Let P be a plane containing the line x-13=y+64=z+52 and parallel to the line x-34=y-2-3=z+…

Question

Let $P$ be a plane containing the line $\dfrac{x-1}{3}=\dfrac{y+6}{4}=\dfrac{z+5}{2}$ and parallel to the line $\dfrac{x-3}{4}=\dfrac{y-2}{-3}=\dfrac{z+5}{7}$. If the point $(1,-1,\alpha)$ lies on the plane $P$, then the value of $|5\alpha|$ is equal to:

Answer: 38

Accepted Answer

Normal $=(3,4,2)\times(4,-3,7)=(34,-13,-25)$.
$$P:\ 34(x-1)-13(y+6)-25(z+5)=0 \implies 34x-13y-25z-237=0.$$
Point $(1,-1,\alpha)$:
$$34+13-25\alpha-237=0 \implies -25\alpha=190 \implies \alpha=-\frac{38}{5}.$$
$$|5\alpha|=38.$$

Solutions · 1