Reflection image of a point across a plane

Geometry Level 3

Given the point (1, 3, 5), you want to find its image after a reflection across the plane 2 x − 3 y + 5 z = 10 2 x - 3 y + 5 z = 10 .

If the coordinates of the image are ( x , y , z ) (x, y, z) , and x + y + z = a b x + y + z = \dfrac{a}{b} , where a a and b b are coprime positive integers, then find a + b a + b .


The answer is 158.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

David Vreken
Oct 6, 2018

A vector perpendicular to the plane 2 x − 3 y + 5 z = 10 2x - 3y + 5z = 10 is ( 2 , − 3 , 5 ) (2, -3, 5) .

A vector equation through point ( 1 , 3 , 5 ) (1, 3, 5) at t = 0 t = 0 with vector ( 2 , − 3 , 5 ) (2, -3, 5) is ( x , y , z ) = ( 1 , 3 , 5 ) + t ( 2 , − 3 , 5 ) (x, y, z) = (1, 3, 5) + t(2, -3, 5) , so x = 1 + 2 t x = 1 + 2t , y = 3 − 3 t y = 3 - 3t , and z = 5 + 5 t z = 5 + 5t .

Substituting these values into the plane equation gives 2 ( 1 + 2 t ) − 3 ( 3 − 3 t ) + 5 ( 5 + 5 t ) = 10 2(1 + 2t) - 3(3 - 3t) + 5(5 + 5t) = 10 , which solves to t = − 4 19 t = -\frac{4}{19} .

So at t = − 4 19 t = -\frac{4}{19} the image is on the plane, and in another t = − 4 19 t = -\frac{4}{19} , or at t = − 8 19 t = -\frac{8}{19} , the image will be past the plane at the same distance it started away from it on a perpendicular vector, which would be the same as its reflected image. Using x = 1 + 2 t x = 1 + 2t , y = 3 − 3 t y = 3 - 3t , and z = 5 + 5 t z = 5 + 5t from above, and t = − 8 19 t = -\frac{8}{19} , we get x = 3 19 x = \frac{3}{19} , y = 81 19 y = \frac{81}{19} , and z = 55 19 z = \frac{55}{19} , so x + y + z = 139 19 x + y + z = \frac{139}{19} , which means a = 139 a = 139 and b = 19 b = 19 , and 139 + 19 = 158 139 + 19 = \boxed{158} .

5 Helpful 0 Interesting 1 Brilliant 0 Confused

Thanks for a well-explained solution.

Hosam Hajjir - 2 years, 8 months ago

I had learnt a direct formula to find the image of the point P ( x 0 , y 0 , z 0 ) P(x_0,y_0,z_0) about reflection from the plane π : a x + b y + c z + d = 0 \pi : ax+by+cz+d=0 given by

x − x 0 a = y − y 0 b = z − z 0 c = − 2 ( a x 0 + b y 0 + c z 0 + d ) a 2 + b 2 + c 2 \dfrac{x-x_0}{a} = \dfrac{y-y_0}{b} = \dfrac{z-z_0}{c} = \dfrac{-2 \left(ax_0 + by_0 + cz_0 + d \right)}{a^2+b^2+c^2}

which gives the correct answer. Can you please hint me how to prove this?

In short , why t = − 2 ( a x 0 + b y 0 + c z 0 + d ) a 2 + b 2 + c 2 t = \dfrac{-2 \left(ax_0 + by_0 + cz_0 + d \right)}{a^2+b^2+c^2} ?

Tapas Mazumdar - 2 years, 8 months ago

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That's a useful equation! You can prove it with the same explanation above but replace ( 2 , − 3 , 5 ) (2, -3, 5) with the general case of ( a , b , c ) (a, b, c) and replace ( 1 , 3 , 5 ) (1, 3, 5) with the general case of ( x 0 , y 0 , z 0 ) (x_0, y_0, z_0) :

A vector perpendicular to the plane a x + b y + c z = d ax + by + cz = d is ( a , b , c ) (a, b, c) .

A vector equation through point ( x 0 , y 0 , z 0 ) (x_0, y_0, z_0) at t = 0 t = 0 with vector ( a , b , c ) (a, b, c) is ( x , y , z ) = ( x 0 , y 0 , z 0 ) + t ( a , b , c ) (x, y, z) = (x_0, y_0, z_0) + t(a, b, c) , so x = x 0 + a t x = x_0 + at , y = y 0 + b t y = y_0 + bt , and z = z 0 + c t z = z_0 + ct . (Rearranging, we get t = x − x 0 a = y − y 0 b = z − z 0 c t = \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} ).

Substituting these values into the plane equation gives a ( x 0 + a t ) + b ( y 0 + b t ) + c ( z 0 + c t ) = d a(x_0 + at) + b(y_0 + bt) + c(z_0 + ct) = d , which solves to t = − ( a x 0 + b y 0 + c z 0 − d ) a 2 + b 2 + c 2 t = \frac{-(ax_0 + by_0 + cz_0 - d)}{a^2 + b^2 + c^2} .

So at t = − ( a x 0 + b y 0 + c z 0 − d a 2 + b 2 + c 2 ) t = \frac{-(ax_0 + by_0 + cz_0 - d}{a^2 + b^2 + c^2}) the image is on the plane, and in another t = − ( a x 0 + b y 0 + c z 0 − d ) a 2 + b 2 + c 2 t = \frac{-(ax_0 + by_0 + cz_0 - d)}{a^2 + b^2 + c^2} , or at t = − 2 ( a x 0 + b y 0 + c z 0 − d ) a 2 + b 2 + c 2 t = \frac{-2(ax_0 + by_0 + cz_0 - d)}{a^2 + b^2 + c^2} , the image will be past the plane at the same distance it started away from it on a perpendicular vector, which would be the same as its reflected image.

Therefore, t = x − x 0 a = y − y 0 b = z − z 0 c = − 2 ( a x 0 + b y 0 + c z 0 − d ) a 2 + b 2 + c 2 t = \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = \frac{-2(ax_0 + by_0 + cz_0 - d)}{a^2 + b^2 + c^2} .

(Note that it is − d -d in the equation, not + d +d ).

David Vreken - 2 years, 8 months ago

The value of t is wrong.Solving the equation in t we get t= -8/39

Siddharth Banerjee - 2 years, 7 months ago

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Solving for t t gives t = − 4 19 t = -\frac{4}{19} :

2 ( 1 + 2 t ) − 3 ( 3 − 3 t ) + 5 ( 5 + 5 t ) = 10 2(1 + 2t) - 3(3 - 3t) + 5(5 + 5t) = 10

2 + 4 t − 9 + 9 t + 25 + 25 t = 10 2 + 4t - 9 + 9t + 25 + 25t = 10

18 + 38 t = 10 18 + 38t = 10

38 t = − 8 38t = -8

t = − 8 38 = − 4 19 t = -\frac{8}{38} = -\frac{4}{19}

David Vreken - 2 years, 7 months ago

Why not the normal to the plane is (-2, 3, -5) ?

A Former Brilliant Member - 2 years, 3 months ago

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( − 2 , 3 , − 5 ) (-2, 3, -5) is also a normal to the plane that can be used to get the same answer. The parametric equations using ( − 2 , 3 , − 5 ) (-2, 3, -5) are ( x , y , z ) = ( 1 , 3 , 5 ) + t ( − 2 , 3 , − 5 ) (x, y, z) = (1, 3, 5) + t(-2, 3, -5) , so x = 1 − 2 t x = 1 - 2t , y = 3 + 3 t y = 3 + 3t , and z = 5 − 5 t z = 5 - 5t , which substituted into the plane equation gives 2 ( 1 − 2 t ) − 3 ( 3 + 3 t ) + 5 ( 5 − 5 t ) = 10 2(1 - 2t) - 3(3 + 3t) + 5(5 - 5t) = 10 , which solves to t = 4 19 t = \frac{4}{19} , which makes the reflected image at t = 8 19 t = \frac{8}{19} , and using this value gives x = 3 19 x = \frac{3}{19} , y = 81 19 y = \frac{81}{19} , and z = 55 19 z = \frac{55}{19} once again.

David Vreken - 2 years, 3 months ago

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