Write the line lying in 2x+3y-z+1=0 , passing through the…

Write the line lying in $2x+3y-z+1=0$, passing through the point $M$ where the plane meets the $z$-axis, and making a minimal angle with $\tfrac{x-3}1=\tfrac{y-3}2=\tfrac z1$.

Step 1: Solution

$M(0,0,1)$. The smallest angle between $\vec d=\{1,2,1\}$ and a direction in the plane is attained by the orthogonal projection of $\vec d$ onto the plane: \[\vec u=\vec d-\frac{\vec d\cdot\vec n}{|\vec n|^2}\vec n=\{1,2,1\}-\tfrac{7}{14}\{2,3,-1\}=\{0,\tfrac12,\tfrac32\}\parallel\{0,1,3\}.\]

\[x=0,\qquad \frac y1=\frac{z-1}{3}\]

The minimal angle is the angle between $\ell$ and the plane: $\cos\varphi=\tfrac{5}{\sqrt{60}}=\tfrac{\sqrt{15}}{6}$, so $\varphi\approx49.80^\circ\approx49^\circ48'$. Equivalently, $\sin\varphi=\tfrac{7}{\sqrt6\sqrt{14}}$.