Question

Find the area of a triangle having the points $A(1,1,1),B(1,2,3)$ and $C(2,3,1)$ as its vertices.

• A. $\frac{\sqrt{21}}{5}$sq.units
• B. $-\frac{\sqrt{21}}{2}$sq.units
• C. $\frac{\sqrt{21}}{2}$sq.units
• D. None of these

Answer 1

The correct option is C $\frac{\sqrt{21}}{2}$sq.units

$\overrightarrow{AB}=\hat{j}+2\hat{k}$

$\overrightarrow{AC}=\hat{i}+2\hat{j}$

Area of triangle = $\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$

$=\frac{1}{2}|(\hat{j}+2\hat{k})\times (\hat{i}+2\hat{j})|$

$=\frac{1}{2}|-4\hat{i}+2\hat{j}-\hat{k}|$

$=\frac{1}{2}\sqrt{(4{)}^2+(2{)}^2+(1{)}^2}=\frac{\sqrt{21}}{2}$ sq. units