3 d geometry lecture notes

8/11/2019 3 d Geometry Lecture Notes

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3D Geometry

August 31, 2014

We will learn 3D Geometry with the help of vectors. All derivation willgo in sync with our understanding with Vectors. So I propose to do Vectorsfirst before starting to prepare 3D Geometry.

NOTE : Another version of these notes is coming along whereproblems for each theory part will be included. Here right nowonly theory is being discussed.

Here we discuss the flow of topics which will be covered inthis supplement 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

concepts.Second part surfaces the relation in angles made with the axes

and concept of direction in 3D geometry, (corresponding to slopeconcept in 2D geometry)

Third part talks about the plane equation and some interestingcombinations of planes with line and points and more than oneplanes

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

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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 . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.2 Direction Ratio . . . . . . . . . . . . . . . . . . . . . . . . . 73.2.1 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 . . . . . . . . . . . 83.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4 Plane 94.1 Equation of a plane in different forms . . . . . . . . . . . . . 9

4.1.1 General form . . . . . . . . . . . . . . . . . . . . . . 9

4.1.2 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

4.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

5 Point & a Plane 115.1 Position of a point with respect to a plane . . . . . . . . . . . 11

5.1.1 same side of a plane . . . . . . . . . . . . . . . . . . 115.1.2 opposite side of a plane . . . . . . . . . . . . . . . . 115.1.3 Origin side of the plane . . . . . . . . . . . . . . . . 11

5.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115.3 Point outside a plane . . . . . . . . . . . . . . . . . . . . . . 11

5.3.1 Distance of a point from the plane . . . . . . . . . . . 125.3.2 Perpendicular foot of a point onto the plane . . . . . 125.3.3 Image of a point in a plane . . . . . . . . . . . . . . . 12

5.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

6 Two planes 12

6.1 Angle between two planes . . . . . . . . . . . . . . . . . . . 126.1.1 Special Case . . . . . . . . . . . . . . . . . . . . . . . 13

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6.1.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . 136.2 Distance between planes . . . . . . . . . . . . . . . . . . . . 13

6.2.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . 136.3 Plane bisector of two planes . . . . . . . . . . . . . . . . . . 14

6.3.1 Direction of normal to a plane . . . . . . . . . . . . . 146.3.2 Origincontaining region between two intersecting planes 146.3.3 Bisector containing the origin . . . . . . . . . . . . . 156.3.4 Bisector of acute/obtuse angle . . . . . . . . . . . . . 15

7 Projection of an Area 16

8 Linear combination of planes 178.1 Non-parallel planes . . . . . . . . . . . . . . . . . . . . . . . 178.2 Parallel planes . . . . . . . . . . . . . . . . . . . . . . . . . . 17

8.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 Plane sects a line joining two points 17

9.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

10 Three planes 1710.1 Normals are coplanar . . . . . . . . . . . . . . . . . . . . . . 1710.2 Normals are non-coplanar . . . . . . . . . . . . . . . . . . . 18

11 Equation of a line 1811.1 Two point form of a line . . . . . . . . . . . . . . . . . . . . 1811.2 Slope point form . . . . . . . . . . . . . . . . . . . . . . . . 19

11.2.1 Special case . . . . . . . . . . . . . . . . . . . . . . . 1911.2.2 Unsymmetric form . . . . . . . . . . . . . . . . . . . 19

11.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2011.4 Point and a line . . . . . . . . . . . . . . . . . . . . . . . . . 21

11.4.1 Foot of perpendicular . . . . . . . . . . . . . . . . . . 2111.4.2 Perpendicular distance . . . . . . . . . . . . . . . . . 2111.4.3 Image of a point in a line . . . . . . . . . . . . . . . . 2111.4.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . 22

11.5 Two lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2211.5.1 Skew lines . . . . . . . . . . . . . . . . . . . . . . . . 2211.5.2 Coplanar . . . . . . . . . . . . . . . . . . . . . . . . 2211.5.3 Angle between two lines . . . . . . . . . . . . . . . . 23

11.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

12 Line & a Plane 2312.1 Angle between line & a plane . . . . . . . . . . . . . . . . . 2312.2 Projection of a line onto a plane . . . . . . . . . . . . . . . . 2312.3 Image of a line in a plane . . . . . . . . . . . . . . . . . . . . 24

12.3.1 Line is parallel to plane . . . . . . . . . . . . . . . . . 2412.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

13 Techniques in specific Problems 25

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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 posi-tive y-axis. Hence the relation between the angles made by this vector withx & y axes is + = 2 .

If we try to think of something similar relation in 3D geometry. But theproblem 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 betweenthe angles make by a line in 3D space.

2 Points in 3D Geometry2.1 Two points

Given two points A (x 1 , y1 , z1 ) and B (x 2 , y2 , z 2 ) in 3D space.

2.1.1 Distance between A & B

Distance l(AB ) = p (x 1 x 2 )2 + ( y1 y2 )2 + ( z1 z2 )22.1.2 Section formula

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

C =mx 2 nx 1

m n ,

my 2 ny 1m n

, mz 2 nz 1

m n

where corresponds to internal or external section

Problem 2.1. To find the ratio in which the point C (a,b,c ) divides the line joining A (x 1 , y1 , z1 ) & B (x 2 , y2 , z2 )

AC CB

= x1 aa x 2= y1 bb y2

= z1 cc z2

2.1.3 Midpoint of segment AB

midpoint =x 1 + x2

2 ,

y1 + y22

, z1 + z2

2

3 Points are collinearThree points A (x 1 , y1 , z 1 ), B (x 2 , y2 , z 2 ) & C (x 3 , y3 , z3 ) are collinear then

x 1

x 2x 2 x 3 =

y1

y2y2 y3 =

z1

z2z2 z3

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2.1.4 Triangle

1. Area of triangle

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

2. Formula for centroid(G), Incentre(I), Excentres ( I 1 , I 2 , I 3 ), Orthocenter (H )and Circumcenter(O)

use Pm i x iPm i, Pm i yiPm i

, Pm i ziPm ia) Centroid : (m 1 , m 2 , m 3 ) G (1, 1, 1)b) Incenter : (m 1 , m 2 , m 3 ) I (sin A, sin B, sin C )

c) Excenter : (m 1 , m 2 , m 3 ) I 1 (sin A, sin B, sin C ) similarlyfor other excenterd) Orthocenter : (m 1 , m 2 , m 3 ) H (tan A, tan B, tan C )e) Circumcenter : (m 1 , m 2 , m 3 ) O (sin 2A, sin 2B, sin 2C )

3 Direction Cosines & Direction Ratios

3.1 Direction Cosine

Let us start with a point P (x 0 , y0 , z0 ). So the position vector of this point willbe r = x 0 i + y0 j + z0 k . Let us convert this point into polar form. Let r makesan angle , & with x, y & z axis respectively and | r | = r .

Therefore r = x 0 i + y0 j + z0 k = rcos i + rcos j + rcos kNow the unit vector along the line joining origin and point P isr = x 0

q x 20 + y 20 + z 20i + y0

q x 20 + y 20 + z 20 j + z0

q x 20 + y 20 + z 20k = 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 vectoralong line OP. So this unit vector helps in knowing the direction of this line.And any vector has one unique unit vector along its direction. This can beeasily proved.

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

u 1 = u 2 | u 1 | = | u 2 | Taking modulus on both sides.

| | = 1

= 1. So along a line there is one unique unitvector 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 ) l i + m j + n k . In 3D geometry a vector is written as a coordi-nate. So any vector a i + b j + ck in 3D is (a,b,c )

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3.1.1 Identities

1. cos 2 + cos2 + cos2 = 1

2. sin2 + sin2 + sin2 = 2

3. Pcos 2 = 14. Direction cosine of coordinate axes are (1, 0, 0), (0, 1, 0) & (0, 0, 1)3.2 Direction Ratio

Direction Ratio is defined 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 findthe unit vector along this direction. So just divide the direction ratio with thelength of direction ratio.

DC = a

a 2 + b2 + c2 , b

a 2 + b2 + c2 , c

a 2 + b2 + c2 ,

3.3 Summary of Direction cosine and Direction Ratio1. Direction cosine of a line in 3D geometry is Unit vector along that line

in Vectors

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

3.4 Projection of a line segment onto another line

Given a line segment joining points A (x 1 , y1 , z1 ) & B (x 2 , y2 , z2 ) and a linewith direction cosines (l ,m,n ) then the projection of the segment onto the

line is (x 2 x 1 )l + ( y2 y1 )m + ( z2 z1 )n

3.4.1 Projection of AB along coordinate axes

AB along

1. x axis : x2 x 12. y axis : y2 y13. z axis : z2 z1

Example 3.1. What is the direction cosine of

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x-axis

1. To find the direction cosine along x-axis we first find the unit vec-tor along x-axis. i = 1 i + 0 j + 0 k is the unit vector along x-axishence (1, 0, 0) is the direction cosine along x-axis.

2. Another way is to find the angle x-axis makes with positive x,y& z axis. So = 0 , = /2 & = /2 . Hence the directioncosine 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 finding any vector along the direction of this bisector

and then finding unit vector from that vector which then can beclaimed as the direction cosine. Any vector along the bisector =i + k . And unit vector along this direction is

i 2 + 0 j +

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 (x 1 , y1 , z1 ) & Q(x 2 , y2 , z2 ) is avector along this line. So vector P Q = q p = ( x 2 x 1 , y2 y1 , z2 z1 ).This is any vector that is direction ratio.

3.6 Problems

1. Can a directed line have direction ratios 45, 60 and 120 (in degrees)

2. A lines makes an angle of /4 with each of x axis y-axis and z-axisthen find the angle made by it with x axis

3. If the direction cosines of a variable line in two adjacent points bel ,m,n and l+ l, m + m, n + n , show that the small angle betweenthe two positions is given by 2 = l2 + m 2 + n 2

4. If a line OP through the origin O makes angles 30, 45, 60 with x, y, zaxes respectively. then the direction cosines of OP are ?

5. If A (1, 2, 3) , B (2, 4, 1) , C (

1, 2, 4)

andD (1, 0, 3)

arefour points. Findthe projection of CD on AB.

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4 Plane

4.1 Equation of a plane in different 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 the plane from origin d then equation of the plane

ax + by + cz + d = 0

4.1.3 Point Normal form

Given a point (x 0 , 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 (a 1 , a 2 , a 3 ) and(b1 , b2 , b3 ) and passing through the point ( x 0 , y0 , z0 )

x x 1 y y1 z z1a 1 a2 a3b1 b2 b3

= 0

4.1.5 A plane containing two points and parallel to a line

Let (x 1 , y1 , z1 ) & (x 2 , 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 the plane

x x 1 y y1 z z1x 1 x 2 y1 y2 z1 z2a b c= 0

4.1.6 Three point form

Given three points (x 1 , y1 , z1 ), (x 2 , y2 , z 2 ) & (x 3 , y3 , z3 ) lie on a plane whoseequation is

x x 1 y y1 z z1x 1

x 2 y1

y2 z1

z2x 2 x 3 y2 y3 z2 z3

= 0

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4.1.7 Intercept form

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

xa

+ yb

+ zc

= 1

4.1.8 Special cases

1. x = 0 : is the y z plane2. y = 0 : is the x y plane3. z = 0 : is the x y plane4. plane parallel to the x axis is ay + bz = 15. plane parallel to the y axis is ax + bz = 16. plane parallel to the z axis is ax + by = 1

4.2 Problems

1. A variable plane is at a constant distance p from the origin and meetsthe coordinated axes in A,B,C. Show that the locus of the centroid of the tetrahedron OABC is x2 + y2 + z2 = 16 p2

2. Find the equation of the plane passing through the points (2, 2, 1) and(9, 3, 6) and perpendicular to the plane 2x + 6 y + 6 z = 9

3. A variable plane at a constant distance p from origin meets the axesat A, B, C. Through A,B,C planes are drawn parallel to the coordinateplanes. Show that the locus of their point of intersection is given byx2 + y2 + z2 = p2

4. Find the equation of the plane which is perpendicular to the plane5x + 3 y + 6 z + 8 = 0 and which contains the line of intersection of theplanes x + 2 y + 3 z 4 = 0 and 2x + y z + 5 = 0

5. Find the equation of the plane passing through the points (2, 1, 0) , (5, 0, 1)and (4, 1, 1)

6. The equation of the plane passing through the origin and containing thelines whose direction cosines are proportional to 1, 2, 2 and 2, 3, 1

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7. The plane passing through the point (2, 2, 2) and containing theline joining the points (1 , 1, 1) and (1, 1, 2) makes intercepts on thecoordinate axes, the sum of whose lengths is ?

5 Point & a Plane

5.1 Position of a point with respect to a plane

5.1.1 same side of a plane

If two points P (x 1 , y1 , z 1 ) & Q (x 2 , y2 , z2 ) are on the same side of the planeax + by + cz + d = 0 then

(ax 1 + by1 + cz1 + d)(ax 2 + by2 + cz2 + d) > 0

5.1.2 opposite side of a plane

If two points P (x 1 , y1 , z 1 ) & Q (x 2 , y2 , z2 ) are on the other side of the planeax + by + cz + d = 0 then

(ax 1 + by1 + cz1 + d)(ax 2 + by2 + cz2 + d) < 0

5.1.3 Origin side of the plane

To find if a point P (x 1 , y1 , z1 ) is on the origin side of a plane then first stepis to make d positive in the plane equation and substitute (x 1 , y1 , z1 ) in theline equation to see its sign. From above deduction we get

ax 1 + by1 + cz1 + d > 0 (d>0)

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

ax 1 + by1 + cz1 + d < 0 (d>0)

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

5.2 Problems

1. Check if the following points are on the origin side or non-origin sideof the plane x +

y2

+ z3

= 1

P(1, 2, 3), Q (1, 2, 3)

5.3 Point outside a plane

There are three types of interesting problems which can arise if a point is out-

side a plane. Foot of the perpendicular of the point onto the plane, distanceof the point from the plane & Image of the point in the plane.

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5.3.1 Distance of a point from the plane

Distance of a point (x 1 , y1 , z 1 ) from a plane ax + by + cz + d = 0 is

|ax 1 + by1 + cz1 + d| a 2 + b2 + c2

5.3.2 Perpendicular foot of a point onto the plane

Foot of the perpendicular of a point (x 1 , y1 , z1 ) onto the plane ax + by + cz +d = 0 is given by

x x 1a

= y y1

b =

z z1c

= ax 1 + by1 + cz1 + d

a 2 + b2 + c2

5.3.3 Image of a point in a plane

Image of a point (x 1 , y1 , z1 ) in the plane ax + by + cz + d = 0 is given by

x x 1a

= y y1

b =

z z1c

= 2ax 1 + by1 + cz1 + d

a 2 + b2 + c2

5.4 Problems

1. Find the distance of the point (2, 1, 0) from the plane 2x + y+2 z +5 = 0

(solution : 10/3 )2. Find the image of the point (3, 2, 1) in the plane 3x y + 4z = 2(solution : (0, 1, 3) )3. Find the equation of the plane passing through the points (2, 1, 0) , (5, 0, 1)

and (4, 1, 1)

4. If P is a point (2, 1, 6) then find the point Q such that PQ is perpendic-ular to the plane in problem 3 above, and the midpoint of PQ lies onit. (solution : (6, 5, 2) )

5. Find the incentre of the tetrahedron formed by the planes x = 0 , y =0, z = 0 and x + y + z = a

6 Two planes

6.1 Angle between two planes

Angle between two planes a1 x + b1 y + c1 + d1 = 0 and a2 x + b2 y + c2 + d2 = 0is same as angle between their normals

cos = a1 a 2 + b1 b2 + c1 c2

p a21 + b21 + c21p a

22 + b22 + c22

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6.1.1 Special Case

1. If they are parallel

a 1b1

= a2b2

= a3b3

i.e.(a 1 , b1 , c1 ) = (a 2 , b2 , c2 )

As vectors they are collinear vectors

2. If they are perpendicular

a 1 a 2 + b1 b2 + c1 c2 = 0

6.1.2 Problems

1. Find the angle between the lines whose direction cosines are propor-tional to 0,1,1 and 1, 0, 1 respectively.

2. Find the obtuse angle between the lines with direction ratios propor-tional to 3, 6, 2 and 1, 2, 2

3. Show that the join of the points (1, 2, 3) and (2, 3, 5) is parallel to the join of the points (1, 2, 3) and (1, 4, 1)

6.2 Distance between planesTwo 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(as they are parallel )

|d1 d2 | a 2 + b2 + c22. If they are non-parallel

The distance between the planes that intersect is zero.

6.2.1 Problems

1. Find the equations of the planes parallel to the plane x 2y +2 z3 = 0which is at a unit distance from the point (1, 2, 3) (solution x2y+2 z =0 and x 2y + 2 z = 6 )2. Find the distance between the parallel lines x + y z + 4 = 0 andx + y z + 5 = 0 (Solution : 1/ 3 )3. Find the equation of the plane mid parallel between the planes 2x 2y + z + 3 = 0 and 2x

2y + z + 9 = 0

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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 d negative thenwhat is 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 can

be assumed to be negative) can be converted into the vector equation as

(x,y,z ) a

a 2 + b2 + c2 , b

a 2 + b2 + c2 , c

a 2 + b2 + c2 = d

a 2 + b2 + c2

where n = a

a 2 + b2 + c2 , b

a 2 + b2 + c2 , c

a 2 + b2 + c2 and p = d

a 2 + b2 + c2So the direction of

a a 2 + b2 + c2 ,

b a 2 + b2 + c2 ,

c a 2 + b2 + c2 is away

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

So the final conclusion : For the a plane equation, ax + by + cz + d = 0where d is negative the direction of normal (a,b,c ) is away from origintowards the plane.

6.3.2 Origin containing region between two intersecting planes

Given two plane equations P 1 : a1 x + b1 y + c1 z + d1 = 0 and P 2 :a 2 x +b2 y + c2 z + d2 = 0 such that d1 & d2 > 0 then the green region containingthe origin is positive for both the planes P 1 and P 2 and the other verticallyopposite green region is negative for both P 1 and P 2 . So for the green region

P 1 P 2 > 0 (where d1 , d 2 > 0)

Now lets concentrate on the red region. Both the red region are such thateither of P 1 or P 2 is negative2. Hence

P 1 P 2 < 0 (where d1 , d 2 < 0)

2P 1 , P 2 is negative means P 1 planes expression value at (0 , 0, 0) is negative i.e. a 1 (0) +b1 (0) + c1 (0) + d1 < 0 and vice-versa for being positive

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So from the above discussion we know which region will have the originlying in it. Now next we worry to find a method to find the acute anglebisector from problem solving perspective and which angle (acute or obtuse)will contain the origin.

6.3.3 Bisector containing the origin

Here we are interested in knowing which angle bisector contains the origin.Any point on the bisector satisfies

|a 1 x + b1 y + c1 z + d1 |

p a21 + b21 + c21

= |a 2 x + b2 y + c2 z + d2 |

p a22 + b22 + c22

a 1 x + b1 y + c1 z + d1

p a 21 + b21 + c21

= a 2 x + b2 y + c2 z + d2

p a 22 + b22 + c22

The angle containing the origin (above diagram green region) has botha 1 x + b1 y + c1 z + d1 & a 2 x + b2 y + c2 z + d2 positive(in origin region) or bothnegative(vertically opposite to origin). Correspondingly, a1 x + b1 y + c1 z + d1or a 2 x + b2 y + c2 z + d2 one of these is negative in the red region.

So,

a 1 x + b1 y + c1 z + d1

p a21 + b21 + c21

= a2 x + b2 y + c2 z + d2

p a22 + b22 + c22

, (origin containing angle bisector)

a 1 x + b1 y + c1 z + d1

p a21 + b21 + c21

= a 2 x + b2 y + c2 z + d2p a22 + b22 + c22

, (non-origin containing angle bisector)

6.3.4 Bisector of acute/obtuse angle

To find the equation, of the acute angle bisector between the planes a 1 x +b1 y + c1 z + d1 = 0 and a 2 x + b2 y + c2 z + d2 = 0 .

For this problem we first locate the region of origin.Step I : Make d1 > 0 and d2 > 0Step II : (a 1 , b1 , c1 ) and (a 2 , b2 , c2 ) are normals to planes starting from origin

to these respective planes

Step III : Angle between the normals : (a 1 , b1 , c1 ) (a 2 , b2 , c2 ) = a1 a 2 +b1 b2 + c1 c2

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If a 1 a 2 + b1 b2 + c1 c2 > 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 a1 a 2 + b1 b2 + c1 c2 < 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.

a 1 a 2 + b1 b2 + c1 c2 > 0; d1 , d 2 > 0 a1 a 2 + b1 b2 + c1 c2 < 0; d1 , d 2 > 0Origin liesin

Obtuse Angle Acute Angle

OrigincontainingAnglebisector

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

Theequation of angle bisector a 1 x + b1 y + c1 z + d1

p a21 + b21 + c21

= a 2 x + b2 y + c2 z + d2

p a22 + b22 + c22

7 Projection of an Area

Two inclined planes, with an area A lying on one of the planes then we findthe projection of A onto the other plane.

If projection of areas of a plane A in 3D space is Axy , A yz , A zx onto thex-y, y-z & z-x planes then A 2 = A 2xy + A2yz + A

2zx

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8 Linear combination of planes

8.1 Non-parallel planes

Given two planes P 1 : a 1 x + b1 y + c1 z + d1 = 0 & P 2 : a 2 x + b2 y + c2 z + d2 = 0then the equation of planes passing through intersection of these planes isgiven by

P 1 + P 2 = 0

8.2 Parallel planes

Given two planes P 1 : ax + by + cz + d1 = 0 & P 2 : ax + by + cz + d2 = 0then linear combination represents family of parallel planes to these planes.

8.3 Problems1. Find the equation of the plane containing the line of intersection of the

plane x + y + z 6 = 0 and 2x + 3 y + 4 z + 5 = 0 and passing throughthe point (1, 1, 1) (Solution : 20x + 23 y + 26 z 69 = 0 )2. Find the equation of the plane that is perpendicular to the plane 5x +

3y+6 z +8 = 0 and which contains the line of intersection of the planesx + 2y + 3 z 4 = 0 and 2x + y z + 5 = 0 (solution : 51x + 15 y 50z + 173 = 0 )

3. The plane x 2y + 3z = 0 is rotated a right angle about the line of intersection with the plane 2x + 3 y 4z 5 = 0 , find the equation of the plane in its new position. (solution : 22x + 5 y 4z 35 = 0)

9 Plane sects a line joining two points

Ratio in which a plane ax + by+ cz+ d = 0 divides the line joining A (x 1 , y1 , z 1 )& B(x 2 , y2 , z2 ) is given by

mn

= ax 1 + by1 + cz1 + dax 2 + by2 + cy2 + d

9.1 Problems

1. Find the ratio in which the join A (2, 1, 5) and B (3, 4, 3) is divided bythe plane 2x + 2 y 2z = 1 . Also, find the coordinates of the point of division.

10 Three planes

10.1 Normals are coplanar

If the normals to three planes are coplanar then there are twopossibilities.

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

To locate this case we need box product of the normals [N 1 N 2 N 3 ] = 0and show linear combination of any two is producing the third formunique values 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 andshow the any point on the intersection of any two planes is not satisfy-ing the third plane.

10.2 Normals are non-coplanar

If the normals to three planes are not coplanar then the planesintersect in a unique point. see the diagram above

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 wecan proceed to find the 3D equation of a line.

11.1 Two point form of a line

We have a line that passes through two points (x 1 , y1 , z1 ) & (x 2 , y2 , z 2 ) thenthe we deduce the equation of this line as follows.

Let (x,y,z ) be a point on this line (locus point)Therefore (x x 1 , y y1 , z z1 ) = (x 1 x 2 , y1 y2 , z1 z2 )

x x 1x 1 x 2

= y y1y1 y2

= z z1z1 z2

=

This is the equation of the line in two point form

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11.2 Slope point form

We have a line that passes through a point (x 1 , y1 , z 1 ) and parallel to the line

with direction ratios (a,b,c ) then equation of this line is

x x 1

a =

y y1b

= z z1

c = constant

11.2.1 Special case

Equation of line passing through the point (x 1 , y1 , z1 ) and parallel to the linewhose direction cosines are (l ,m,n ) then equation of this line is

x x 1

l =

y y1m

= z z1

n = r

where r is the distance of between the points (x,y,z ) & (x 1 , y1 , z 1 )

11.2.2 Unsymmetric form

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

a 1 x + b1 y + c1 z + d1 = 0 (11.1)

a 2 x + b2 y + c2 z + d2 = 0 (11.2)

Direction of the line of intersection of planes is same as perpendicular to

normals to these planes.Normal to the two planes are N 1 : (a 1 , b1 , c1 ) &N 2 : (a 2 , 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 Directioncosines of the line is

Direction Ratio of Line =i j k

a 1 b1 c1a 2 b2 c2

Example 11.1. Given asymmetric form of equation of a line x + y = 1 , z = 1 .What is symmetric form of the equation of this line? And parametric form of the equation of this line.

Normal to the plane x + y = 1 : N 1 = (1 , 1, 0) & Normal to the secondplane z = 1 : N 2 = (0 , 0, 1).

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Direction cosine of the line of intersection of these two lines is N 1 N 2 =(1, 1, 0)Any point on this line of intersection is (1, 0, 1) (this we get by trial in theabove plane equations)So the symmetric equation of the line of intersection is

x 11

= y 1

0 =

z 01

=

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

11.3 Problems

1. Find the equation of a line passing through the point P (1, 2, 3) and

having direction cosines 23

, 23

, 13

. Also find the coordinates of a

point on the line at a distance of 6 units from P (solution : equation of

the line is : x 1

2 =

y 22

= z 3

1 and required point (5, 2, 5))

2. Find the distance of the point of intersection of the line x 3

1 =

y 42

=z 5

2 and the plane x + y + z = 17 from the point A (3, 4, 5). (solution

: 3 units)

3. Find the direction cosines of the line whose equations are6x 2 = 3y + 1 = 2 z 2

(solution : 1 14 ,

2 14 ,

3 14 )

4. Prove that the lines x = ay + b, z = cy + d and x = a0y + b0,z = c0y + d0are perpendicular if aa 0 + cc0 + 1 = 0

5. Find the equation of the line through the point (1, 3, 2) and perpen-dicular to the lines x1

= y

2 =

z

2and

x + 2

3 =

y 12

= z + 1

5

(solution : x + 1

2 =

y 37

= z + 2

4 )

6. Find the distance of the point P (1, 2, 3) from the plane x y + z = 5measured parallel to the line

x2

= y3

= z

6 (solution : 1)

7. Reduce in symmetrical form the equations of the lines x y + 2 z = 5and 3x + y + z = 6 (solution : 4x 1133

= 4y + 95 = z1 )

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11.4 Point and a line

Given a point P (x 1 , y1 , z 1 ) outside the line x

a =

y b

= z

c

11.4.1 Foot of perpendicular

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

Let x

a =

y b

= z

c =

For a specific the foot of the perpendicular is ( a + , b + , c + )Direction ratio of the perpendicular is ( a + x 1 , b+ y1 , c + z1 )So

(a,b,c ) ( a +

x 1 , b +

y1 , c +

z1 ) = 0

= (a,b,c ) (x 1 , y1 , z1 )

a 2 + b2 + c2

Foot of perpendicular from (x 1 , y1 , z 1 ) onto the line x

a =

y b

= z

cis given by

x a

= y

b =

z c

=(a,b,c ) (x 1 , y1 , z1 )

Pa 2

= (ax 1 + by1 + cz1 ) (a + b + c )a 2 + b2 + c2

11.4.2 Perpendicular distance

You find the foot of the perpendicular from the above formula and then usethe distance formula to find the distance between the foot and the point.

11.4.3 Image of a point in a line

Similarly from the above formula we get if (x,y,z ) is the image of a point(x 1 , y1 , z1 ) then image is given by

x a

= y

b =

z c

= 2(a,b,c ) (x 1 , y1 , z 1 )

Pa2

= 2(ax 1 + by1 + cz1 ) (a + b + c )

a 2 + b2 + c2

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11.4.4 Problems

1. Find the foot of the perpendicular from the point (0, 2, 3) on the linex + 3

5 = y

1

2 = z + 4

3 . Also find the length of the perpendicular.(solution : (2, 3, 1) and 21 )

2. Find the image of the point (1, 6, 3) in the line x1

= y 1

2 =

z 23

(solution : (1, 0, 7) )

3. Find the distance of the point P (3, 8, 2) from the line x 1

2 =

y 34

=z 2

3 measured parallel to the plane 3x + 2 y 2z + 15 = 0 (solution :

7 )

4. Find the foot and the length of the perpendicular from P (5, 7, 3) to the

line x 15

3 =

y 298

= z 5

5 . Find also the equation of the plane

in which the perpendicular and the given straight line lie. (solution :(9, 13, 15) , 14, 9x 4y z = 14 )

11.5 Two lines

11.5.1 Skew lines

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

Such lines are called skew lines.Two lines x x 1a 1

= y y1b1

= z z1c1

and x x 2a 2

= y y2b2

= z z2c2

Shortest distance between them is

[(x 1 x 2 , y1 y2 , z 1 z2 ) (a 1 , b1 , c1 ) (a 2 , b2 , c2 )]|(a 1 , b1 , c1 ) (a 2 , b2 , c2 )|

Vector form of shortest distanceConverting given line equations into vector form : r = ( x 1 i + y1 j + z1 k) +

(a 1 i + b1 j + c1 k) & r = ( x 2 i + y2 j + z2 k) + (a 2 i + b2 j + c2 k)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|

=

x 1 x 2 y1 y2 z1 z2a 1 b1 c1a 2 b2 c2

p (b1 c2 c1 b2 )2 + ( c1 a 2 a 1 c2 )2 + ( b1 c2 c1 b2 )211.5.2 Coplanar

If [(x 1 x 2 , y1 y2 , z 1 z2 ) (a 1 , b1 , c1 ) (a 2 , b2 , c2 )] = 0 then the above men-tioned lines are coplanar

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11.5.3 Angle between two lines

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

between themLet (l1 , m 1 , n 1 ) & (l2 , m 2 , n 2 ) are the direction cosines of two lines then

cos = l1 l2 + m 1 m 2 + n 1 n 2

Let (a 1 , b1 , c1 ) & (a 2 , b2 , c2 ) are the direction ratios of the lines then

cos = a1 a 2 + b1 b2 + c1 c2

p a21 + b21 + c21p a

22 + b22 + c22

11.6 Problems

12 Line & a Plane

12.1 Angle between line & a plane

We make use of the normals direction ratio that is available for us fromequation of the plane. Angle made by the line with the plane is

Angle = 2 cos

1 (a,b,c ) ( p, q, r ) a 2 + b2 + c2p p2 + q 2 + r 2 !

where equation of plane is ax + by + cz + d = 0 and x 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.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 find its foot of perpen-

dicular onto the plane say P 0 (given a point outside a plane find its foot of perpendicular)

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

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12.3 Image of a line in a plane

There are two possibilities with a line and a plane. If the direction cosineof the line is perpendicular to the plane then the line is either parallel to theplane or is contained in the plane.If the line is not parallel to the plane then it must be intersecting.

12.3.1 Line is parallel to plane

Image of a line when parallel to the planeStep 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 find its

image point

Step III : Using these two points write the equation of the line which is theimage of the given line in the given plane.

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12.4 Problems

1. Find the angle between the line x 1

1 =

y + 21

= z 4

0 and the plane

y + z + 2 = 0 (solution : = 6 )

2. find the equation of the line passing through the point (3, 0, 1) and

parallel to the planes x + 2 y = 4 and 3y z = 5 (solution : x 3

2 =

y1

= z 1

3 )

3. Find the equation of the plane passing through the line of intersectionof the planes 2x + y z = 3 and 5x 3y + 4 z + 9 = 0 and parallel tothe line

x 12

= y 3

4 =

z 55

(solution : 7x + 9 y 10z = 27 )

4. Find the projection of the line x 1

3 =

y 24

= z 3

5 on the plane

x y + z = 2 (solution : x + 2

2 13

= y + 2

2 + 10

3

= z + 2

2 + 53

)

5. Find the projection of the line 3x y +2 z 1 = 0 and x +2 y z = 2 onthe plane 3x +2 y + z = 0 (solution : 3x +2 y + z = 0 = 3 x 8y +7 z +4)

13 Techniques in specific Problems

Problem 13.1. Given a line x11 = y22 = z

33 and a point (1, 1, 1) find the

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

Problem 13.2. Find the equation of a plane passing through intersection of planes x + 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 intersec-tion of the given planes is

(x + y + z 1) + (2x + y z 2) = 0Now this plane passes through origin hence (0, 0, 0) hence satisfies this im-plies

1 + (2) = 0

=

12

Substituting we get the equation of the plane.

3 d geometry lecture notes

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