Find distance from line k given parametric equation of: $x=2+t, y=-3+2t, z=2-t, t\in\mathbb{R}$ from plane $\pi:2x+y+4z=0$.

Question

My solution:

$$2(2+t)+(-3+2t)+4(2-t)$$ $$4+2t-3+2t+8-4t$$ $$9=0$$ Contradiction, so no solutions, line and plane are parallel.

Its first time where I have such an example where equation is contradiction, I followed some example from internet and those are my next steps:

$$d(l,\pi)=\frac{|9|}{\sqrt{1+2^2+1}}=\frac{9\sqrt{6}}{6}$$

I am not sure about this method, is this distance value right? Maybe there's better way to calculate it?

Answer 1

Take any point on line and find length of perpendicular dropped from that point onto the plane. Let point is (2,-3,2),(t=0). So the formula for perpendicular distance is $\frac {|ax_0+by_0+cz_0+d_0|} {\sqrt{a^2+b^2+c^2}}$. Using this the distance is $\frac {|2×2+1×(-3)+4×2|} {\sqrt{2^2+1^2+4^2}}$

Answer 2

The direction vector $v$ of the line and the plane normal $n$ are orthogonal: $$\begin{pmatrix}1\\2\\-1 \end{pmatrix} \cdot \begin{pmatrix}2\\1\\4 \end{pmatrix} = 0$$ and thus the line is parallel to the plane and all points on the line have the same distance to the plane.

The formula for the distance from a point $p$ to a plane $\pi:n\cdot x + d = 0$ is $$\frac{|p \cdot n + d|}{||n||}$$ For your plane: $$\frac{\left|\begin{pmatrix}2\\-3\\2 \end{pmatrix} \cdot \begin{pmatrix}2\\1\\4 \end{pmatrix} + 0\right|}{\sqrt{2^2+1^2+4^2}} = \frac{9}{\sqrt{21}}$$

So you accidentally normalized the the wrong vector (line direction vector instead of plane normal). Besides that your calculations are correct.