Question

Find the image of the point 1,2,3 in the plane x + 2y + 4z - 38 =0 Help?

Answer 1

See below

Explanation:

The image of point `P = (1,2,3 )` has co-ordinates `Q = (alpha, beta , gamma)`.

Midpoint of `P " and " Q` sits in plane `bb pi: x + bb2 y + bb4 z = 38`, so:

• `bb1(alpha + 1)/2 + bb2(beta + 2)/2 + bb4(gamma + 3)/2 = 38`

`:. alpha + 2 beta + 4 gamma = 59 qquad square`

And the line joining `P " and " Q` has same direction vector as the normal to the plane, `bbn = (1,2,4)`, so:

`bb(QP) = k \ bbn`

`(alpha - 1)/1 = (beta - 2)/2 = (gamma - 3)/4 = k`

• `implies {(beta = 2 alpha),(gamma = 4 alpha -1):}`

Turn everything in `square` into `alpha`:

`alpha + 2 (2 alpha) + 4 (4 alpha - 1) = 59`

• `implies {(alpha = 3),(beta = 6),(gamma = 11):}`

Check:

`bb(QP) = (2, 4, 8) = 2 \ bbn, qquad :. k = 2`