Finding the symmetry of a point with respect to the plane?
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Please show us what you have tried and exactly where you are stuck.
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Please share your work/thoughts about this problem.
Dr.Peterson
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Have you learned any methods or formulas for this? If not, can you define the point you're looking for, and how to represent that by vectors? In order to help, we need to know what tools you have available, and what goes wrong when you try to use them.
Thank you for your reply,
I tried this but not sure.
[MATH]{D : x-y+2z+4=0}, {P(2,3,-1)}[/MATH]
And normal vector of plane [MATH]{N = (1, -1, 2)}[/MATH]
[MATH]{\frac{x-2}{1}}={\frac{y-3}{-1}={\frac{z+1}{2}}=k}[/MATH]
and parametric equation is
[MATH] {x = k+2, y = -k+3, z = 2k-1} [/MATH]
Then I put x, y, z of type k in the equation
[MATH]{k+2-(-k+3)+2(2k-1)+4 = 0}[/MATH][MATH]{k = \frac{-1}{6}}[/MATH]
Final answer: (symetric equation)
[MATH]{(\frac{-11}{6}, \frac{19}{6}, \frac{-4}{3})}[/MATH]
Last edited:
Dr.Peterson
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You didn't quite explain what each step did; for instance, the equation you wrote is for the line though P normal to D, right? Then you found k for the intersection of the line and the plane; then you put that k into the equation of the line.
Won't that find a point on the plane, rather than symmetrical with respect to the plane?
I think there's also a sign error in your work, because the point you got doesn't satisfy the equation of the plane.
oh Yes, the last point I wrote is a point on the plane. Now I have to find the symmetry of the point P given in the question with respect to the plane using the coordinates of this point.
That is:
[MATH]{(\frac{-17}{3},\frac{10}{3},\frac{-5}{3})}[/MATH]Of course, even if I take it according to the x-axis or the y or z-axis (change the signs), unfortunately it doesn't provide the equation of the plane.
HallsofIvy
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The plane x−y+2z+4=0 has normal vector <1, -1, 2> (the coefficients of x, y, z). A line with that direction vector, through the point (2, 3, -1) is given by parametric equations x= t+ 2, y= -t+ 3, z= 2t- 1.
That line intersects the plane where (t+ 2)-(-t+ 3)+ 2(2t- 1)+ 4= 6t- 3+4= 6t+ 1= 0 so t= -1/6. That is, x= -1/6+ 2=11/6, y= 1/6+ 3= 19/6, z= -2/6- 1= -1/3- 1= -4/3. The line intersects the plane at (11/6, 19/6, -4/3). You can check that that point IS on the plane- x- y+ 2z= 11/6- 19/6- 8/3+ 4= (11- 19- 16)/6+ 4= (11- 35)/6+ 4= -24/6+ 4= -4+ 4= 0.
The distance from (2, -3, 1) to (11/6, 19/6, -4/3) is \(\displaystyle \sqrt{(2-11/6)^2+ (-3- 19/6)^2+ (1+ 4/3)^2}\)\(\displaystyle = \frac{\sqrt{1566}}{6}\)
Now, the symmetric point to (2, -3, 1) lies on that line at the same distance, on the other side of the plane- so lies at the intersection of the line with the sphere centered at (11/6, 19/6, -4/3) with radius \(\displaystyle \frac{\sqrt{1566}}{6}\) so \(\displaystyle (x- 11/6)^2+ (y- 19/6)^2+ (z+ 4/3)^2= \frac{1566}{6}= 261\).
Since the line is given by x= t+ 2, y= -t+ 3, z= 2t- 1, You need to solve \(\displaystyle (t+ 2- 11/6)^2+ (-t+3-29/6)^2+ (2t- 1+ 4/3)^2= (t+ 1/6)^2+ (t+ 11/6)^2+ (2t+ 1/3)^2= 261. Of course, t= -1/6 is a solution because the initial point, (2, 3, -1) is on the sphere and line. The point you want is the other solution to that quadratic equation.\)
That line intersects the plane where (t+ 2)-(-t+ 3)+ 2(2t- 1)+ 4= 6t- 3+4= 6t+ 1= 0 so t= -1/6. That is, x= -1/6+ 2=11/6, y= 1/6+ 3= 19/6, z= -2/6- 1= -1/3- 1= -4/3. The line intersects the plane at (11/6, 19/6, -4/3). You can check that that point IS on the plane- x- y+ 2z= 11/6- 19/6- 8/3+ 4= (11- 19- 16)/6+ 4= (11- 35)/6+ 4= -24/6+ 4= -4+ 4= 0.
The distance from (2, -3, 1) to (11/6, 19/6, -4/3) is \(\displaystyle \sqrt{(2-11/6)^2+ (-3- 19/6)^2+ (1+ 4/3)^2}\)\(\displaystyle = \frac{\sqrt{1566}}{6}\)
Now, the symmetric point to (2, -3, 1) lies on that line at the same distance, on the other side of the plane- so lies at the intersection of the line with the sphere centered at (11/6, 19/6, -4/3) with radius \(\displaystyle \frac{\sqrt{1566}}{6}\) so \(\displaystyle (x- 11/6)^2+ (y- 19/6)^2+ (z+ 4/3)^2= \frac{1566}{6}= 261\).
Since the line is given by x= t+ 2, y= -t+ 3, z= 2t- 1, You need to solve \(\displaystyle (t+ 2- 11/6)^2+ (-t+3-29/6)^2+ (2t- 1+ 4/3)^2= (t+ 1/6)^2+ (t+ 11/6)^2+ (2t+ 1/3)^2= 261. Of course, t= -1/6 is a solution because the initial point, (2, 3, -1) is on the sphere and line. The point you want is the other solution to that quadratic equation.\)
Dr.Peterson
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I'm not sure what you mean by "it doesn't provide the equation of the plane". Didn't you use that equation?
There are many ways you can solve the problem. Your work mostly uses equations of lines and planes, but you mentioned a vector. I would tend to use vectors more heavily, if you know enough about them. But you can use a vector easily for the final step.
Call the (corrected) point you found Q. Consider vector PQ. How can you use that to find the symmetric point R?
It looks like you are claiming that [MATH]R={(\frac{-17}{3},\frac{10}{3},\frac{-5}{3})}[/MATH], though you didn't say how you got it. That's almost correct; but you didn't correct the sign error I told you about!
Sorry I'm late for an answer. I found the solution, as the "HallsofIvy" member said, but as you said in your previous post, there is a sign error in the plane equation. I checked the question again, and exactly as I wrote it, the question may have been misspelled, but I'm still not sure. I could not find any other solution.
Dr.Peterson
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I didn't say the error was in the equation in the problem! I said it was in your work. Check your work in post #5 where you found the x component!
Then compare that to what Halls said in post #8 without commenting on your error:
Moreover, I checked 3 times and made a simple mistake on three of them. I also say to myself "where is my fault?" There are questions I ask on other topics and some other questions, I get tired of dealing with all of them, unfortunately I missed it
Thank you again.