Analytical Solid Geometry

1 / 44

# Analytical Solid Geometry - PowerPoint PPT Presentation

Analytical Solid Geometry. Concept of Octant. Z. P(x 1 , y 1 , z 1 ). . M. X. . . Y. O. Z. P(x 1 , y 1 , z 1 ). . Y. M. X. ELEMENTARY RESULTS. r. O. COS  ?. Let direction cosines of OP be l, m, n. Then OM = y 1. COS  = m = OM/OP  cos  = y 1 / r

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Analytical Solid Geometry' - matana

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Analytical Solid Geometry

Z

P(x1, y1, z1)

M

X

Y

O

Z

P(x1, y1, z1)

Y

M

X

ELEMENTARY RESULTS

r

O

COS  ?

Let direction cosines of OP be l, m, n.

Then OM = y1 .

COS  = m = OM/OP  cos  = y1 / r

y1 = r cos  = rm

Similarly x1 = rl ; z1 = rn

Hence coordinates of P (lr, mr, nr)

Note :

When r = 1 , coordinates of P becomes (l, m, n).

We have , r2 = (x1)2 +(y1)2+(z1)2

=(lr)2 + (mr)2 + (nr)2

= (l2 + m2 + n2)r2

l2 + m2 + n2 = 1

If a, b, c are the direction ratios a, b, c are the direction ratios of a straight line and

l, m, n are the corresponding direction cosines then,

l2/a2 = m2/b2 = n2/c2

= (l2 +m2 + n2) / (a2 + b2 + c2)

= 1 / (a2 + b2 + c2)

l2 = a2/(a2 + b2 + c2)  l =  a/((a2+b2+c2))

m2 = b2/(a2 + b2 + c2)  m =  b/((a2+b2+c2))

n2 = c2/(a2 + b2 + c2)  n =  c/((a2+b2+c2))

Z

Q(x2, y2, z2)

P(x1, y1, z1)

R

O

Y

X

COS  ?

Let direction cosines of the line PQ are l, m, n and length PQ = r.

QPR =  and PR = y2 – y1

We have from  PQR , cos = ( y2– y1) / r

y2 – y1 = rcos 

Hence direction ratios of the line joining P(x1, y1, z1) and Q(x2, y2, z2)

are x 2 - x1 , y2 – y1 , z2 – z1 .

L 1

l1, m1 , n1

Z

L 2

P

l2 , m2 , n2

Q

Y

O

Cosine rule ?

X

Draw OP = OQ = 1thatare parallel to L1 and L2 respectively as shown in the fig.

Then coordinates P are (l1, m1, n1) and that of Q are (l2, m2, n2).

By cosine rule applying to the  OPQ we get ,

PQ2 = OP2 + OQ2 – 2(OP.OQ) cos 

= 1+1-2cos 

= 2 - 2cos  ()

Also PQ2 = (l2 – l1)2 + (m2 - m1)2 + (n2 – n1)2

= (l12 + m12 + n12) + (l22 + m22 + n22) – 2(l1l2 + m1m2+ n1n2)

= 1 + 1 - 2(l1l2 + m1m2 + n1n2) (**)

By () and () we have

Cos  = l1l 2 + m1m2 + n1n2

ratios of the lines L1 & L2 respectively then

ii. If two lines with these direction cosines l1, m1 ,n1 and

l2, m2, n2 are perpendicular to each other then,

cos  = cos /2 = l1l2 + m1m2 + n1n2

i.e., l1l2 + m1m2+ n1n2 = 0

Exercises:

1. If the lines are parallel to each other then prove

l1 = l2 , m1 = m2 and n1 = n2

2. If the lines have the direction ratios a1, b1, c1 and a2, b2, c2

then

they are parallel when a1/a2 = b1/b2 = c1/c2 and

they are perpendicular when

a1a2 + b1b2 + c1c2 = 0

Q

R

P

M

N

L

Definition: If PQ is the line segment and  is the angle made by PQ with L then,

projection of PQ on L = MN = PR = PQ. cos  = rcos 

 In general projection = rcos 

Q(x2, y2, z2) on a line with direction cosines l, m, n.

R

M

N

L

direction cosines ?

Q(x2, y2, z2)

P(x1, y1, z1)

l, m, n

direction cosines ?

Direction ratios of the line segment PQ are

x 2 - x1 , y 2 - y1 , and z 2 - z1.

The corresponding direction cosines are

(x 2 - x1)/ r, (y 2 - y1)/r , (z2 - z1)/r

where

The angle between PQ and the line L is given by

The projection is given by

= rcos  = l (x2- x1) + m (y2 – y1 ) + n (z2- z1 )

direction ratios ?

direction ratios ?

Z

B’(0,0,a)

A’(0,a,a)

C’(a,0,a)

P (a, a, a)

C (0,0,a)

O(0,0,0)

Y

A(a,0,0)

B (a,a,0)

X

Equation of a plane (normal form) :

Z

P(x, y, z)

Q

Y

O

X

Consider a plane with OQ as normal to the plane from the origin O.

Let l, m, n be the direction cosines of the normal OQ.

Let length OQ = p.

Let P (x, y, z) be a general point on the plane.

Observe that projection OP on OQ is OQ.

i.e., p = l (x- 0) + m (y - 0) + n( z – 0)

i.e. lx + my + nz = p

Then equation of the plane is

Or

Thus a general linear equation in x, y, z namely ax + by + cz + d = 0 represents a plane with a, b, c as direction ratios of a normal to the plane. Thus for the plane ax + by + cz + d = 0 the perpendicular distance from the origin is given by

Z

(0,0,c)

c

b

Y

a

(0,b,0)

(a,0,0)

X

2. Equation of a plane in the intercept form :

Consider a plane Ax +By + Cz + d = 0 which makes

intercept a with x-axis ,

b with y-axis and c with z-axis.

As (0, b, 0) is a point on the plane, B b + d = 0 B = - d/b

As (0, 0, c) is a point on the plane, C c + d = 0 C = - d/c

Substituting we get, x/a + y/b + z/c = 1

Consider a plane ax + by + cz + d = 0. If it passes through the point P (x1, y1, z1).

Then ax1 + by1 + cz1 + d = 0 or d = -ax1 –by1 -cz1

Equation of the plane becomes

ax + by + cz +d – (ax1 + by1 + cz1 +d) = 0

i.e., a(x-x1)+ b(y-y1)+ c(z-z1) = 0

Thus equation of a plane which passes through the point

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

(x1, y1, z1), (x2 ,y2, z2) & (x3, y3, z3)

If the plane passes through P1 then its equation is given by

a(x-x1) + b(y-y1) + c(z-z1) = 0 (1)

If P2 is on the plane, then,

a(x2-x1) + b(y2-y1) + c(z2-z1) = 0 (2)

If P3 is a point on the plane then

a(x3-x1) + b(y3-y1) + c(z3-z1) = 0 (3)

are coplanar if

P(x1, y1, z1)

Q(f, g, h)

R

ax+by+cz=d

Example :

Find the perpendicular distance from the point (x1 , y1, z1) to the plane ax + by + cz + d = 0

From the figure

PR = projection of PQ on PR

The direction cosines of PQ are

Hence

( because (f, g, h) is a point on ax + by + cz = 0, af + bg + ch = -d)

Eg: A variable plane is at a constant distance p from the origin and cuts the axis at A, B and C. Prove that the locus of the centroid of the triangle ABC is

1/x2 + 1/y2 + 1/z2 = q/p2

Solution :

Let (a, 0, 0), (0, b, 0) and (0, 0, c) be the coordinates of A, B, C

respectively. The equation of the plane is given by

x/a + y/b + z/c = 1

The perpendicular distance p from the origin is given by

• 1/a2 +1/b2 +1/c2 = 1/p2 (1)

Substituting in (1) we get, 1/(9 2) + 1/(9 2) +1/(9 2 ) = 1/P2 1/2 + 1/ 2 +1/2 = 9/P2Hence locus of (, , ) is given by 1/x2 + 1/ y2 +1/z2 = 9/P2

Let (, , ) be the centroid of the triangle ABC. Then  = (a + 0 + 0)/3 = a/3,  = b/3 and  = c/3i.e., a = 3  , b = 3  and c = 3 