# 3d geometry     Post on 04-Sep-2014

54 views

Category:

## Documents

Tags:

• #### straight line

Embed Size (px)

TRANSCRIPT

THREE-DIMENSIONAL GEOMETRYRECTANGULAR COORDINATE SYSTEM IN SPACELet O be any point in space and be three lines perpendicular to each other. These lines are known as coordinate axes and O is called origin. The planes XY, YZ, ZX are known as the coordinate planes.

Coordinates of a Point in Space:Consider a point P in space. The position of the point P is given by triad (x, y, z) where x, y, z are perpendicular distance from YZ-plane, ZX-plane and XY-plane respectively. If We assume i, j, k unit vectors along OX, OY, OZ respectively, then position vector of point P is xi + yj + zk or simply (x, y, z).

x-axis = {( x, y, z) | y = z = 0} y-axis = {(x, y, z) | x = z = 0} z-axis = {(x, y, z) | x = y = 0} xy plane = {(x, y, z) | z = 0} yz plane = {(x, y, z) | x = 0} zx plane = {(x, y, z) | y = 0} 2 2 2 OP = x + y + z

hifting the Origin:

Shifting the origin to another point without changing the directions of the axes is called the translation of axes.

Let the origin O be shifted to another point O' (x', y', z') without changing the direction of axes. Let the new coordinate frame be O'X'Y'Z'. Let P (x, y, z) be a point with respect to the coordinate frame OXYZ.

Then, coordinate of point P w.r.t. new coordinate frame O'X'Y'Z' is (x1, y1, z1), where

x1 = x x', y1 = y y', z1 = z z'

Example -1: If the origin is shifted (1, 2, 3) without changing the directions of the axes then find the new coordinates of the point (0, 4, 5) with respect to new frame. Solution: x' = x x1, where (x1, y1, z1) is the shifted origin y' = y y1 z' = z z1 x' = 0 1 = 1 y' = 4 2 = 2 z' = 5 + 3 = 8 The coordinates of the point w.r.t. to new coordinate frame is (-1, 2, 8). Note: Distance between the points P(x1, y1, z1) and Q (x2, y2, z2) is 2 2 2 (x1 x2) + (y1 y2) + (z1 z2) The point dividing the line joining P(x1, y1, z1) and Q(x2, y2, z2) in m : n ratio is (mx2 nx1 / m + n, my2 ny1 / m + n, mz2 nz1 / m + n) where m + n 0 ., The coordinates of centroid of a triangle having vertices A (x 1, y1, z1), B (x2, y2, z2) and C (x3, y3, z3) is G (x1 + x2 + x3 / 3, y1 + y2 + y3 / 3, z1 + z2 + z3 / 3). Example -2: Find the coordinates of the point which divides the line joining points (2, 3, 4) and (3, 4, 7) in ratio 3 : 5. Solution: Let the coordinates of the required point be (x, y, z), then

x = 2(3) + 3(5) / 3 + 5 = 21/8 y = 3(3) 4(5) / 3 + 5 = 11/8 z = 4(3) + 7(5) / 3 + 5 = 47/8 Hence the required point is (21/8, 11/8, 47/8). Example -3: Prove that the three points A (3, 2, 4), B (1, 1, 1) and C (1, 4, 2) are collinear. Solution: The general coordinates of a point R which divides the line joining A (3, 2, 4) and B (1, 1, 1) in the ratio : 1 are ( + 3 / + 1, 2 / + 1, + 4 / + 1) (1) If C (1, 4, 2) lies on the line AB, then for some value of m the coordinates of the point R will be the same as those of C. Let x-coordinate of point R = x-coordinate of point C. Then, + 3 / + 1 = 1 => = 2 Putting = 2 in (1) the coordinates of R are (1, 4, 2) which are also the coordinates of C. Hence the points A, B, C are collinear.

DIRECTION COSINES OF A LINEIf , , be the angles which a given directed line makes with the positive directions of the co-ordinate axes, then cos, cos, cos are called the direction cosines of the given line and are generally denoted by l, m, n respectively. Thus, l = cos, m = cos and n = cos By the definition it follows that the direction cosine of the axis of x are respectively cos0, cos90, cos 90 i.e. (1, 0, 0). Similarly direction cosines of the axes of y and z are respectively (0, 1, 0) and (0, 0, 1).

Relation between the Direction Cosines:Let OP be any line through the origin O which has direction cosines l, m, n. Dig: Relation Let P (x, y, z) and OP = r 2 2 2 2 2 cosines Then OP = x + y + z = r . (1) From P draw PA, PB, PC perpendicular on the coordinate axes, so that OA = x, OB = y, OC = z. Also, POA = , POB = and POC = . From triangle AOP, l = cos = x/r => x = lr between direction

Similarly y = mr and z = nr Hence from (1) 2 2 2 2 2 2 2 2 2 2 2 r (l + m + n ) = x + y + z = r => l + m + n = 1 Note: If the coordinates of any point P be (x, y, z) and l, m, n be the direction cosines of the line OP, O being the origin, then (lr, mr, nr) will give us the co-ordinates of a point on the line OP which is at a distance r from (0, 0, 0).

Direction Ratios:If a, b, c are three numbers proportional to the direction cosine l, m, n of a straight line, then a, b, c are called its direction ratios. They are also called direction numbers or direction components. Hence by definition, we have 1/a = m/b = n/c = k (say) 2 2 2 2 2 2 2 => l = ak, m = bk, n = ck => k (a + b + c ) = l + m + n = 1 2 2 2 2 => k = 1 / a + b + c = 1/a l = a/a . Similarly m = b/a and n = n/a2 2 2

where the same sign either positive or negative is to be chosen throughout. Example: If 2, 3, 6 be the direction ratios, then the actual direction cosines are 2/7, 3/7, 6/7. Note: Direction cosines of a line are unique but direction ratios of a line in no way unique but can be infinite.

Parallel Lines:Since parallel lines have the same direction, it follows that the direction cosines of two or more parallel straight lines are the same. So in case of lines, which do not pass through the origin, we can draw a parallel line passing through the origin and direction cosines of that line can be found. CXmhaU -4: Find the direction cosines of two lines which are connected by the relations l5m + 2 2 2 3n = 0 and 7l + 5m 3n = 0. hb : The given relations are l5m + 3n = 0 => l = 5m 3n (1) 2 2 2 and 7l + 5m 3n = 0 (2) Putting the value of l from (1) in (2), we get 2 2 2 7(5m 3n) + 5m 3n = 0 2 2 or, 180m 210mn + 60n = 0 or, (2m n)(3m 2n) = 0 m/n = 1/2 or 2/3 when m/n = 1/2 i.e. n = 2m l = 5m 3n = m or 1/m = 1 thus and = 1 giving 2 2 2 2 2 2 or, 1/1 = m/1 = n/2 = (l + m + n ) / {(1) + l + 2 } = 1/v6 So, direction cosines of one line are 1/v6, 1/v6, 2/v6 Again when m/n = 2/3 or n = 3m/2 l = 5m 3.3m/2 = m/2 or 1/m = 1/2 2 2 2 Thus, m/n = 2/3 and i/m = 1/2 giving i/1 = m/2 = n/3 = 1/v1 + 2 + 3 = 1/v14 The direction cosines of the other line are 1/v14, 2/v14, 3/v14.

Direction Cosine of a Line joining two given Points:The direction ratios of line PQ joining P (x 1, y1, z1) and Q(x2, y2, z2) are x2 x1 = a(say), y2 y1 = b (say) and z2 z1 = c (say). Then direction cosines are 2 l = (x2 x1) / (x2 x1) , m = (y2 y1) / (x2 2 2 x1) , n = (z2 z1) / (x2 x1)

Alt txt: direction cosine of a line CXmhaU -5: Find the direction ratios and direction cosines of the line joining the points A(6, 7, 1) and B(2, 3, 1). hb : Direction ratios of AB are (4, 4, 2) = (2, 2, 1) 2 2 2 a +b +c =9 Direction cosines are ( 2/3, 2/3, 1/3).

Angle between two Lines:

Let be the angle between two straight lines AB and AC whose direction cosines are given whose direction cosines are l 1, m1, n1and l2, m2, n2 respectively, is given by cosq = l1l2 + m1m2 + n1n2 If direction ratios of two lines are a 1, b1, c1 and a2, b2, c2 are given, then angle between two lines is given by cos = Particular Results: We have, sin = 1 cos 2 2 2 2 2 2 2 = (i1 + m1 + n1 )(i2 + m2 + n2 ) (l1l2 + m1m2 + n1n2) = (l1m2 l2m1) + (m1n2 m2n1) + (n1l2 n2l1) 2 => sin = (l1m2 l2m1) . Condition of perpendicularity: If the given lines are perpendicular, then = 90 i.e. cos = 0 => l1l2 + m1m2 + n1n2 = 0 or a1a2 + b1b2 + c1c2 = 0 . Condition of parallelism: If the given lines are parallel, then q = 0 i.e. sin = 0 (l1m2 l2m1) + (m1n2 m2n1) + (n1l2 n2l1)2 2 2 0 0 2 2 2 2 2

alt txt: angle between two lines

which is true, only when

l1m2 l2m1 = 0, m1n2 m2n1 = 0 and n1l2 n2l1 = 0 => I1/l2 = m1/m2 = n1/n2

Similarly, a1/a2 = b1/b2 = c1/c2.

CXmhaU -6:

Show that two lines having direction ratios 1, 3, 2 and 2, 2, 2 are perpendicular.

hb :

a1a2 + b1b2 + c1c2 = (1)(2) + (3)(2) + (2)(2) = 2 + 6 4 = 0

lines are perpendicular

Projection of a Line:Projection of the line joining two point P (x1, y1, z1) and Q (x2, y2, z2) on another line whose direction cosines are l, m, n is AB = l(x2 x1) + m(y2 y1) + n(z2 z1)

Perpendicular Distance of a Point from a Line:Let AB is straight line passing through point A (a, b, c) and having direction cosines l, m, n. AN = projection of line AP on straight line AB = l(x a) + m(y b) + n(z c) 2 2 2 and AP = (xa) + (yb) + (zc) perpendicular distance of point P 2 2 PN = AP AN

CXmhaU -7: Find out perpendicular distance of point P (0, 1, 3) from straight line passing through A (1, 3, 2) and having direction ratios 1, 2, 2.

hb :

Direction cosines of the line is 1/1 + 2 + 2 , 2/1 + 2 + 2 , 2/1 + 2 + 2

2

2

2

2

2

2

2

2

2

i.e. .1/3, 2/3, 2/3

PN = l(x a) + m(y b) + n(z c)

1/3 = (0 1) + 2/3 (1 + 3) + 2/3 (3 2) = 5/3

AP = (01) + (1 + 3) + (32) = 6

2

2

2

Perpendicular distance PN = AP PN = 625/9 = 29/3.

2

2

Area of a Triangle

So, area of ABC is given by the relation = x + y + z

2

2

2

2

THE

PLANE

Definition: Consider the locus of a point P(x, y, z). If x, y, z are allowed to vary without anyrestriction for their different combinations, we have a set of points like P. The surface on which these points lie, is called the locus of P. It may be a plane or any curved surface. If Q be any other point on its locus and all points of the straight line PQ lie on it, it is a plane. In other words if the straight line PQ, however small and in whatever direction it may be, lies completely on the locus, it is a plane, otherwise any curved surface. Equation of Plane in Different Forms:

Recommended