The vertices B and C of a ABC lie on the line x+23=y-10= z4 such that BC=5 units.

Question

The vertices $B$ and $C$ of a $	riangle ABC$ lie on the line $\dfrac{x+2}3=\dfrac{y-1}0=\dfrac z4$ such that $BC=5$ units. Then the area (in sq. units) of this triangle, given that the point $A(1,-1,2)$, is:

Accepted Answer

Let $D=(3\lambda-2,1,4\lambda)$ be the foot of the perpendicular from $A$ to line $BC$ (direction $(3,0,4)$):
$$\overrightarrow{AD}=(3\lambda-3,2,4\lambda-2),\qquad \overrightarrow{AD}\cdot(3,0,4)=0 \implies 25\lambda-17=0\implies\lambda=\tfrac{17}{25}.$$
$$D=\left(\tfrac1{25},1,\tfrac{68}{25}\right),\qquad AD^2=\left(\tfrac{24}{25}\right)^2+4+\left(\tfrac{18}{25}\right)^2=\tfrac{136}{25} \implies AD=\tfrac{2\sqrt{34}}5.$$
$$\text{Area}=\tfrac12\cdot BC\cdot AD=\tfrac12\cdot5\cdot\tfrac{2\sqrt{34}}5=\sqrt{34}.$$

Solutions · 1