A vector n is inclined to the x-axis at 45^, to the y-axis at 60^ and at an acute angle to…

Question

A vector $\vec n$ is inclined to the $x$-axis at $45^{\circ}$, to the $y$-axis at $60^{\circ}$ and at an acute angle to the $z$-axis. If $\vec n$ is a normal to a plane passing through the point $(\sqrt2,-1,1)$, then the equation of the plane is:

Accepted Answer

Direction cosines: $l=\cos45^{\circ}=\tfrac1{\sqrt2},\ m=\cos60^{\circ}=\tfrac12$; from $l^2+m^2+n^2=1$: $n^2=1-\tfrac12-\tfrac14=\tfrac14\Rightarrow n=\tfrac12$ (acute).
So $\vec n\parallel\left(\tfrac1{\sqrt2},\tfrac12,\tfrac12\right)\parallel(\sqrt2,1,1)$, giving plane
$$\sqrt2(x-\sqrt2)+(y+1)+(z-1)=0\implies \sqrt2x+y+z=2.$$
(Testing this against $(\sqrt2,-1,1)$: $2-1+1=2$ . The source's listed option B, $2x+y+2z=2\sqrt2+1$, uses a differently-scaled normal $(2,1,2)$ but also passes through the given point; per the source answer key, B is marked correct, though the direction-cosine--consistent normal derived above is $(\sqrt2,1,1)$.)

Solutions · 1