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Math Help - Distance from a point to a line.

  1. #1
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    Distance from a point to a line.

    How can I find

    Distance from a point P=(-3,4,2) to a line Lx,y,z)=(3,-2,-1)+(1,-2,2)t
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  2. #2
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    If Q+tD is a line the the distance from the point P to the line is \frac{{\left| {\overrightarrow {QP}  \times D} \right|}}{{\left\| D \right\|}}.
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  3. #3
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    How can I find q?

    My answer was 3 square root 8.

    Is this right?
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  4. #4
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    I get \sqrt{65}.
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  5. #5
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    Here's how to do it without that formula. Any plane perpendicular to the line (x, y, z)= )=(3,-2,-1)+(1,-2,2)t is for the form x- 2y+ 2z= C for some constant C. (-3, 4, 2) will be on that plane if -3- 2(4)+ 2(2)= -7= C.

    That is, (-3, 4, 2) lies on the plane x- 2y+ 2z= -7 which is perpendicular to the given line. Now a point is on both the line and the plane (i.e. the line passes through the plane) when (3+ t)- 2(-2-t)+ 2(-1+ 2t)= -7. That gives 5+ 9t= -7, 9t= -12, t= -4/3. That tells us, then, that the line passes through the plane at \left(-3-\frac{4}{3}, -2+ \frac{8}{3},-1-\frac{8}{3}\right)= \left(\frac{5}{3}, \frac{2}{3}, -\frac{11}{3}\right). The line from (-3, 4, 2) to that point is perpendicular to the line and so is the shortest distance to the line. And, of course, the distance form (-3, 4, 2) to \left(\frac{5}{3}, \frac{2}{3}, -\frac{11}{3}\right) is \sqrt{65} as Plato says.
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