^2+k^2+l^2}}\right)y+\left(\frac{l}{\sqrt{h^2+k^2+l^2}}\right)z=\sqrt{h^2+k^2+l^2}$

$\Rightarrow hx+ky+lz=h^2+k^2+l^2=p^2$

$\therefore A\equiv (\frac{p^2}{h},0,0);B\equiv (0,\frac{p^2}{k},0);C\equiv (0,0,\frac{p^2}{l})$

$\Rightarrow A_{xy}=\frac{1}{2}\frac{p^4}{\left|hk\right|};A_{yz}=\frac{1}{2}\frac{p^4}{\left|kl\right|};A_{zx}=\frac{1}{2}\frac{p^4}{\left|hl\right|}$

$\Rightarrow △=\sqrt{{\left(A_{xy}\right)}^2+{\left(A_{yz}\right)}^2+{\left(A_{zx}\right)}^2}=\sqrt{\frac{p^8}{4h^2{k}^2{l}^2}(h^2+k^2+l^2)}$

$=\sqrt{\frac{p^8.p^2}{4h^2{k}^2{l}^2}}=\frac{p^5}{2hkl}$