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 $\left|\overrightarrow{A}\right|=\sqrt{{A_x}^2+{A_y}^2+{A_z}^2}\Rightarrow A^2={A_{xy}}^2+{A_{yz}}^2+{A_{zx}}^2$

A plane makes intercepts $a,b,c$ on $x,y,z$ axes respectively, then twice the area of $△ABC$ including both its sides $(a,b,c$ are +ve$)$ is

• A. $\sqrt{a^2+b^2+c^2}$
• B. $\sqrt{a^2{b}^2+b^2{c}^2+a^2{c}^2}$
• C. $2(ab+bc+ca)$
• D. None of these

Answer 1

The correct option is B $\sqrt{a^2{b}^2+b^2{c}^2+a^2{c}^2}$

Coordinates of points $A,B,C$ are $(a,0,0);(0,b,0)$ and $(0,0,c)$.

Now $A_x,A_y,A_z$ are projection area of $△ABC$ on plane $x=0,y=0,z=0$ respectively.

$\therefore A_x=A_{△OBC}=\frac{1}{2}\left|\overrightarrow{OB}\right|.\left|\overrightarrow{OC}\right|=\frac{1}{2}bc$

Similarly, $A_y=\frac{1}{2}ac;A_z=\frac{1}{2}ab$

$\Rightarrow A_{△ABC}=\frac{1}{2}\sqrt{a^2{b}^2+b^2{c}^2+a^2{c}^2}$