Vectors and 3D

NeoClassical

Vector

▪ Head – Tail

Head

Tail

NeoClassical

Dot Product

▪ Dot product tells how dependent vectors are i.e angle between them

NeoClassical

Uses of Dot Product

▪ Finding Angle between vectors

▪ Checking whether vectors ae perpendicular of not . For perpendicular vectors dot product is zero

▪ Projections

NeoClassical

▪ Dot product can be used to find magnitude of a vector due to the following property

a . a = |a|2

Example to find |A+B| we calculate

|A+B|2 = (A+B) .(A+B) = |A|2 + |B|2 + 2 A . B

Similarly for |A+B+C| we have to its dot product with itself

|A+B+C|2 = (A+B+C) .(A+B+C)

NeoClassical

CROSS PRODUCT

▪ AxB : Result is a vector ,whose magnitude is |A| |B| sin θ and direction is given by right hand thumb rule

“(AxB) vector is perpendicular to both A and B It is perpendicular to plane containing A and B”

Whenever it is given to find out a vector perpendicularto two vectors , cross product will be used

Example : Find a plane perpendicular to two given planes , then the normal vector of third plane will be the cross product of normal vectors of given planes

NeoClassical

▪ Order is Important (AXB) ≠ (BXA)

▪ (AXB) = -(BXA)

▪ If vectors are give in rectangular coordinate system i.e in terms of i,j,k

NeoClassical

Cross Product for calculating area

▪ Area of parallelogram :

Area of Triangle = ½ (A X B )

For calculation area of any other figure like quadrilateral , Pentagon just break them into triangles

NeoClassical

▪ For parallel vectors , Cross product is zero

If two vectors A and B are parallel : A = λ B

Example :

NeoClassical

Scalar Triple Product

▪ STP or box product = A.(BXC) denoted as [ A B C ]

▪ First cross product then dot product

▪ Geometric interpretation : Volume of Parallelepiped = [A B C ]

▪ Tetrahedron = (1/6) [A B C]

▪ A ,B ,C are adjacent sides

NeoClassical

Properties :

▪ Position of Dot and cross can be interchanged

A.(BXC) = (AXB).C

Can be cyclically permuted [a b c] =[b c a] = [c a b]

If order is not maintained it becomes negative [ a b c] = - [ a c b]

Important : If Box product of three vectors A,B,C is zero , then they are coplanar . Reverse is also true

Coplanartiy ::: Think of Box product

NeoClassical

Example

▪ Prove [ (a+b) (b+c) (c+a) ] = 2 [a b c ]

NeoClassical

Plane

▪ To define plane we need

1 point and normal vector

3 points

Normal vector is perpendicular to plane . There can be two normal vectors for a plane , one up and one down

NeoClassical

Equation of plane

▪ Point on plane and normal vector is known

( r – a) . n = 0 r.n = a.n

In rectangular coordinate system Given point (x1,y1,z1)

a(x-x1) + b(y-y1) + c(z-z1) = 0

Coefficient of x , y and z give normal vector

Example : Plane = 3x+ 4y + 7z = 8

Normal vector 3 i + 4 j + 7 k

NeoClassical

If two vectors in a plane are known

▪ As we know that normal vector is a vector perpendicular to plane .

So , if two vectors in plane are known , normal vector will be a vector perpendicular to both these vectors

In the given figure , two vectors(Band C) in the plane (or parallel

to the plane) are known , So normal vector will be the cross

Product of Band C

n = (BXC)

NeoClassical

If three points are known

▪ Three points Plane is fixed

Write any two vectors (AB , AC or BC )

These are in plane

Normal vector will be perpendicular to all of

these

n = (AB X BC) or (ACXBC) or (AB X AC)

Normal vector is known and point on plane is also known .

So we can equation of plane easily

NeoClassical

Perpendicular Distance of a point from plane

▪ General Equation of plane ax +by + cz =d

▪ Whenever we have to assume a plane , we will use general equation of plane

NeoClassical

Foot of perpendicular

Equation of plane ax+by+cz=d

Normal Vector : a i + b j +c k

Point Q (X1 , Y1 , Z1 )

Foot of perpendicular : X ( h,k,l )

Now QX is parallel to normal vector QX = λ n

(h-X1) i + ( k-Y1 ) j + (l-Z1 ) = λ(a i + b j +c k)

h = X1 + λ a , k = Y1 + λ b , l = Z1 + λ c

Put h,k,l in equation of plane and find λ

NeoClassical

Mirror Image in plane

▪ Foot of perpendicular is the midpoint

Of image and point

So first find foot of perpendicular

Parallel Planes distance

AB sin (theta )

= (AB X n ) / |n|

NeoClassical

Line

▪ For defining line we need

▪ Two points

▪ One point and slope

Direction vector is parallel to line

b direction vector

NeoClassical

Equation of line

▪ One point and direction vector

▪ Direction vector = a i + b j + c k

▪ (Xo , Y0 , Z0 ) Point on line

NeoClassical

If two points are given

▪ If two points are given , direction vector can be found out easily

A and B are given points , So AB is direction vector

Unit direction vector give direction cosine

Direction vector : a i + b j + c k

Unit direction vector)/ )

AB

NeoClassical

▪ General point on line = (Xo +λa, Y0 + λb, Z0 + λc)

NeoClassical

Line and Line

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▪ Parallel : If their Direction vectors are parallel

Condition for Parrallel lines : = =

▪ A is point on Line 1

▪ B is point on line 2

▪ Distance = |AB sin θ |

|AB sin θ | is the magnitude of cross

Product of AB and unit direction vector

ABsin(

NeoClassical

Distance between skew lines

▪ Skew lines are neither parallel nor intersecting

▪ Skew Lines have perpendicular direction vectors

▪ They form parallelepiped

▪ A is point on first line , B point on 2nd line

▪ Volume of parallelepiped = | [ u AB v ] |

▪ Volume = area X distance

▪ Area = |u X v|

NeoClassical

Point of Intersection

▪ First check whether they are parallel or not

▪ Assume general points on line and equate them

▪ Example :General points on Lines

Line 1: x= 4+t, y= 19+6t, z= 12+5t Line 2: x= -3+2p, y= -15+8p, z= -19+8p

4+t= -3+2p 19+6t = -15+8p 12+5t = -19+8p

Solve any two to find t and p

But don’t forget to check the third equations .Lines may be skew

NeoClassical

▪ 4 + 2t = 2 + s -5 + 4t = -1 + 3s 1 + 3t = 2s

▪ Solving 1 and 2 we get t=-5 and s =-8

▪ Putting this is third … LHS = -14 RHS = -16 (not satisfied)

▪ Lines are not parallel they are skew

1) : x = 4 + 2t, y = -5 + 4t, z = 1 + 3t 2) : x = 2 + s, y = -1 + 3s, z = 2s

NeoClassical

Distance of point from distance

▪ Distance = AP sin (θ)

▪ d==

▪ Foot of perpendicular

▪ Foot of perpendicular will lie on line. So assume general points on line .

▪ Put condition that PF . r = O to find lambda

F

NeoClassical

Plane and Line

3 cases :

Intersect at a point

Parallel

Line lies in the plane

NeoClassical

▪ If parallel Normal vector of plane and

Direction vector of plane will be perpendicular

▪ If line lies in plane then direction vector will be perpendicular to normal vector plus point on line will lie on plane as well

▪ Example : Plane : 3x+y-z=4 Line : Direction : 1 i -2j + k Point (1,2,1)

Normal vector = 3i + j –k (1 i -2j + k). ( 3i + j –k ) = O

Also (1,2,1 ) satisfies 3x+y-z =4 Line lies in plane

n

NeoClassical

Point of intersection

▪ Assume General point on line

▪ Point of intersection will be the point common to both line and plane

, so it will satisfy the equation of plane also .

Put general point in plane’s equation and find parameter , λ

NeoClassical

Family of Planes

▪ Any plane passing through intersection of two planes P1 and P2 can be written as P1 + λ P2 =O

▪ Whenever it is given that a plane passes through a intersection of planes take it as P1 + λ P2 =O

▪ Angle b/w planes Angle b/w their normal vectors

▪ Angle b/w lines Angle b/w Their direction vectors

▪ Angle b/w line and plane 90 – (Angle b/w Normal and direction vector)

NeoClassical

Distance of a point from plane measured along line

Check this :

http://www.meritnation.com/ask-answer/question/find-the-distance-of-the-point-1-2-3-from-the-plane-x-y-z/three-dimensional-geometry/1811158