Let the coordinates of one vertex of ABC be A(0,2,), and the other two vertices lie on the…

Question

Let the coordinates of one vertex of $	riangle ABC$ be $A(0,2,\alpha)$, and the other two vertices lie on the line $\dfrac{x+\alpha}{5}=\dfrac{y-1}{2}=\dfrac{z+4}{3}$. For $\alpha\in\mathbb Z$, if the area of $	riangle ABC$ is $21$ sq. units and the line segment $BC$ has length $2\sqrt{21}$ units, then $\alpha^2$ is equal to:

Answer: 9

Accepted Answer

Since $\text{Area}=\frac12\cdot BC\cdot h$: $21=\frac12(2\sqrt{21})h \implies h=\sqrt{21}$.

With line point $(-\alpha,1,-4)$, direction $(5,2,3)$: $\vec{(\text{point})A}=(\alpha,1,\alpha+4)$.
$$\vec{(\text{point})A}\times(5,2,3)=(-2\alpha-5,\ 2\alpha+20,\ 2\alpha-5).$$
$$h^2=\frac{(2\alpha+5)^2+(2\alpha+20)^2+(2\alpha-5)^2}{38}=21 \implies 12\alpha^2+80\alpha+450=798.$$
$$3\alpha^2+20\alpha-87=0 \implies \alpha=\frac{-20\pm38}{6} \implies \alpha=3\ (\text{integer root}).$$
$$\alpha^2=9.$$

Solutions · 1