Find the angle between the line tfrac…

Find the angle between the line $\tfrac x1=\tfrac{y+1}3=\tfrac{z-1}{-3/2}$ and the plane $2x+y-z-4=0$.

Explanation of Method

The angle between a line and a plane.

Step 1: Solution

$\vec d\parallel\{2,6,-3\}$ with $|\vec d|=7$, and $\vec n=\{2,1,-1\}$ with $|\vec n|=\sqrt6$. The angle with the plane is the complement of the angle with the normal: \[\sin\varphi=\frac{|4+6+3|}{7\sqrt6}=\frac{13}{7\sqrt6}=\frac{13\sqrt6}{42}\approx0.7581 .\]

\[\varphi=\arcsin\frac{13\sqrt6}{42}\approx49.30^\circ\approx49^\circ18'\]

$\varphi$

The angle between a line and a plane is the angle between the line and its orthogonal projection onto the plane (teal): $\sin\varphi=\tfrac{|\vec d\cdot\vec n|}{|\vec d||\vec n|}$.

Result

\[\boxed{\,\varphi=\arcsin\frac{13\sqrt6}{42}\approx49.30^\circ\approx49^\circ18'\,}\]

Remark

Correction: the source gives $49^\circ15'$. The correct value is $\arcsin 0.7581=49.304^\circ=49^\circ18'$. The piercing point is at $t=\tfrac{6}{13}$, i.e. $\left(\tfrac{6}{13},\tfrac{5}{13},\tfrac{4}{13}\right)$.