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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 $7/\sqrt{17}$

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Looking at your question again, I am not sure that I understand what the line is. It looks like your "line" is given by the equations of two planes. –  Thomas Oct 9 '12 at 16:00
    
@Thomas any we can descibe lines as intersections of non-parallel planes –  no identity Oct 13 '12 at 15:09
    
@Norbert: Sure, I just wasn't sure that that was what the OP meant... –  Thomas Oct 13 '12 at 17:11

3 Answers 3

up vote 1 down vote accepted

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.

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Thanks, I just made a simple mathematical error which resulted in an incorrect answer. –  mike Oct 9 '12 at 16:06

If $z=t,$ for the given line $x=z+3=t+3$ and

$x+2y+4z=6$ or, $t+3+2y+4(t)=6$ or, $2y=3-5t$

So, any point on the given line is $(t+3,t,\frac{3-5t}2)$

The distance of $(t+3,t,\frac{3-5t}2)$ from $3x+2y+2z=5$ is $$\left|\frac{3(t+3)+(3-5t)+2(t)-5}{\sqrt{3^2+2^2+2^2}}\right|=\frac 7{\sqrt {17}}$$

Had it not been a constant(which is due to the parallelism), we had to find $t$such that the distance is minimum.

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Put $(P_1): x - z -3 = 0$, $(P_2): x+2y+4z=6$, $(P_3): 3x + 2y +2z -5 = 0$ and $\Delta$ is intersection of two plane $(P_1)$ and $(P_2)$. We have, a normal vector of the plane $(P_1)$ is $a = (1,0,-1)$, a normal vector of the plane $(P_2)$ is $b = (1,2,4)$. A direction vector $v$ of $\Delta$ is cross product of $a$ and $b$, therefore $v = (3, 2, 2)$. Another way, $M=(3, \dfrac32, 0)$ lies on $\Delta$ and not belongs to $(P_3)$, thus $\Delta$ is parallel to the plane $(P_3)$. We have, the distance between the line $\Delta$ to the plane $(P_3)$ is the distance from the point $M$ to the plane $(P_3)$. From here, we have answer is $$\dfrac{|3\cdot 3 + 2\cdot\dfrac{3}{2} + 2\cdot 0 - 5|}{\sqrt{3^2 + 2^2 + 2^2}} = \dfrac{7\sqrt{17}}{17}.$$

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