Reflect on this!
Let L be the line in space that runs through the origin and through the point ( 2 , 1 , − 2 ) . If you reflect the point ( 3 , 6 , 1 5 ) across L , you end up with some point ( a , b , c ) . Enter a + b + c as your answer.
The answer is -28.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
Yes, very nicely explained! (+1)
Cool.. :D ^
The reflection of x across L is given by T ( x ) = 2 ( u ⋅ x ) u − x , where u is a unit vector parallel to L .(This works because ( u ⋅ x ) u is the orthogonal projection of x onto L ). With u = 3 1 ( 2 , 1 , − 2 ) we quickly find T ( 3 , 6 , 1 5 ) = ( − 1 1 , − 1 0 , − 7 ) . The answer is − 2 8
The plane which passes through (0, 0, 0), (2, 1, -2) and (3, 6, 15) is: 3x - 4y + z = 0
Its normal vector is: (27, -36, 9)
The plane perpendicular to this plane, passing through line L is: 7x + 8y + 11z = 0
Now, we reflect (3, 6, 15) through plane 7x + 8y + 11z = 0, resulting in (-11, -10, -7).
Answer is, therefore, -28.
Just to clarify: Can you show us how you reflect the point ( 3 , 6 , 1 5 ) across the plane 7 x + 8 y + 1 1 z = 0 ?
Log in to reply
The reflection of a point through a plane a x + b y + c z = 0 in 3D space can be calculated by this transformation matrix:
a 2 + b 2 + c 2 1 ⎣ ⎡ − a 2 + b 2 + c 2 − 2 a b − 2 a c − 2 a b a 2 − b 2 + c 2 − 2 b c − 2 a c − 2 b c a 2 + b 2 − c 2 ⎦ ⎤
In our case, a = 7 , b = 8 , c = 1 1 :
7 2 + 8 2 + 1 1 2 1 ⎣ ⎡ − 7 2 + 8 2 + 1 1 2 − 2 ∗ 7 ∗ 8 − 2 ∗ 7 ∗ 1 1 − 2 ∗ 7 ∗ 8 7 2 − 8 2 + 1 1 2 − 2 ∗ 8 ∗ 1 1 − 2 ∗ 7 ∗ 1 1 − 2 ∗ 8 ∗ 1 1 7 2 + 8 2 − 1 1 2 ⎦ ⎤ = 2 3 4 1 ⎣ ⎡ 1 3 6 − 1 1 2 − 1 5 4 − 1 1 2 1 0 6 − 1 7 6 − 1 5 4 − 1 7 6 − 8 ⎦ ⎤
1 1 7 1 ⎣ ⎡ 6 8 − 5 6 − 7 7 − 5 6 5 3 − 8 8 − 7 7 − 8 8 − 4 ⎦ ⎤ ⎣ ⎡ 3 6 1 5 ⎦ ⎤ = ⎣ ⎡ − 1 1 − 1 0 − 7 ⎦ ⎤
By the way, thank you for this problem. I had never faced reflection of a point through a line in 3D space. It made me think a lot.
Log in to reply
True! But it is actually easier to reflect across the line L directly; see my solution.
Let D represent the given point i.e ( a , b , c ) . Equation of line in Cartesian form: 2 x − 2 = 1 y − 1 = − 2 z + 2 Any point on this line is of the form and P ( 2 r + 2 , r + 1 , − 2 r − 2 ) . Let this be the coordinates of foot of perpendicular from A ( 3 , 6 , 1 5 ) to the given line. Now, since given line and AP are perpendicular hence dot of their direction ratios must be 0. Hence ( 2 r − 1 ) ( 2 ) + ( r − 5 ) ( 1 ) + ( − 2 r − 1 7 ) ( − 2 ) = 0 ⇒ r = − 3 Now since P is the mid point of A D , we get by distance formula its coordinates as ( 4 r + 1 , 2 r − 4 , − 4 r − 1 9 ) sum of which is 2 r − 2 2 . Placing r = − 3 we get the required answer as 2 ( − 3 ) − 2 8 = − 2 8 .