Question

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{1}=1$ meets the co - ordinate axes in the points $A,B,C$ respectively. The area of triangle $ABC$ in square units is :

• A. $\frac{5}{2}$
• B. $\frac{7}{2}$
• C. $\frac{9}{2}$
• D. None of these

Answer 1

The correct option is B $\frac{7}{2}$

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{1}=1$ meets the co - ordinat axes in the points $A,B,C$ respectively.
Therefore, $\overrightarrow{OA}=2i$, $\overrightarrow{OB}=3j$, $\overrightarrow{OC}=k$
Area of triangle $ABC$ $=\Delta$

$=\left|\frac{\overrightarrow{AB}\times \overrightarrow{AC}}{2}\right|$

$=\left|\frac{(3j-2i)\times (k-2i)}{2}\right|$

$=\left|\frac{3i+2j+6k}{2}\right|$

$=\frac{7}{2}$
Ans: B