3D Geometry

September 26, 2009

We will learn 3D Geometry with the help of vectors. All derivation will goin sync with our understanding with Vectors. So I propose to do Vectors �rstbefore starting to prepare 3D Geometry.

Abstract

NOTE : Another version of these notes is coming along where

problems for each theory part will be included. Here right now

only theory is being discussed.

Here we discuss the �ow of topics which will be covered in this sup-plement and in what order.

We divide the whole topic of 3D geometry into four parts.

First part deals with the points in 3D geometry and associated con-cepts.

Second part surfaces the relation in angles made with the axes andconcept of direction in 3D geometry, (corresponding to slope concept in2D geometry)

Third part talks about the plane equation and some interesting com-binations of planes with line and points and more than one planes

Fourth part, comes with lines and its combinations with many linesand in reference with plane.

1

2 CONTENTS

Contents

1 Analogy of 2D with 3D 5

2 Points in 3D Geometry 52.1 Two points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.1.1 Distance between A & B . . . . . . . . . . . . . . . . . . . 52.1.2 Section formula . . . . . . . . . . . . . . . . . . . . . . . . 52.1.3 Midpoint of segment AB . . . . . . . . . . . . . . . . . . 52.1.4 Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Direction Cosines & Direction Ratios 63.1 Direction Cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3.1.1 Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Direction Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.2.1 Creating direction cosine from direction ratio . . . . . . . 73.3 Summary of Direction cosine and Direction Ratio . . . . . . . . . 73.4 Projection of a line segment onto another line . . . . . . . . . . . 7

3.4.1 Projection of AB along coordinate axes . . . . . . . . . . 73.5 Direction ratio of line joining two points . . . . . . . . . . . . . . 8

4 Plane 94.1 Equation of a plane in di�erent forms . . . . . . . . . . . . . . . 9

4.1.1 General form . . . . . . . . . . . . . . . . . . . . . . . . . 94.1.2 Normal distance form . . . . . . . . . . . . . . . . . . . . 94.1.3 Point Normal form . . . . . . . . . . . . . . . . . . . . . . 94.1.4 A plane parallel to two lines and passing through a point 94.1.5 A plane containing two points and parallel to a line . . . 94.1.6 Three point form . . . . . . . . . . . . . . . . . . . . . . . 94.1.7 Intercept form . . . . . . . . . . . . . . . . . . . . . . . . 104.1.8 Special cases . . . . . . . . . . . . . . . . . . . . . . . . . 10

5 Point & a Plane 105.1 Position of a point with respect to a plane . . . . . . . . . . . . . 10

5.1.1 same side of a line . . . . . . . . . . . . . . . . . . . . . . 105.1.2 opposite side of a line . . . . . . . . . . . . . . . . . . . . 105.1.3 Origin side of the line . . . . . . . . . . . . . . . . . . . . 10

5.2 Point outside a plane . . . . . . . . . . . . . . . . . . . . . . . . . 115.2.1 Distance of a point from the plane . . . . . . . . . . . . . 115.2.2 Perpendicular foot of a point onto the plane . . . . . . . . 115.2.3 Image of a point in a plane . . . . . . . . . . . . . . . . . 11

6 Two planes 116.1 Angle between two planes . . . . . . . . . . . . . . . . . . . . . . 11

6.1.1 Special Case . . . . . . . . . . . . . . . . . . . . . . . . . 116.2 Distance between planes . . . . . . . . . . . . . . . . . . . . . . . 126.3 Plane bisector of two planes . . . . . . . . . . . . . . . . . . . . . 13

CONTENTS 3

6.3.1 Direction of normal to a plane . . . . . . . . . . . . . . . 136.3.2 Origin containing region between two intersecting planes . 136.3.3 Bisector containing the origin . . . . . . . . . . . . . . . 146.3.4 Bisector of acute/obtuse angle . . . . . . . . . . . . . . . 15

7 Projection of an Area 15

8 Linear combination of planes 168.0.5 Non-parallel planes . . . . . . . . . . . . . . . . . . . . . . 168.0.6 Parallel planes . . . . . . . . . . . . . . . . . . . . . . . . 16

9 Plane sects a line joining two points 16

10 Three planes 1610.1 Normals are coplanar . . . . . . . . . . . . . . . . . . . . . . . . . 1610.2 Normals are non-coplanar . . . . . . . . . . . . . . . . . . . . . . 17

11 Equation of a line 1811.1 Two point form of a line . . . . . . . . . . . . . . . . . . . . . . . 1811.2 Slope point form . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

11.2.1 Special case . . . . . . . . . . . . . . . . . . . . . . . . . . 1811.2.2 Unsymmetric form . . . . . . . . . . . . . . . . . . . . . . 18

11.3 Point and a line . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1911.3.1 Foot of perpendicular . . . . . . . . . . . . . . . . . . . . 1911.3.2 Perpendicular distance . . . . . . . . . . . . . . . . . . . . 2011.3.3 Image of a point in a line . . . . . . . . . . . . . . . . . . 20

11.4 Two lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2011.4.1 Skew lines . . . . . . . . . . . . . . . . . . . . . . . . . . . 2011.4.2 Coplanar . . . . . . . . . . . . . . . . . . . . . . . . . . . 2111.4.3 Angle between two lines . . . . . . . . . . . . . . . . . . 21

12 Line & a Plane 2112.1 Angle between line & a plane . . . . . . . . . . . . . . . . . . . . 2112.2 Projection of a line onto a plane . . . . . . . . . . . . . . . . . . 2112.3 Image of a line in a plane . . . . . . . . . . . . . . . . . . . . . . 22

12.3.1 Line is parallel to plane . . . . . . . . . . . . . . . . . . . 2312.4 Techniques in speci�c Problems . . . . . . . . . . . . . . . . . . . 23

4 CONTENTS

5

1 Analogy of 2D with 3D

Lets start with few learning from 2D Geometry, cartesian coordinate system.Any point at a distance of r units from origin and making an angle � withpositive x axis. The polar form of the point is (r cos�, r sin�). Line joiningthis point with origin makes an angle � with positive x axis and � with positivey-axis. Hence the relation between the angles made by this vector with x & yaxes is �+ � = �

2 .If we try to think of something similar relation in 3D geometry. But the

problem here lies that the angles are in 3D and hence may not form a supple-mentary or complementary. But surely there will be some relation between theangles make by a line in 3D space.

2 Points in 3D Geometry

2.1 Two points

Given two points A(x1, y1, z1) and B(x2, y2, z2) in 3D space.

2.1.1 Distance between A & B

Distance l(AB) =√

(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2

2.1.2 Section formula

If a point C divided line joining segment AB internally or externally in the ratiom : n then

C =

(mx2 ± nx1m± n

,my2 ± ny1m± n

,mz2 ± nz1m± n

)where ± corresponds to internal or external section

Problem 1. To �nd the ratio in which the point C(a, b, c) divides the linejoining A(x1, y1, z1) & B(x2, y2, z2)

AC

CB=x1 − aa− x2

=y1 − bb− y2

=z1 − cc− z2

2.1.3 Midpoint of segment AB

midpoint =

(x1 + x2

2,y1 + y2

2,z1 + z2

2

)3 Points are collinearThree points A(x1, y1, z1), B(x2, y2, z2) & C(x3, y3, z3) are collinear then

x1 − x2x2 − x3

=y1 − y2y2 − y3

=z1 − z2z2 − z3

6 3 DIRECTION COSINES & DIRECTION RATIOS

2.1.4 Triangle

1. Area of triangle

Use vector cross product better or we will see another formula to get itgoing using 3D geometry

2. Formula for centroid(G), Incentre(I), Excentres (I1, I2, I3), Orthocenter(H)and Circumcenter(O)

use

(∑mixi∑mi

,

∑miyi∑mi

,

∑mizi∑mi

)(a) Centroid : (m1,m2,m3) ≡ G ≡ (1, 1, 1)

(b) Incenter : (m1,m2,m3) ≡ I ≡ (sinA, sinB, sinC)

(c) Excenter : (m1,m2,m3) ≡ I1 ≡ (− sinA, sinB, sinC) similarly forother excenter

(d) Orthocenter : (m1,m2,m3) ≡ H ≡ (tanA, tanB, tanC)

(e) Circumcenter : (m1,m2,m3) ≡ O ≡ (sin 2A, sin 2B, sin 2C)

3 Direction Cosines & Direction Ratios

3.1 Direction Cosine

Let us start with a point P (x0, y0, z0). So the position vector of this point will

be r = x0i+ y0j + z0k. Let us convert this point into polar form. Let r makesan angle �, � & with x, y & z axis respectively and ∣r∣ = r.

Therefore r = x0i+ y0j + z0k = rcos�i+ rcos�j + rcos kNow the unit vector along the line joining origin and point P isr = x0√

x20+y

20+z

20

i+ y0√x20+y

20+z

20

j + z0√x20+y

20+z

20

k = cos�i+ cos�j + cos k

Point P lies on the line passing through origin and point P. We have createda unit along this line.

Now the terms which we have got (cos�, cos�, cos )1 is the unit vector alongline OP. So this unit vector helps in knowing the direction of this line. And anyvector has one unique unit vector along its direction. This can be easily proved.

Let two unit vectors be along the same lines. Hence they arecollinear unit vectors.

u1 = �u2 ⇒ ∣u1∣ = ∣�u2∣ Taking modulus on both sides.⇒ ∣�∣ = 1 ⇒ � = ±1. So along a line there is one unique unit

vector the other is just its opposite direction.Notation : A direction cosine is denoted as (l,m, n)

So we claim, direction cosine is unit vector along a line.

1Direction cosine (l,m, n) ≡ li + mj + nk. In 3D geometry a vector is written as a

coordinate. So any vector ai+ bj + ck in 3D is (a, b, c)

3.2 Direction Ratio 7

3.1.1 Identities

1. cos2 �+ cos2 � + cos2 = 1

2. sin2 �+ sin2 � + sin2 = 2

3.∑

cos 2� = −1

4. Direction cosine of coordinate axes are (1, 0, 0), (0, 1, 0) & (0, 0, 1)

3.2 Direction Ratio

Direction Ratio is de�ned as any scalar multiple of the direction cosine. Invector terms, as direction cosine is the unit vector similarly direction ratio isany vector along a particular line. Note: Any vector ≡direction ratio and

unit vector ≡ direction cosine

3.2.1 Creating direction cosine from direction ratio

If we know the direction ratio say (a, b, c) then we have any vector along itsdirection. To create the direction cosine along the same direction is to �ndthe unit vector along this direction. So just divide the direction ratio with thelength of direction ratio.

DC =

(a√

a2 + b2 + c2,

b√a2 + b2 + c2

,c√

a2 + b2 + c2

),

3.3 Summary of Direction cosine and Direction Ratio

1. Direction cosine of a line in 3D geometry is Unit vector along that line inVectors

2. Direction ratio along a line in 3D geometry is Any vector along that linein Vectors.

3.4 Projection of a line segment onto another line

Given a line segment joining points A(x1, y1, z1) & B(x2, y2, z2) and a line withdirection cosines (l,m, n) then the projection of the segment onto the line is

(x2 − x1)l + (y2 − y1)m+ (z2 − z1)n

3.4.1 Projection of AB along coordinate axes

AB along

1. x− axis : x2 − x1

2. y − axis : y2 − y1

3. z − axis : z2 − z1

8 3 DIRECTION COSINES & DIRECTION RATIOS

Example 2. What is the direction cosine of

∙ x-axis

1. To �nd the direction cosine along x-axis we �rst �nd the unit vectoralong x-axis. i = 1i + 0j + 0k is the unit vector along x-axis hence(1, 0, 0) is the direction cosine along x-axis.

2. Another way is to �nd the angle x-axis makes with positive x,y & zaxis. So � = 0, � = �/2 & = �/2. Hence the direction cosine is(cos 0, cos�2 , cos

�2 ) = (1, 0, 0)

∙ bisector of x-z axes

1. The bisector of x-z axes will lie in the x-z plane. So we solve thisproblem by �nding any vector along the direction of this bisector andthen �nding unit vector from that vector which then can be claimedas the direction cosine. Any vector along the bisector = i + k. And

unit vector along this direction is i√2

+ 0j + k√2. Hence the direction

cosine = ( 1√2, 0, 1√

2)

2. Another way is through (cos�, cos�, cos ). Here the bisector makesan angle �/4 with both x & z axes. It makes a right angle with yaxis. Hence direction cosine = (cos�4 , cos 0, cos�4 ) = ( 1√

2, 0, 1√

2)

3.5 Direction ratio of line joining two points

Direction ratio of line joining two points P(x1, y1, z1) & Q(x2, y2, z2) is a vector

along this line. So vector−−→PQ = q − p = (x2 − x1, y2 − y1, z2 − z1). This is any

vector that is direction ratio.

9

4 Plane

4.1 Equation of a plane in di�erent forms

4.1.1 General form

General form of a plane equation is

ax+ by + cz + d = 0

4.1.2 Normal distance form

Given the direction cosine of the normal to a plane (a, b, c) and distance of theplane from origin −d then equation of the plane

ax+ by + cz + d = 0

4.1.3 Point Normal form

Given a point (x0, y0, z0) lying on the plane and normal to the plane (a, b, c)then the equation of the plane

a(x− xo) + b(y − y0) + c(z − z0) = 0

4.1.4 A plane parallel to two lines and passing through a point

A plane parallel to two lines with direction cosines/ratios (a1, a2, a3) and (b1, b2, b3)and passing through the point (x0, y0, z0)∣∣∣∣∣∣

x− x1 y − y1 z − z1a1 a2 a3b1 b2 b3

∣∣∣∣∣∣ = 0

4.1.5 A plane containing two points and parallel to a line

Let (x1, y1, z1) & (x2, y2, z2) be two points on the required plane and a lineparallel to the plane with direction cosine/ratio (a, b, c) then the equation of theplane ∣∣∣∣∣∣

x− x1 y − y1 z − z1x1 − x2 y1 − y2 z1 − z2

a b c

∣∣∣∣∣∣ = 0

4.1.6 Three point form

Given three points (x1, y1, z1), (x2, y2, z2) & (x3, y3, z3) lie on a plane whoseequation is ∣∣∣∣∣∣

x− x1 y − y1 z − z1x1 − x2 y1 − y2 z1 − z2x2 − x3 y2 − y3 z2 − z3

∣∣∣∣∣∣ = 0

10 5 POINT & A PLANE

4.1.7 Intercept form

A plane that makes intercepts a, b & c with the coordinate axes is

x

a+y

b+z

c= 1

4.1.8 Special cases

1. x = 0 : is the y − z plane

2. y = 0 : is the x− y plane

3. z = 0 : is the x− y plane

4. plane perpendicular to the x− axis is ay + bz = 1

5. plane perpendicular to the y − axis is ax+ bz = 1

6. plane perpendicular to the z − axis is ax+ by = 1

5 Point & a Plane

5.1 Position of a point with respect to a plane

5.1.1 same side of a line

If two points P (x1, y1, z1) & Q(x2, y2, z2) are on the same side of the planeax+ by + cz + d = 0 then

(ax1 + by1 + cz1 + d)(ax2 + by2 + cz2 + d) > 0

5.1.2 opposite side of a line

If two points P (x1, y1, z1) & Q(x2, y2, z2) are on the other side of the planeax+ by + cz + d = 0 then

(ax1 + by1 + cz1 + d)(ax2 + by2 + cz2 + d) < 0

5.1.3 Origin side of the line

To �nd if a point P (x1, y1, z1) is on the origin side of a line then �rst step isto make d positive in the line equation and substitute (x1, y1, z1) in the lineequation to see its sign. From above deduction we get

ax1 + by1 + cz1 + d > 0 (d>0)

then the point is on the origin side of the line else

ax1 + by1 + cz1 + d < 0 (d>0)

then the point lies on the non-origin side of the line.

5.2 Point outside a plane 11

5.2 Point outside a plane

There are three types of interesting problems which can arise if a point is outsidea plane. Foot of the perpendicular of the point onto the plane, distance of thepoint from the plane & Image of the point in the plane.

5.2.1 Distance of a point from the plane

Distance of a point (x1, y1, z1) from a plane ax+ by + cz + d = 0 is

∣ax1 + by1 + cz1 + d∣√a2 + b2 + c2

5.2.2 Perpendicular foot of a point onto the plane

Foot of the perpendicular of a point (x1, y1, z1) onto the plane ax+by+cz+d = 0is given by

x− x1a

=y − y1b

= −(ax1 + by1 + cz1 + d

a2 + b2 + c2

)5.2.3 Image of a point in a plane

Image of a point (x1, y1, z1) in the plane ax+ by + cz + d = 0 is given by

x− x1a

=y − y1b

= −2

(ax1 + by1 + cz1 + d

a2 + b2 + c2

)

6 Two planes

6.1 Angle between two planes

Angle between two planes a1x+ b1y + c1 + d1 = 0 and a2x+ b2y + c2 + d2 = 0is same as angle between their normals

cos � =a1a2 + b1b2 + c1c2√

a21 + b21 + c21√a22 + b22 + c22

6.1.1 Special Case

1. If they are parallel

a1b1

=a2b2

=a3b3

i.e.(a1, b1, c1) = �(a2, b2, c2)

As vectors they are collinear vectors

2. If they are perpendicular

a1a2 + b1b2 + c1c2 = 0

12 6 TWO PLANES

6.2 Distance between planes

Two plane in a 3D space always intersect.

1. If they are parallelDistance between ax+ by + cz + d1 = 0 and ax+ by + cz + d2 = 0 is (asthey are parallel )

∣d1 − d2∣√a2 + b2 + c2

2. If they are non-parallelThe distance between the planes that intersect is zero.

6.3 Plane bisector of two planes 13

6.3 Plane bisector of two planes

6.3.1 Direction of normal to a plane

Given a plane ax+ by + cz + d = 0 if we make the constant positive then whatis the direction of the normal to this plane.

Lets think in terms of vector geometry

r ⋅ n = p

where p is the distance of the plane from the origin.

The above equation ax + by + cz + d = 0 (without loss of generality, d canbe assumed to be negative) can be converted into the vector equation as

(x, y, z) ⋅(

a√a2 + b2 + c2

,b√

a2 + b2 + c2,

c√a2 + b2 + c2

)=

−d√a2 + b2 + c2

where n =

(a√

a2 + b2 + c2,

b√a2 + b2 + c2

,c√

a2 + b2 + c2

)and p =

−d√a2 + b2 + c2

So the direction of

(a√

a2 + b2 + c2,

b√a2 + b2 + c2

,c√

a2 + b2 + c2

)is away

from origin. i.e. (a, b, c) is away from the origin.

So the �nal conclusion : For the a plane equation, ax+ by + cz + d = 0 onmaking d positive the direction of normal (a, b, c) is away from origin towards theplane.

6.3.2 Origin containing region between two intersecting planes

Given two plane equations P1 : a1x+ b1y+ c1z+d1 = 0 and P2 :a2x+ b2y+c2z+d2 = 0 such that d1& d2 > 0 then the green region containing the origin ispositive for both the planes P1 and P2 and the other vertically opposite greenregion is negative for both P1 and P2. So for the green region

P1 ⋅ P2 > 0 (where d1, d2 > 0)

14 6 TWO PLANES

Now lets concentrate on the red region. Both the red region are such thateither of P1 or P2 is negative

2. Hence

P1 ⋅ P2 < 0 (where d1, d2 < 0)

So from the above discussion we know which region will have the origin lyingin it. Now next we worry to �nd a method to �nd the acute anlge bisector fromproblem solving perspective and which angle (acute or obtuse) will contain theorigin.

6.3.3 Bisector containing the origin

Here we are intereseted in knowing which angle bisector contains the origin.Any point on the bisector satis�es

∣a1x+ b1y + c1z + d1∣√a21 + b21 + c21

=∣a2x+ b2y + c2z + d2∣√

a22 + b22 + c22a1x+ b1y + c1z + d1√

a21 + b21 + c21= ±a2x+ b2y + c2z + d2√

a22 + b22 + c22

The angle containing the origin (above diagram green region) has botha1x + b1y + c1z + d1 & a2x + b2y + c2z + d2 positive(in origin region) or bothnegative(vertically opposite to origin). Correspondingly, a1x+ b1y+ c1z+d1 ora2x+ b2y + c2z + d2 one of these is negative in the red region.

So,

a1x+ b1y + c1z + d1√a21 + b21 + c21

=a2x+ b2y + c2z + d2√

a22 + b22 + c22, (origin containing angle bisector)

a1x+ b1y + c1z + d1√a21 + b21 + c21

= −a2x+ b2y + c2z + d2√a22 + b22 + c22

, (non-origin containing angle bisector)

2P1, P2 is negative means P1 plane's expression value at (0, 0, 0) is negative i.e. a1(0) +b1(0) + c1(0) + d1 < 0 and vice-versa for being positive

15

6.3.4 Bisector of acute/obtuse angle

To �nd the equation, of the acute angle bisector between the planes a1x+ b1y+c1z + d1 = 0 and a2x+ b2y + c2z + d2 = 0.

For this problem we �rst locate the region of origin.

Step I : Make d1 > 0 and d2 > 0

Step II : (a1, b1, c1) and (a2, b2, c2) are normals to planes starting fromorigin to these respective planes

Step III : Angle between the normals : (a1, b1, c1) ⋅ (a2, b2, c2) = a1a2 +b1b2 + c1c2

If a1a2 + b1b2 + c1c2 > 0 then the normal between the planes is acute andthat implies the angle between the planes not containing the origin is acutemeans angle containing the origin is obtuse.

If a1a2 + b1b2 + c1c2 < 0 then the normal between the planes is obtuse andthat implies the angle between the planes not containing the origin is obtusemeans angle containing the origin is acute.

a1a2 + b1b2 + c1c2 > 0; d1, d2 > 0 a1a2 + b1b2 + c1c2 < 0; d1, d2 > 0

Origin lies in Obtuse Angle Acute AngleOrigincontainingAnglebisector

put + in the below formula put - in the below formula

The equation of angle bisectora1x+ b1y + c1z + d1√

a21 + b21 + c21= ±a2x+ b2y + c2z + d2√

a22 + b22 + c22

7 Projection of an Area

Two inclined planes, with an area A lying on one of the planes then we �nd theprojection of A onto the other plane.

16 10 THREE PLANES

If projection of areas of a plane A in 3D space is Axy, Ayz, Azx onto the x-y,y-z & z-x planes then A2 = A2

xy +A2yz +A2

zx

8 Linear combination of planes

8.0.5 Non-parallel planes

Given two planes P1 : a1x+ b1y + c1z + d1 = 0 & P2 : a2x+ b2y + c2z + d2 = 0then the equation of planes passing through intersection of these planes is givenby

P1 + �P2 = 0

8.0.6 Parallel planes

Given two planes P1 : ax+ by + cz + d1 = 0 & P2 : ax+ by + cz + d2 = 0 thenlinear combination represents family of parallel planes to these planes.

9 Plane sects a line joining two points

Ratio in which a plane ax+ by+ cz+ d = 0 divides the line joining A(x1, y1, z1)& B(x2, y2, z2) is given by

m

n= −

(ax1 + by1 + cz1 + d

ax2 + by2 + cy2 + d

)

10 Three planes

10.1 Normals are coplanar

∙ If the normals to three planes are coplanar then there are two

posibilities.

10.2 Normals are non-coplanar 17

1. All the three pass through a common line of intersection( this will form apencil of planes)

To locate this case we need box product of the normals [−→N1−→N2−→N3] = 0 and

show linear combination of any two is producing the third form uniquevalues of the scalar

2. One of the planes is not passing through intersection of the other two(they will form a prism kind of structure)

To locate this case we need box product of normals to be zero and showthe any point on the intersection of any two planes is not satisfying thethird plane.

10.2 Normals are non-coplanar

∙ If the normals to three planes are not coplanar then the planes

intersect in a unique point. see the diagram below

18 11 EQUATION OF A LINE

11 Equation of a line

If two vectors are along the same line then they are collinear and hence theyare scalar multiple of each other. If you know and can prove this then we canproceed to �nd the 3D equation of a line.

11.1 Two point form of a line

We have a line that passes through two points (x1, y1, z1) & (x2, y2, z2) then thewe deduce the equation of this line as follows.

Let (x, y, z) be a point on this line (locus point)Therefore (x− x1, y − y1, z − z1) = �(x1 − x2, y1 − y2, z1 − z2)

⇒ x− x1x1 − x2

=y − y1y1 − y2

=z − z1z1 − z2

= �

This is the equation of the line in two point form

11.2 Slope point form

We have a line that passes through a point (x1, y1, z1) and parallel to the linewith direction ratios (a, b, c) then equation of this line is

⇒ x− x1a

=y − y1b

=z − z1c

= constant

11.2.1 Special case

Equation of line passing through the point (x1, y1, z1) and parallel to the linewhose direction cosines are (l,m, n) then equation of this line is

⇒ x− x1l

=y − y1m

=z − z1n

= r

where r is the distance of between the points (x, y, z) & (x1, y1, z1)

11.2.2 Unsymmetric form

In 3D a line is uniquely determined by intersection of two planes.

a1x+ b1y + c1z + d1 = 0 (1)

a2x+ b2y + c2z + d2 = 0 (2)

Direction of the line of intersection of planes is same as perpendicular tonormals to these planes.

Normal to the two planes are−→N1 : (a1, b1, c1) &

−→N2 : (a2, b2, c2), hence the

vector along the line of intersection of the two planesis the vector that is perpendicular to the two normals. So the Direction cosines

11.3 Point and a line 19

of the line is

Direction Ratio of Line =

∣∣∣∣∣∣i j ka1 b1 c1a2 b2 c2

∣∣∣∣∣∣Problem. Given asymmetric form of equation of a line x+y = 1, z = 1. Whatis symmetric form of the equation of this line? And parametric form of theequation of this line.

Normal to the plane x+ y = 1:−→N1 = (1, 1, 0) & Normal to the second plane

z = 1:−→N2 = (0, 0, 1).

Direction cosine of the line of intersection of these two lines is−→N1×

−→N2 = (1, 1, 0)

Any point on this line of intersection is (1, 0, 1) (this we get by trial in the aboveplane equations)So the symmetric equation of the line of intersection is

x− 1

1=y − 1

0=z − 0

1= �

(lambda is a constant)So the parametric form of any point on this line is (�+ 1, 1, �)

11.3 Point and a line

Given a point P (x1, y1, z1) outside the linex− �a

=y − �b

=z − c

11.3.1 Foot of perpendicular

Direction ratio of the line is (a, b, c).

Letx− �a

=y − �b

=z − c

= �

For a speci�c � the foot of the perpendicular is (�a+ �, �b+ �, �c+ )Direction ratio of the perpendicular is (�a+ �− x1, �b+ � − y1, �c+ − z1)So

(a, b, c) ⋅ (�a+ �− x1, �b+ � − y1, �c+ − z1) = 0

� =(a, b, c) ⋅ (x1 − �, y1 − �, z1 − )

a2 + b2 + c2

Foot of perpendicular from (x1, y1, z1) onto the linex− �a

=y − �b

=z − c

is

given by

x− x1a

=y − y1b

=z − z1c

=

((a, b, c) ⋅ (x1 − �, y1 − �, z1 − )∑

a2

)=

((ax1 + by1 + cz1)− (a�+ b� + c )

a2 + b2 + c2

)

20 11 EQUATION OF A LINE

11.3.2 Perpendicular distance

You �nd the foot of the perpendicular from the above formula and then use thedistance formula to �nd the distance between the foot and the point.

11.3.3 Image of a point in a line

Similarly from the above formula we get if (x, y, z) is the image of a point(x1, y1, z1) then image is given by

x− x1a

=y − y1b

=z − z1c

= 2

((a, b, c) ⋅ (x1 − �, y1 − �, z1 − )∑

a2

)= 2

((ax1 + by1 + cz1)− (a�+ b� + c )

a2 + b2 + c2

)

11.4 Two lines

11.4.1 Skew lines

In 3D geometry, two lines can be non-parallel as well as non-intersecting. Suchlines are called skew lines.

Two linesx− x1a1

=y − y1b1

=z − z1c1

andx− x2a2

=y − y2b2

=z − z2c2

Shortest distance between them is

[(x1 − x2, y1 − y2, z1 − z2) (a1, b1, c1) (a2, b2, c2)]

∣(a1, b1, c1)× (a2, b2, c2)∣

Vector form of shortest distanceConverting given line equations into vector form : r = (x1i + y1j + z1k) +

�(a1i+ b1j + c1k) & r = (x2i+ y2j + z2k) + �(a2i+ b2j + c2k)Now from vectors we know if two vectors r = a+ �b & r = c+ �d then the

shortest distance between this skew lines is

=[a− c b d]

∣b× d∣

=

∣∣∣∣∣∣x1 − x2 y1 − y2 z1 − z2a1 b1 c1a2 b2 c2

∣∣∣∣∣∣√(b1c2 − c1b2)2 + (c1a2 − a1c2)2 + (b1c2 − c1b2)2

21

11.4.2 Coplanar

If [(x1−x2, y1−y2, z1−z2) (a1, b1, c1) (a2, b2, c2)] = 0 then the above mentionedlines are coplanar

11.4.3 Angle between two lines

If direction cosines or direction ratios of two lines are known then angle betweenthem

Let (l1,m1, n1) & (l2,m2, n2) are the direction cosines of two lines then

cos� = l1l2 +m1m2 + n1n2

Let (a1, b1, c1) & (a2, b2, c2) are the direction ratios of the lines then

cos� =a1a2 + b1b2 + c1c2√

a21 + b21 + c21√a22 + b22 + c22

12 Line & a Plane

12.1 Angle between line & a plane

We make use of the normal's direction ratio that is available for us from equationof the plane. Angle made by the line with the plane is

Angle =�

2− cos−1

((a, b, c) ⋅ (p, q, r)

√a2 + b2 + c2

√p2 + q2 + r2

)

where equation of plane is ax+ by + cz + d = 0 andx− �p

=y − �q

=z − r

12.2 Projection of a line onto a plane

Projection of a line onto a plane is worked out in the following steps.

22 12 LINE & A PLANE

Step I : Find the intersection point of intersection of the plane and the line.say point M (intersection of a line and a plane from above)

Step II : Now take the point on the given line and �nd its foot of per-pendicular onto the plane say P ′ (given a point outside a plane �nd its foot ofperpendicular)

Step III : Now get the equation of the projection using the two point formof a line in 3D

12.3 Image of a line in a plane

There are two posibilities with a line and a plane. If the direction cosine of theline is perpendicular to the plane then the line is either parallel to the plane oris contained in the plane.If the line is not parallel to the plane then it must be intersecting.

12.4 Techniques in speci�c Problems 23

12.3.1 Line is parallel to plane

Image of a line when parallel to the plane

Step I : Take two points on the line and get their images in the plane.Step II : Using these two points write the equation of the image line using

two point form.Image of a line when intersecting the plane

Step I : Find the intersection of the line with the planeStep II : Take the point which lies on the line from the formula and �nd

its image pointStep III : Using these two points write the equation of the line which is the

image of the given line in the given plane.

12.4 Techniques in speci�c Problems

Problem 3. Given a line x−11 = y−2

2 = z−33 and a point (1, 1, 1) �nd the foot

of the perpendicular & image of this point in the given line

Problem 4. Find the equation of a plane passing through intersection of planesx+ y + z = 1 & 2x+ y − z = 2 and passing through (0, 0, 0) ?

If we have two planes u = 0 & v = 0 then equation of family of planespassing through the intersection of these two planes is u+ �v = 0

Using the above fact, the equation of the plane passing through intersectionof the given planes is

(x+ y + z − 1) + �(2x+ y − z − 2) = 0

Now this plane passes through origin hence (0, 0, 0) hence satis�es this implies

−1 + �(−2) = 0

⇒ � = −1

2

Substituing we get the equation of the plane.