Determine the distance of the point $(3,8,2)$ from the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ measured parallel to the plane $3x+2y-2z+15=0$.
I am getting two different answers using two different approaches:
The first approach is by determining the parametric point on the line $(2r+1,4r+3,3r+2)$. Then according to the condition the vector $(2r+1-3,4r+3-8,3r+2-2)$ will be perpendicular to the normal of the plane. Doing dot product,we get $r=2$ and so the point is $(5,11,8)$ and the distance comes out to be $7$.
But the second approach is my first calculating the perpendicular distance $d$ from the point to the line which comes out to be $\sqrt{\frac{265}{29}}$. And if we let $\theta$ to be the angle between the line and the inclined line parallel to the plane,the required distance would be $\frac{d}{\sin \theta}$. Now,by observation $\theta$ is actually the angle between the plane and the vector $2i+4j+3k$ since the vector is parallel to the line and the inclined line is parallel to the plane. So,$90-\theta$ is the angle between the vector and the normal to the plane.Doing dot product,$\cos (90-\theta)=\sin \theta=\frac{8}{\sqrt{29}\sqrt{17}}$. But substituting the values of $d$ and $\sin \theta$ we get,the distance is $\frac{\sqrt{265}\sqrt{17}}{8}$ which isn't equal to $7$.
1st approach should always give valid answer since this is very elementary. However,I also cross checked the calculations in my 2nd approach multiplie times to see if there is any error but I couldn't find any. So,what is the problem with my 2nd approach. It will be very helpful if my mistake is pointed out.
The first approach is a good one. The problem in the second approach (I've checked all your calculations and get the same results ) has been pointed out by @Bob Dobbs: You reason as if you were in one and the same plane when you have non-coplanar vectors.
To make this clear, let's take a simpler similar situation:
Let $D=(-1,0,0)$ , the line $l:x=y=z$ and the plane $\mathcal P:z=0$ where a normal vector is $\vec{n}=(0,0,1)$.
The "distance of the point $D$ from the line $l$ parallel to $\mathcal P$" is of course $1$.
On the other hand, if we were to follow your second approach, we would calculate the distance $d$ from D to its orthogonal projection $E=\frac13(-1,-1,-1)$ on $l$, which is $\frac13\sqrt6$. And then we'd look to use the angle $\theta=\widehat{EOD}$.
But there, with the evidence of the figure, we would not make your mistake, which my explanations have enabled you to understand, I hope.