If vertices A and C of a \( \Delta ABC \) lie along the line \( \displaystyle\frac{x-3}{3}=\displaystyle\frac{y+7}{-1}=\displaystyle\frac{z+2}{5} \) and vertex B has coordinates (3, -8, 1) and line segment AC has length \( \sqrt{35} \), then the area of \( \Delta ABC \) is:-
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{2}\sqrt{94} \) Sq. units
Since A and C lie on the line, the height of the triangle is the distance of B from the line. The line passes through P(3, −7, −2) with direction \( \vec{d}=(3,-1,5) \), \( |\vec{d}|=\sqrt{35} \).
\( \overrightarrow{PB}=(0,-1,3) \)
\( \displaystyle \overrightarrow{PB}\times\vec{d}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\0 & -1 & 3\\3 & -1 & 5\end{vmatrix} \)
\( =-2\hat{\imath}+9\hat{\jmath}+3\hat{k} \)
\( \displaystyle \text{Height}=\frac{|\overrightarrow{PB}\times\vec{d}|}{|\vec{d}|} \)
\( \displaystyle =\frac{\sqrt{4+81+9}}{\sqrt{35}} \)
\( \displaystyle =\frac{\sqrt{94}}{\sqrt{35}} \)
\( \displaystyle \text{Area}=\frac{1}{2}\times\sqrt{35}\times\frac{\sqrt{94}}{\sqrt{35}} \)
\( \displaystyle =\frac{1}{2}\sqrt{94}\ \text{sq. units} \)