The vertices B and C of a triangle ABC lie on the line (x)/(1)=(1-y)/(-2)=(z-2)/(3).

Question

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{z-2}{3}$. The coordinates of A and $B$ are $(1,6,3)$ and $(4,9,\alpha )$ respectively and $C$ is at a distance of 10 units from $B$. The area (in sq. units) of $	riangleABC$ is :

Accepted Answer

$L:\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda $

Any point $(\lambda ,2\lambda +1,3\lambda +2)$

Also $B$ lies on $L$

$\begin{gathered} \Rightarrow \frac{4}{1}=\frac{\alpha -2}{3}\Rightarrow \alpha =14\, 	herefore B(4,9,14) \ C\equiv \langle 4,9,14angle \pm \frac{10}{\sqrt{14}}\langle 1,2,3angle \end{gathered}$

Area of $	riangleABC=\frac{1}{2}|\overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{AC}|$

$\begin{gathered} =\overset{	o \leftarrow }{AB}=\langle 3,3,11angle \ \overset{	o \leftarrow }{AC}=\overset{	o \leftarrow }{AB}+\frac{10}{\sqrt{11}}\langle 1,2,3angle \end{gathered}$

Cross product of $\overset{	o \leftarrow }{AB}$ to eliminate parallel component

So, area $=\frac{1}{2}	imes \frac{10}{\sqrt{14}}(\overset{	o \leftarrow }{AB}	imes \langle 1,2,3angle )$

$\overset{	o \leftarrow }{AB}	imes \langle 1,2,3angle \begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 3 \quad  3 \quad  11 \ 1 \quad  2 \quad  3 \end{vmatrix}$

$\begin{gathered} =\langle -13,2,3angle \ =\frac{1}{2}	imes \frac{10}{\sqrt{14}}\cdot \sqrt{182}=5\sqrt{13} \end{gathered}$

Solutions · 1