The equation of the plane containing the straight line x2= y3= z4 and perpendicular to the…

Question

The equation of the plane containing the straight line $\dfrac x2=\dfrac y3=\dfrac z4$ and perpendicular to the plane containing the straight lines $\dfrac x3=\dfrac y4=\dfrac z2$ and $\dfrac x4=\dfrac y2=\dfrac z3$ is:

Accepted Answer

Normal to the plane containing $(3,4,2)$ and $(4,2,3)$-direction lines: $(3,4,2)\times(4,2,3)=(12-4,\,8-9,\,6-16)=(8,-1,-10)$.
Let required plane's normal be $(l,m,n)$: perpendicular to this, and containing direction $(2,3,4)$:
$$-8l+m+10n=0,\qquad 2l+3m+4n=0.$$
Solving: $(l,m,n)=(1,-2,1)$ (up to scale). Plane: $x-2y+z=0$.

Solutions · 1