Let in a ABC, the length of the side AC be 6, the vertex B be (1,2,3), and the vertices A,…

Question

Let in a $	riangle ABC$, the length of the side $AC$ be $6$, the vertex $B$ be $(1,2,3)$, and the vertices $A,C$ lie on the line $\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}$. Then the area (in sq. units) of $	riangle ABC$ is:

Accepted Answer

Let $M(3\lambda+6,\ 2\lambda+7,\ -2\lambda+7)$ be the foot of perpendicular from $B$ to line $AC$. With $\vec{AC}\cdot\vec{BM}=0$:
$$3(3\lambda+5)+2(2\lambda+5)-2(-2\lambda+4)=0 \implies 17\lambda+17=0 \implies \lambda=-1.$$
$$\vec{BM}=(2,3,6),\qquad |\vec{BM}|=7.$$
$$\text{Area}=\frac12\times AC\times BM=\frac12\times6\times7=21.$$

Solutions · 1