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Vector reflection in a plane.Find the equation of the image of a line `(x,y,z) =...

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utilityfan | Student, College Freshman | Valedictorian

Posted January 24, 2012 at 10:02 PM via web

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Vector reflection in a plane.

Find the equation of the image of a line `(x,y,z) = (1,2,3) + t(1,-1,0)` reflected in the plane `x+y+z-3=0`

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txmedteach | High School Teacher | (Level 3) Associate Educator

Posted January 25, 2012 at 12:04 AM (Answer #1)

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To start, let's figure out what it means to reflect the line in the plane. We can use an affine transformation matrix (see link below), but this does not illustrate what's going on conceptually. To think of this visually, remember what it means to reflect. For any point P (or its associated vector from the origin `vecP`) on the line, we take a vector normal to the plane that reaches P. We then take the negative vector from the surface of the plane to find the image of P (P' or its associated vector from the origin `vec(P')` ) based on reflection. Put another way, supoose we have a normal, `vecn` , and P is a perpendicular distance `d/|vecn|` away from a point M (represented by a vector from the origin `vecM`) on the plane. P' can be expressed in the following way:

`vec(P') = vecM - dvecn = vecP - 2dvecn`

The last result follows from our knowledge that `vecM = vecP-dvecn` . So, we can perform this sort of calculation from each point on the line we're given to a point on the plane.

We do this by sutracting the vector components of the normal vector from each of the components of the points on the line. In other words, there will be a given point on the plane `M_t` for any given `t` such that:

`vec(M_t) = <1,2,3>+t<1,-1,0> - dvecn`

Now, recall that we are given the normal vector in the equation for the plane. Also, recall that we are given how to find all points on the plane. Therefore, we know:

`vec(M_t) = <1,2,3>+t<1,-1,0> + d<-1,-1,-1>`

giving us:

`vec(M_t) = <1+t-d, 2-t-d, 3-d>`

Therefore, our points on the plane are have the same arguments for x, y, and z. We can now get `d` as a function of `t` by using the equation for the plane:

`(1+t-d) + (2-t-d) + (3-d) = 3`

Simplifying:

`6-3d = 3`

`-3d = -3`

`d = 1`

Therefore, our line is a constand displacement of `vecn` away from the plane. Recall, now our equation to find the image of the line across the plane:

`vec(P') = vecP - 2dvecn`

Now, we substitute our equation for the line for `vecP`, 1 for `d`, and <1,1,1> for `vecn` :

`vec(P') = <1,2,3> + t<1,-1,0> - 2*1*<1,1,1>`

Simplifying, we get our solution!

`vec(P') = <-1,0,1> + t<1,-1,0>`

There is your reflection. I hope this helps in terms of the conceptual aspect of the problem!

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