Image of a Point in a Plane
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Direct answer
To reflect a point P(x1, y1, z1) in the plane ax + by + cz + d = 0, compute the signed ratio s = (ax1 + by1 + cz1 + d)/(a² + b² + c²). The foot of the perpendicular is Q = P − s(a, b, c), and the image is P′ = P − 2s(a, b, c), because the foot is the midpoint of P and its reflection. The distance from P to the plane is |s|√(a² + b² + c²), and PP′ is exactly double. The same machinery extends to mirror images of lines and to foot-of-perpendicular questions, including the foot on a line, where the condition becomes PQ·b = 0.
What you must remember
- One substitution does everything: s = (ax1 + by1 + cz1 + d)/(a² + b² + c²) is the only computation; foot, image and distances are scalar multiples of it.
- Foot: Q = P − s(a, b, c); image: P′ = P − 2s(a, b, c) — subtract along the normal because the normal points from the plane toward larger values of the left side.
- Midpoint property: Q is the midpoint of PP′, the fastest self-check after any computation.
- Distance pair: P to plane is |ax1 + by1 + cz1 + d|/√(a² + b² + c²); P to P′ is exactly double.
- Direction only reflects its normal component: the component of any vector along the normal flips sign; the component inside the plane survives — which is why image directions can be written without computing points.
- Foot on a line (the sibling skill): for a line through A with direction b, the foot Q = A + tb satisfies (P − Q)·b = 0, giving t = (P − A)·b/|b|².
- Degenerate tells: if s = 0 the point already sits on the plane and is its own image.
From foot to image
Find the image of P(1, 2, 3) in the plane x + y + z = 9. The left side at P is 6, so the signed ratio is s = (6 − 9)/(1 + 1 + 1) = −1. The foot is P − s(1, 1, 1) = (1, 2, 3) + (1, 1, 1) = (2, 3, 4) — and indeed 2 + 3 + 4 = 9, on the plane. The image doubles the step: P′ = (3, 4, 5), with midpoint (2, 3, 4) confirming the construction. Now the same one-liner on a less friendly plane: reflect the origin in 2x + y − 2z + 9 = 0. Here s = 9/9 = 1, so the image is (0, 0, 0) − 2(2, 1, −2) = (−4, −2, 4). The origin sits at distance |9|/3 = 3 from the plane; the image sits at distance |−9|/3 = 3 on the opposite side, since substituting gives 2(−4) + (−2) − 2(4) + 9 = −9. The verification that closes the case is the midpoint test: (−2, −1, 2) satisfies the plane exactly, because −4 − 1 − 4 + 9 = 0.
Mirror questions across levels
JEE Main asks the point-reflection directly as a numerical — the losses come from sign errors in s, which is why computing the foot first and doubling is safer than applying the image formula blind. JEE Advanced escalates to images of lines and planes: the image of a line in a plane is found by reflecting one point of it and flipping the direction's normal component; the image of one plane in another is the locus of reflected points, handled through two foot computations. The trap to rehearse: distance uses the modulus, but the image needs the signed s; candidates who take |s| early mirror the point to the wrong side whenever the point's left-side value is negative, and the options include exactly that wrong-side point. End every such question with the midpoint check — it costs five seconds and catches every sign slip.
Frequently asked questions
What is the image of (x1, y1, z1) in ax + by + cz + d = 0?
P′ = P − 2s(a, b, c) with s = (ax1 + by1 + cz1 + d)/(a² + b² + c²); the foot of the perpendicular is P − s(a, b, c).
Why is the foot the midpoint of the point and its image?
Reflection is symmetric about the mirror, so the plane bisects the perpendicular segment PP′ at the foot.
How far is the image from the original point?
Twice the point-to-plane distance: 2|ax1 + by1 + cz1 + d|/√(a² + b² + c²).
How do you reflect a line in a plane?
Reflect one point of it and flip the direction's normal component; the join gives the image line.
How is the foot on a line found?
Take Q = A + tb on the line and impose (P − Q)·b = 0, solving t = (P − A)·b/|b|².