Find the distance between the line $x-z=3, x+2y+4z=6$ and the plane $3x+2y+2z=5$.
What I have done so far,
Found the vector by crossing $1,0-1$ and $1,2,4$ from the first line
The line is parallel to plane $\langle 2,-5,2\rangle \cdot \langle 3,2,2 \rangle = 0$
I think my next step is to find a point on Line 1 which satisfies both equations and then insert those values into the plane $3(x)+2(y)+2(z)=5$ and use the formula, $$
\frac{(3(x)+2(y)+2(z)-5)}{(3^2+2^2+2^2)}$$ to find the distance.
The values that I calculuate do not match the posted answer of $7/\sqrt{17}$
Best Answer
Hint: The line and the plane (as you have noted) are parallel. The distance from the plane to the line is therefore the distance from the plane to any point on the line. So just pick any point on the line and use "the formula" to find the distance from this point to the plane.