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

The area of $△$ whose vertices are $(1,2,3),(2,4,1),(3,4,5)$ respectively is

• A. $10$
• B. $16$
• C. $\sqrt{26}$
• D. $212$

Answer 1

Answer: (C)

$\sqrt{26}$

Let $A\equiv (1,2,3);B\equiv (2,4,1);C\equiv (3,4,5)$

$\therefore$ Area of $△ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$

$=\frac{1}{2}\left|\begin{array}{ccc}i& j& k\\ 1& 2& -2\\ 2& 2& 2\end{array}\right|=\sqrt{26}$ square units.