Question

If a plane cuts off intercepts OA = a, OB = b, OC = c from the coordinate axes (where 'O' is the origin). then the area of the triangle ABC is equal to

• A. $\frac{1}{2}\sqrt{b^2{c}^2+c^2{a}^2+a^2{b}^2}$
• B. $\frac{1}{2}$ $(bc+ca+ab)$
• C. $\frac{1}{2}$ abc
• D. $\frac{1}{2}\sqrt{(b-c{)}^2+(c-a{)}^2+(a-b{)}^2}$

Answer 1

Answer: (A)

$\frac{1}{2}\sqrt{b^2{c}^2+c^2{a}^2+a^2{b}^2}$

$\overrightarrow{AC}=-a\hat{i}+c\hat{k}$

$\overrightarrow{AB}=-a\hat{i}+b\hat{j}$

Area of $△ABC=\frac{1}{2}\mid \overrightarrow{A}B\times \overrightarrow{A}C\mid$

$\overrightarrow{AB}\times \overrightarrow{AC}=\left[\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ -a0c\\ -ab0\end{array}\right]$

$=-\left(bc\right)\hat{i}-\left(ac\right)\hat{j}-\left(ab\right)\hat{k}$

$\mid \overrightarrow{AB}\times \overrightarrow{AC}\mid =\sqrt{b^2{c}^2+a^2{c}^2+a^2{b}^2}$

Area$=\frac{1}{2}\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}$