The vertices B and C of a triangle ABC lie on the line x1=1-y-2=z-23.

Question

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

Accepted Answer

Rewrite the line as $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}=\lambda$, general point $(\lambda,\ 2\lambda+1,\ 3\lambda+2)$. Setting $x=4\Rightarrow \lambda=4$, so $y=9$ (consistent) and $z=3(4)+2=14\Rightarrow\alpha=14$. Thus $B=(4,9,14)$.

$C$ lies on the line at distance $10$ from $B$: $C=B\pm\dfrac{10}{\sqrt{14}}(1,2,3)$.

$\vec{AB}=B-A=(3,3,11)$. Since $\vec{AC}=\vec{AB}\pm\dfrac{10}{\sqrt{14}}(1,2,3)$,
$$\vec{AB}\times\vec{AC}=\pm\frac{10}{\sqrt{14}}\big(\vec{AB}\times(1,2,3)\big).$$
$$\vec{AB}\times(1,2,3)=\hat i(9-22)-\hat j(9-11)+\hat k(6-3)=(-13,2,3),\qquad |(-13,2,3)|=\sqrt{182}.$$
$$\text{Area}=\frac12\cdot\frac{10}{\sqrt{14}}\cdot\sqrt{182}=5\sqrt{\frac{182}{14}}=5\sqrt{13}.$$

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