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

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Answer 1 · 2018-06-11T21:45:17

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

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

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

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

$bb(QP) = k \ bbn$

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

Turn everything in $square$ into $alpha$ :

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

Check:

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