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    Hi everyone, can anyone explain to me that what does the reflection mean in Q6 c) ?
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    (Original post by 123321123)
    Hi everyone, can anyone explain to me that what does the reflection mean in Q6 c) ?
    Name:  ImageUploadedByStudent Room1460451678.888932.jpg
Views: 94
Size:  129.3 KB
    Thanks


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    A reflection of a point over a plane is the point which lies at the end of the perpendicular line starting from the point and being bisected by the plane.
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    it means that you extend the perpendicular from P to the plane an equal distance the other side of the plane to arrive at P'
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    (Original post by oShahpo)
    A reflection of a point over a plane is the point which lies at the end of the perpendicular line starting from the point and being bisected by the plane.
    Thanks I understand this now, but how do I get the coordinate of the reflected point? not just changing the sign, is it? could you explain it further for me plz. thanks so much. Also, could you tell me that how do I calculate the distance from a point to a plane if i know the equation of the plane and the coordination of the point.
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    (Original post by the bear)
    it means that you extend the perpendicular from P to the plane an equal distance the other side of the plane to arrive at P'
    yeah you are right, i get this now. but still not sure how to get the coordinate of the reflected point. any idea? thanks
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    suppose P was ( 2,5,6) and the foot of the perpendicular was ( 5,-1,8 ).... then you look at the x coordinates... 2 has increased by 3 to make 5, so add on another 3

    5 has decreased to -1 on the y coordinates so subtract another 6

    ....

    so in this made up example the new point will be at ( 8, -7, 10 )
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    Well, to find the point on the other end of the perpendicular, simply write an equation of the perpendicular line, then find the point on the line whose distance from the original point is equal to twice the distance of the original point from the plane. That's the normal way of doing it, however, it must be stated that if you actually draw the whole thing on a diagram, using vector triangles there is always an easier way of doing it.
 
 
 
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