Question

A plane make intercept $OA,OB,OC$ whose measurements are $a,b,c$ on axes $OX,OY,OZ,$ then area of triangle $ABC$ is

• A. $\frac{1}{2}\sum ab$
• B. $\frac{1}{2}{\left[\right(ab{)}^2+(bc{)}^2+(ca{)}^2]}^{1/2}$
• C. $\frac{abc}{2}\sum a$
• D. $\frac{1}{2}[\sum a\sum ab]$

Answer 1

Answer: (B)

$\frac{1}{2}{\left[\right(ab{)}^2+(bc{)}^2+(ca{)}^2]}^{1/2}$

Using fact
Area $=△=\sqrt{(△x{)}^2+(△y{)}^2+(△z{)}^2}$
$=\frac{1}{2}\sqrt{(bc{)}^2+(ac{)}^2+(ab{)}^2}$
where ${△}_x=\frac{1}{2}\left|\begin{array}{ccc}{y}_1& y_2& y_3\\ z_1& z_2& z_3\\ 1& 1& 1\end{array}\right|$
$=\frac{1}{2}\left|\begin{array}{ccc}0& b& 0\\ 0& 0& c\\ 1& 1& 1\end{array}\right|$
$=\frac{1}{2}\left(bc\right)$
Area of $△ABC$
$=\frac{1}{2}|\left(\overline{AB}\right)\times \left(\overline{AC}\right)|$
$=\frac{1}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -a& b& 0\\ a& 0& -c\end{array}\right|$
$=\frac{1}{2}|-bc\hat{i}-ac\hat{j}-ab\hat{k}$t
$=\frac{1}{2}\sqrt{(-bc{)}^2+(-ac{)}^2+(ab{)}^2}$