Here you will learn how to find image of a point in a plane formula with examples.
Let’s begin –
Image of a Point in a Plane
Let P and Q be two points and let \(\pi\) be a plane such that
(i) line PQ is perpendicular to the plane \(\pi\) and,
(ii) mid-point of PQ lies on the plane \(\pi\).
Then, either of the point is the image of the other in the plane \(\pi\).
In order to find the image of a points \((x_1, y_1, z_1)\) in the plane ax + by + cz + d = 0, we may use the following algorithm.
Algorithm :
1). Write the equations of the line passing through P and normal to the given plane as \(x – x_1\over a\) = \(y – y_1\over b\) = \(z – z_1\over c\).
2). Write the coordinates of image Q as \((x_1 + ar, y_1 + br, z_1 + cr)\).
3). Find the coordinates of the mid-point R of PQ.
4). Obtain the value of r by substituting the coordinates of R in the equation of the plane.
5). Put the value of r in the coordinates of Q.
Example : Find the image of the point P (3, -2, 1) in the plane 3x – y + 4z = 2.
Solution : Let Q be the image of P(3, -2, 1) in the plane 3x – y + 4z = 2. Then PQ is normal to the plane. Therefore, direction ratios of PQ are proportional to 3, -1, 4. Since PQ passes through P(3, -2, 1) and has direction ratios proportional to 3, -1, 4.
Therefore, equation of PQ is
\(x – 3\over 3\) = \(y + 2\over -1\) = \(z – 1\over 4\) = r (say)
Let the coordinates of Q be (3r + 3, -r – 2, 4r + 1). Let R be the mid-point of PQ. Then, R lies on the plane 3x – y + 4z = 2.
The coordinates of R are (\({3r + 3 + 3\over 3}, {-r – 2 – 2\over 2}, {4r + 1 +1\over 2}\))
or, (\({3r + 6\over 3}, {-r – 4\over 2}, {2r + 1}\))
Since r lies on the plane 3x – y + 4z = 2.
\(3({3r + 6\over 2}) – ({-r – 4\over 2})\) + 4(2r + 1) = 2
\(\implies\) 13r = -13 \(\implies\) r = – 1
Putting r = -1 in (3r + 3, -r – 2, 4r + 1), we obtain the coordinates of Q as (0, – 1, -3).
Hence, the image of (3, -2, 1) in the plane 3x – y + 4z = 2 is (0, -1, -3).
