JEE Mains Maths · Three Dimensional Geometry

Distance, Foot and Image in a Plane

Measuring from a point to a plane: the perpendicular distance, the distance between parallel planes and which side a point lies on, the foot and the mirror image of a point, and projections onto a plane.

All Three Dimensional Geometry notes JEE Mains Maths strategy

Why this matters

Twenty-seven PYQs, most of them the same computation: put the point into the plane's equation and divide by the normal's length. The image of a point in a plane is the most common single question. Three ideas cover the page.

Concept 1 of 3: Distance from a plane, parallel planes and sides of a plane

Put a point into the left side of

ax+by+cz+d

: the number you get, divided by

|\vec n|

, is its signed distance from the plane. Its size is the perpendicular distance and its sign tells which side the point is on. Parallel planes, once written with the same normal, are

\frac{|d_1-d_2|}{|\vec n|}

apart.

Definition

Distance: $\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$ .
Parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ : $\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$ (same $a,b,c$ first).
Same side: the two values have the same sign; opposite signs mean opposite sides.
Bisector planes: $\frac{P_1}{|\vec n_1|}=\pm\frac{P_2}{|\vec n_2|}$ .

Distance of a point from a plane

d=\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}

Worked example

Find the distance of

(1,2,3)

from

2x-y+2z=3

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 9 · Q86Moderate

Example 1 · Three Dimensional Geometry · Distance, Foot and Image in a Plane

The equation of the planes parallel to the plane

x

- 2y+ 2z- 3 = 0

which are at unit distance from the point

(1,2,3)

ax+by+cz+d= 0

. If

(b-d) =K(c-a)

, then the positive value of

K

Match the normals before subtracting constants

2x+y-2z=1

and

4x+2y-4z=11

are parallel, but

|11-1|/3

is wrong: rescale one equation so both have the same

a,b,c

first.

Drill 5 more on distance from a plane, parallel planes and sides of a plane

Concept 2 of 3: The foot and the image of a point in a plane

From

P

, walk along the normal until you hit the plane: that is the foot

F

. Keep going the same distance again: that is the image

Q

. Both are

P

minus a multiple of

\vec n

, and the multiple is the plane's value at

P

divided by

|\vec n|^2

, once for the foot and twice for the image.

Definition

Let $k=\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$ .
Foot: $F=P-k\,\vec n$ .
Image: $Q=P-2k\,\vec n$ .
$F$ is the midpoint of $P$ and $Q$ , and $\vec{PQ}\parallel\vec n$ .
With the image given and unknowns in the plane: $\vec{PQ}\parallel\vec n$ and the midpoint lies on the plane.

Foot and image

F=P-k\,\vec n,\quad Q=P-2k\,\vec n,\quad k=\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}

Worked example

Find the foot and the image of

(1,2,3)

x+y+z=3

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 11 April 2023 · Q77Moderate

Example 2 · Three Dimensional Geometry · Distance, Foot and Image in a Plane

Let

(\alpha,\beta,\gamma)

be the image of the point

P(2,3,5)

in the plane

2x + y - 3z = 6

. Then

\alpha + \beta + \gamma

is equal to

Keep the sign of the plane's value

k

can be negative, which moves the point along

+\vec n

. Taking

|k|

sends the foot and image to the wrong side.

Drill 16 more on the foot and the image of a point in a plane

Concept 3 of 3: Projections onto a plane

The projection of a segment onto a plane is what is left after removing its component along the normal. So its length is

\sqrt{|\vec v|^2-(\vec v\cdot\hat n)^2}

. When two points have equal values in the plane's equation, the segment joining them is parallel to the plane and projects to its own length.

Definition

Length of the projection of $\vec v$ : $\sqrt{|\vec v|^2-(\vec v\cdot\hat n)^2}$ .
Projection of a segment $PQ$ : the segment joining the two feet.
Equal plane-values at $P$ and $Q$ : $PQ$ is parallel to the plane and the projection has length $PQ$ .
The projection of a curve: take the foot of its general point, then remove the parameter.

Length of a projection onto a plane

\sqrt{|\vec v|^2-(\vec v\cdot\hat n)^2}

Worked example

Find the length of the projection of

\vec v=(1,2,2)

on the plane

x+y+z=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 June 2022 · Q83Moderate

Example 3 · Three Dimensional Geometry · Distance, Foot and Image in a Plane

Let

d

be the distance between the foot of perpendiculars of the points

P(1,2,-1)

and

Q(2, -

1,3) on the plane

-x+y+z= 1

. Then

d^{2}

is equal to

Subtract the normal component, not the whole vector

The projection keeps the part of

\vec v

that lies in the plane. Using

\vec v\cdot\hat n

as the answer gives the part that was removed.

Drill 3 more on projections onto a plane

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Distance from a plane, parallel planes and sides of a plane
Distance of a point from a plane
$d=\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$
The foot and the image of a point in a plane
Foot and image
$F=P-k\,\vec n,\quad Q=P-2k\,\vec n,\quad k=\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$
Projections onto a plane
Length of a projection onto a plane
$\sqrt{|\vec v|^2-(\vec v\cdot\hat n)^2}$

Watch out for (3)

Match the normals before subtracting constants
→ Distance from a plane, parallel planes and sides of a plane
Keep the sign of the plane's value
→ The foot and the image of a point in a plane
Subtract the normal component, not the whole vector
→ Projections onto a plane

Test yourself on Three Dimensional Geometry

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