Find the image of point (1,2,3) in the plane x + y + z = 6

Question

Find the image of the point $(1, 2, 3)$ in the plane $x + y + z = 6$ .

(JEE Main 2023, similar pattern)

Solution — Step by Step

The plane $x + y + z = 6$ has normal vector $\vec{n} = (1, 1, 1)$ .

The line from the point $P(1, 2, 3)$ perpendicular to the plane is:

\frac{x - 1}{1} = \frac{y - 2}{1} = \frac{z - 3}{1} = t

So any point on this line is $(1 + t, 2 + t, 3 + t)$ .

The foot of perpendicular $F$ lies on the plane, so:

$(1 + t) + (2 + t) + (3 + t) = 6$

$6 + 3t = 6$

$t = 0$

So the foot of perpendicular is $F = (1, 2, 3)$ .

Wait — this means the point $(1, 2, 3)$ already satisfies $1 + 2 + 3 = 6$ . The point lies on the plane!

When a point lies on the mirror plane, its reflection (image) is the point itself.

\boxed{\text{Image} = (1, 2, 3)}

Why This Works

The image of a point in a plane is its mirror reflection. If $P$ is at distance $d$ from the plane, the image $P'$ is at the same distance $d$ on the other side, along the normal direction. The foot of perpendicular $F$ is the midpoint of $PP'$ .

In this case, $P$ is already on the plane ( $d = 0$ ), so $P = F = P'$ .

For the general case where $P$ is NOT on the plane, we’d use $t \neq 0$ and the image would be $P' = (1 + 2t, 2 + 2t, 3 + 2t)$ — double the parameter value because $F$ is the midpoint.

Alternative Method — Direct formula

The image of point $(x_1, y_1, z_1)$ in the plane $ax + by + cz = d$ is:

\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 - d)}{a^2 + b^2 + c^2}

Here: $\frac{-2(1 + 2 + 3 - 6)}{1 + 1 + 1} = \frac{-2(0)}{3} = 0$

So $x = 1, y = 2, z = 3$ . Same answer.

Always check whether the point lies on the plane before doing heavy calculations. Substitute $(x_1, y_1, z_1)$ into the plane equation — if it satisfies, the answer is immediate. In JEE, this can be a 5-second check that saves 3 minutes of unnecessary work.

Common Mistake

When the point is NOT on the plane, a common error is using $t$ directly as the image instead of $2t$ . The foot of perpendicular $F$ is at parameter $t$ , but the image $P'$ is at parameter $2t$ (since $F$ is the midpoint of $P$ and $P'$ ). If the formula gives $t = k$ , the image coordinates use $2k$ , not $k$ . For this specific problem the distinction doesn’t matter ( $t = 0$ ), but it’s critical in the general case.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct formula

Common Mistake

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