Mathematics-23.Unit-19 Three Dimensional Geometry
THREE-DIMENSIONAL GEOMETRY
RECTANGULAR COORDINATE SYSTEM IN SPACE
Let ‘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 unit vectors along OX, OY, OZ respectively, then position vector of point P is or simply (x, y, z).
Note:
• 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}
• OP =
Shifting 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 OXYZ. 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 OXYZ 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
• The point dividing the line joining P(x1, y1, z1) and Q(x2, y2, z2) in m : n ratio is
where m + n 0 .
• The coordinates of centroid of a triangle having vertices A (x1, y1, z1), B (x2, y2, z2) and C (x3, y3, z3) is G .
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 =
y =
z =
Hence the required point is .
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 ……(1)
If C (–1, 4, –2) lies on the line AB, then for some value of 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, = –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 LINE
If , , 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.
Let P (x, y, z) and OP = r
Then OP2 = x2 + y2 + z2 = r2 …. (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 = lr
Similarly y = mr and z = nr
Hence from (1)
r2(l2 + m2 + n2) = x2 + y2 + z2 = r2 l2 + m2 + n2 = 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
(say)
l = ak, m = bk, n = ck k2(a2 + b2 + c2) = l2 + m2 + n2 = 1
k = =
l = . Similarly m = and n =
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 .
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
l–5m + 3n = 0 and 7l2 + 5m2 – 3n2 = 0.
hb : The given relations are
l–5m + 3n = 0 l = 5m – 3n ……(1)
and 7l2 + 5m2 – 3n2 = 0 ……(2)
Putting the value of l from (1) in (2), we get
7(5m – 3n)2 + 5m2 – 3n2 = 0
or, 180m2 – 210mn + 60n2 = 0 or, (2m – n)(3m – 2n) = 0
or
when i.e. n = 2m
l = 5m – 3n = –m or = –1
thus and = –1 giving
or,
So, direction cosines of one line are – , ,
Again when or n =
l = 5m – 3. or
Thus, and giving
The direction cosines of the other line are .
Direction Cosine of a Line joining two given Points:
The direction ratios of line PQ joining P (x1, 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
l = , m = , n =
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)
a2 + b2 + c2 = 9
Direction cosines are .
Angle between two Lines:
Let be the angle between two straight lines AB and AC whose direction cosines are given whose direction cosines are l1, m1, n1 and l2, m2, n2 respectively, is given by cos = l1l2 + m1m2 + n1n2
If direction ratios of two lines are a1, b1, c1 and a2, b2, c2 are given, then angle between two lines is given by
cos =
Particular Results:
• We have, sin2 = 1 – cos2
= – (l1l2 + m1m2 + n1n2)2
= (l1m2 – l2m1)2 + (m1n2 – m2n1)2 + (n1l2 – n2l1)2
sin = .
• Condition of perpendicularity:
If the given lines are perpendicular, then = 900 i.e. cos = 0
l1l2 + m1m2 + n1n2 = 0 or a1a2 + b1b2 + c1c2 = 0 .
• Condition of parallelism:
If the given lines are parallel, then = 00 i.e. sin = 0
(l1m2 – l2m1)2 + (m1n2 – m2n1)2 + (n¬¬1l2 – n2l1)2 = 0
which is true, only when
l1m2 – l2m1 = 0, m1n2 – m2n1 = 0 and n¬¬1l2 – n2l1 = 0
Similarly, .
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)
and AP =
perpendicular distance of point P
PN =
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
i.e. .
PN = l(x – a) + m(y – b) + n(z – c) = (0 – 1) + (–1 + 3) + (3 – 2) =
AP =
Perpendicular distance PN = .
Area of a Triangle
x =
y =
z =
So, area of ABC is given by the relation 2 =
THE PLANE
Definition:
Consider the locus of a point P(x, y, z). If x, y, z are allowed to vary without any restriction 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 it’s 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:
• General equation of a plane is ax + by + cz + d = 0
• Equation of the plane in Normal form is lx + my + nz = p where p is the length of the normal from the origin to the plane and (l, m, n) be the direction cosines of the normal.
• The equation to the plane passing through P(x1, y1, z1) and having direction ratios
(a, b, c) for its normal is a(x – x1) + b(y – y1) + c (z – z1) = 0
• The equation of the plane passing through three non-collinear points (x1, y1, z1),
(x2, y2, z2) and (x3, y3 , z3) is = 0
• The equation of the plane whose intercepts are a, b, c on the x, y, z axes respectively is = 1 (a b c 0)
• Equation of YZ plane is x = 0, equation of plane parallel to YZ plane is x = d.
Equation of ZX plane is y = 0, equation of plane parallel to ZX plane is y = d.
Equation of XY plane is z = 0, equation of plane parallel to XY plane is z = d.
• Four points namely A (x1, y1, z1), B (x2, y2, z2), C (x3, y3, z3) and D (x4, y4, z4) will be coplanar if one point lies on the plane passing through other three points.
CXmhaU -8: Find the equation to the plane passing through the point (2, -1, 3) which is the foot of the perpendicular drawn from the origin to the plane.
hb : The direction ratios of the normal to the plane are 2, -1, 3.
The equation of required plane is 2(x –2) –1 (y + 1) + 3 (z –3) = 0
2x – y + 3z –14 = 0
Angle between the Planes
Angle between the planes is defined as angle between normals of the planes drawn from any point to the planes.
Angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is
Note:
• If a1a2 +b1b2 +c1c2 = 0, then the planes are perpendicular to each other.
• If then the planes are parallel to each other.
CXmhaU -9: Find angle between the planes 2x – y + z = 11 and x + y + 2z = 3.
hb : cos =
cos = = .
CXmhaU -10: Find the equation of the plane passing through (2, 3, –4), (1, –1, 3) and parallel to x-axis.
hb : The equation of the plane passing through (2, 3, –4) is
a(x – 2) + b(y – 3) + c(z + 4) = 0 ……(1)
since (1, –1, 3) lie on it, we have
a + 4b – 7c = 0 ……(2)
since required plane is parallel to x-axis i.e. perpendicular to YZ plane i.e.
1.a + 0.b + 0.c = 0 a = 0 4b – 7c = 0
Equation of required plane is 7y + 4z = 5.
Perpendicular Distance:
The length of the perpendicular from the point P(x1, y1, z1) to the plane ax + by + cz + d = 0 is .
Family of Planes:
Equation of plane passing through the line of intersection of two planes u = 0 and v = 0 is
u + v = 0.
Intersection of a Line and Plane:
If equation of a plane is ax + by + cz + d = 0, then direction cosines of normal to this plane are a, b, c. So angle between normal to the plane and a straight line having direction cosines l, m ,n is given by cos = .
Then angle between the plane and the straight line is .
• Plane and straight line will be parallel if al + bm + cn = 0
• Plane and straight line will be perpendicular if .
CXmhaU -11: Find the equation of plane passing through the intersection of planes
2x – 4y + 3z + 5= 0, x + y + z = 6 and parallel to straight line having direction cosines (1, –1, –1).
hb : Equation of required plane be
(2x – 4y + 3z + 5) + (x + y – z – 6) = 0
i.e. (2 + )x + (–4 + )y + z(3 – ) + (5 – 6) = 0
This plane is parallel to a straight line. So, al + bm + cn = 0
1(2 + ) + (–1)(–4 + ) + (–1)(3 – ) = 0 i.e. = –3
Equation of required plane is –x – 7y + 6z + 23 = 0
i.e. x + 7y – 6z – 23 = 0.
Bisector Planes of Angle between two Planes:
The equation of the planes bisecting the angles between two given planes a1x +b1y +c1z +d1 = 0 and a2x + b2y + c2z +d2 = 0 is
THE STRAIGHT LINE
Straight line in three dimensional geometry is defined as intersection of two planes. So general equation of straight line is stated as the equations of both plane together i.e. general equation of straight line is a1x + b1y + c1z + d1 = 0, a2x + b2y + c2z + d2 = 0 ……(1)
So, equation (1) represents straight line which is obtained by intersection of two planes.
Equation of Straight Line in Different Forms:
Symmetrical Form:
• Equation of straight line passing through point P (x1, y1, z1) and whose direction cosines are l, m, n is
• Equation of straight line passing through two points P (x1, y1, z1) and Q (x2, y2, z2) is
.
Note:
• The general coordinates of a point on a line is given by (x1 + lr, y1 + mr, z1 + nr) where r is distance between point (x1, y1, z1) and point whose coordinates is to be written.
CXmhaU 10. Find the equations of the straight lines through the point (a, b, c) which are
(a) parallel to z-axis
(b) perpendicular to z-axis
10. (i) Equation of straight lines parallel to zaxis have = 900, = 900, = 00
l = 0, m = 0, n = 1
Therefore equation of straight line is parallel to zaxis and passing through (a, b, c) is
(ii) equation of straight lines perpendicular to zaxis
let they make , angle with x and y axes respectively.
Then equation of straight lines perpendicular to z axis and passing through (a, b, c) is
.
CXmhaU -12: Find the coordinates of the point where the line joining the points (2, –3, 1) and (3, –4, –5) cuts the plane 2x + y + z = 7.
hb : The direction ratios of the line are 3 – 2, –4 – (–3), –5 – 1 i.e. 1, –1, –6
Hence equation of the line joining the given points is
= r (say)
Coordinates of any point on this line are (r + 2, –r – 3, –6r + 1)
If this point lies on the given plane 2x + y + z = 7, then
2(r + 2) + (–r – 3) + (–6r + 1) = 7 r = –1
Coordinates of the point are (–1 + 2, –(–1) – 3, –6(–1) + 1) i.e. (1, –2, 7).
Note:
• If equation of straight line is given in general form, it can be changed into symmetrical form. The method is described in following Example.
CXmhaU -13: Find in symmetrical form the equations of the line
3x + 2y – z – 4 = 0= 4x + y – 2z + 3.
hb : The equation of the line in general form are
3x + 2y – z – 4 = 0, 4x + y – 2z + 3 = 0 ……(1)
Let l, m, n be the direction cosines of the line. Since the line is common to both the planes, it is perpendicular to the normals to both the planes.
Hence 3l + 2m – n = 0, 4l + m – 2n = 0
Solving these we get,
i.e.
So, direction cosines of the line are – , , –
Now to find the coordinates of a point on a line. Let us find out the point where it meets the plane z = 0. Putting z = 0 in the equation given by (1), we have
3x + 2y – 4 = 0, 4x + y + 3 = 0
solving these, we get x = –2, y = 5
So, one point of the line is (–2, 5, 0)
equation of the line in symmetrical form is
i.e. .
Shortest Distance between two non Intersecting Line:
Two lines are called non intersecting lines if they do not lie in the same plane. The straight line which is perpendicular to each of non-intersecting lines is called the line of shortest distance. And length of shortest distance line intercepted between two lines is called length of shortest distance.
Method: Let the equation of two non-intersecting lines be
= r1 (say) ……(1)
And = r2 (say) ……(2)
Any point on line (1) is P (x1 + l1r1, y1 + m1r1, z1 + n1r1) and on line (2) is
Q (x2 + l2r2, y2 + m2r2, z2 + n2r2).
Let PQ be the line of shortest distance. Its direction ratios will be
[(l1r1 + x1– x2– l2r2), (m1r1 + y1– y2– m2r2), (n1r1 + z1– z2– n2r2)]
This line is perpendicular to both given line. By using condition of perpendicularity we obtain 2 equations in r1 and r2.
So by solving these, values of r1 and r2 can be found. And subsequently point P and Q can be found. The distance PQ is shortest distance.
The shortest distance can be found by PQ = .
Note:
• If any straight line is given in general form then it can be transformed into symmetrical form and we can further proceed.
CXmhaU -14: Find the shortest distance between the lines , . Also find the equation of line of shortest distance.
hb : Given lines are = r1 (say) ……(1)
= r2 (say) ……(2)
Any point on line (1) is P (3r1 + 3, 8 – r1, r1 + 3) and on line (2) is
Q (–3 – 3r2, 2r2 – 7, 4r2 + 6).
If PQ is line of shortest distance, then direction ratios of PQ
= (3r1 + 3) – (–3 – 3r2), (8 – r1) – (2r2 – 7), (r1+ 3) – (4r2 + 6)
i.e. 3r1 + 3r2 + 6, –r1 – 3r2 + 15, r1 – 4r2 – 3
As PQ is perpendicular to liens (1) and (2)
3(3r1 + 3r2 + 6) – 1(–r1 – 2r2 + 15) + 1(r1 – 4r2 + 3) = 0
11r1 + 7r2 = 0 ……(3)
and –3(3r1 + 3r2 + 6) + 2(–r1 – 2r2 + 15) + 4(r1 – 4r2 + 3) = 0
i.e. 7r1 + 11r2 = 0 ……(4)
On solving equations (3) and (4), we get r1 = r2= 0.
So, point P (3, 8,3) and Q (–3, –7, 6)
Length of shortest distance PQ =
Direction ratios of shortest distance line is 2, 5, –1
Equation of shortest distance line .
THE SPHERE
A sphere is a locus of a point which moves in space such that its distance from a fixed point is constant. Fixed point is called centre of sphere and constant distance is called radius of sphere.
Equation of Sphere in Different Forms:
• If centre of sphere is (a, b, c) and radius is r, then equation of sphere is
(x – a)2 + (y – b)2 + (z – c)2 = r2.
• If centre of sphere is origin and radius is r, then x2 + y2 + z2 = r2.
• General form: The general equation of a sphere is x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0
Centre of sphere = (–u, –v, –w), radius = .
• Diameter form: Equation of a sphere whose extremities of diameter are A (x1, y1, z1) and B (x2, y2, z2) is (x – x1) (x – x2) + (y – y1) (y – y2) + (z – z1) (z – z2) = 0.
CXmhaU 13. Find the equation of the sphere which passes through the points (1, –3, 4), (1, –5, 2) and (1, –3, 0) and whose centre is on the plane x + y + z = 0.
hb : Let equation of the sphere be
x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0
its centre is ( u, v, w) which is on x + y + z = 0
u + v + w = 0 … (1)
it passes through (1, 3, 4) 2u 6v + 8w + d = 26 … (2)
(1, 5, 2) 2 u 10 v + 4 w + d = 30 … (3)
and it passes through (1, 3, 0) 2 u 6 v + d = 10 … (4)
solving these four equations we get,
u = 1, v = 3, w = 2 and d = 10
Therefore required equation of the sphere is
x2 + y2 + z2 2 x + 6 y 4 z + 10 = 0.
CXmhaU -15: Find the equation of the sphere whose centre is (2, –3, 4) and which passes through the point (1, 2, –1).
hb : Radius of sphere =
Equation of the sphere is (x – 2)2 + (y + 3)2 + (z – 4)2 = ( )2
i.e. x2 + y2 + z2 – 4x + 6y – 8z – 22 = 0.
SOLVED PROBLEMS
OBJECTIVE
1. The angle between two lines whose direction cosines are given by the equation l + m + n = 0, l2 + m2 + n2 = 0 is
(A) (B)
(C) (D) None of these
hb : Eliminating n between the two relations, we have
l2 + m2 – (l + m)2 = 0 or 2lm = 0 either l = 0 or m = 0
if l = 0, then m + n = 0 i,e. m = – n
, giving the direction ratios of one line.
If m = 0, then l + n = 0 i.e. l = – n
, giving direction ratios of the other lines.
The angles between these lines is
cos-1 = cos–1 .
2. The equation of the plane which contains the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and perpendicular to the xy plane is:
(A) x – 2y + 11 = 0 (B) x + 2y + 11 = 0
(C) x + 2y – 11 = 0 (D) x – 2y – 11 = 0
hb : Equation of the required plane is (x + y + z – 6) + (2x + 3y + z + 5) = 0
i.e. (1 + 2)x + (1 + 3)y + (1 + )z + (–6 + 5) = 0
This plane is perpendicular to xy plane whose equation is z = 0
i.e. 0 . x + 0 . y + z = 0
By condition of perpendicularity
0.(1 + 2) + 0. (1 + 3) + (1 + ) .1 = 0 i.e. = –1
Equation of required plane is
(1 – 2)x + (1 – 3)y + (1 – 1)z + (–6 – 5) = 0 or x + 2y + 11 = 0.
3. The coordinates of the foot of the perpendicular drawn from the origin to the plane 3x + 4y – 6z + 1 = 0 are :
(A) (B)
(C) (D)
hb : The equation of the plane is 3x + 4y – 6z + 1 = 0 ……(1)
The direction ratios of the normal to the plane (1) are 3, 4, –6. So equation of the line through (0, 0, 0) and perpendicular to the plane (1) are
= r (say) ……(2)
The coordinates of any point P on (2) are (3r, 4r, –6r). If this point lie on the plane (1), then
3(3r) + 4(4r) – 6(–6r) + 1 = 0 i.e. r = –
Putting the value of r coordinates of the foot of the perpendicular P are .
4. The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line is :
(A) 7 unit (B) 4 unit
(C) 1 unit (D) 2 unit
hb : Here we are not to find perpendicular distance of the point from the plane but distance measured along with the given line. The method is as follow:
The equation of the line through the point (1, –2, 3) and parallel to given line is
= r (say)
The coordinate of any point on it is (2r + 1, 3r – 2, –6r + 3).
If this point lies in the given plane then
2r + 1 – (3r – 2) + (–6r + 3) = 5 –7r = –1 or r =
point of intersection is
The required distance
= the distance between the points (1, –2, 3) and
= = 1 unit.
5. The image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0 is :
(A) (–3, 5, 2) (B) (3, 2, 5)
(C) (–5, 3, –2) (D) (–2, 5, 3)
hb : As it is clear from the figure that PQ will be perpendicular to the plane and foot of this perpendicular is mid point of PQ i.e. N.
So, direction ratios of line PQ is 2, –1, 1
Equation of line PQ = r (say)
Any point on line PQ is (2r + 1, –r + 3, r + 4)
If this point lies on the plane, then
2(2r + 1) – (–r + 3) + (r + 4) + 3 = 0 r = –1
coordinate of foot of perpendicular N = (–1, 4, 3)
As N is middle point of PQ
–1 = , 4 = , 3 =
x1 = –3, y1 = 5, z1 = 2
Image of point P (1, 3, 4) is the point Q (–3, 5, 2).
6. The equation of the sphere which passes through the points (1, 0, 0), (0, 1, 0) and (0, 0, 1) and has its radius as small as possible is :
(A) 3(x2 + y2 + z2) + 2(x + y + z) – 1= 0
(B) 3(x2 + y2 + z2) – 2(x + y + z) – 1= 0
(C) (x2 + y2 + z2) – 2(x + y + z) – 1= 0
(D) (x2 + y2 + z2) + 2(x + y + z) – 1= 0
hb : Let equation of sphere be given by
x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 ……(1)
As sphere passes through points (1, 0, 0), (0, 1, 0) and (0, 0, 1). So we have
1 + 2u + d = 0, 1 + 2v + d = 0, 1 + 2w + d = 0
On solving u = v = w = – (d + 1)
If r is the radius of the sphere, then
r =
r2 = (d + 1)2 – d = (say)
for r to be minimum
= 0 .2(d + 1) – 1 = 0 or d = –
Also, = positive at d = –
Hence is minimum at d = –
So, substituting value of d we have u = v = w = –
equation of the sphere
x2 + y2 + z2 – (x + y + z) – = 0 3(x2 + y2 + z2) – 2(x + y + z) – 1= 0.
7. A point moves so that the ratio of its distances from two fixed points is constant. Its locus is a:
(A) plane (B) st. lines
(C) circle (D) sphere
hb : Let the coordinates of moving point P be (x, y, z). Let A (a, 0, 0) and B (–a, 0, 0) be two fixed points. According to given condition
= constant = k (say) AP2 = k2BP2
or, (x – a)2 + (y – 0)2 + (z – 0)2 = k2{(x + a)2 + (y – 0)2 + (z – 0)2}
(1 – k2)(x2 + y2 + z2) – 2ax(1 + k2) + a2(1 – k2) = 0
required locus is x2 + y2 + z2 – + a2 = 0. Which is a sphere.
8. The ratio in which yz–plane divides the line joining (2, 4, 5) and (3, 5, 7)
(A) -2 : 3 (B) 2 : 3
(C) 3 : 2 (D) -3 : 2
hb : Let the ratio be : 1
x-coordinate is = 0 = -2/3 .
Hence (A) is the correct answer.
9. A line makes angles , , , with the four diagonals of a cube then
cos2 + cos2 + cos2 + cos2 =
(A) 1 (B) 4/3
(C) 3/4 (D) 4/5
hb The direction ratios of the diagonal
(1, 1, 1)
Direction cosine are
Similarly direction cosine of are
are
are
Let l, m, n be direction cosines of the line
cos = , cos = , cos = , cos =
cos2 + cos2 + cos2 + cos2 = = ( since l2 + m2 + n2 = 1)
Hence (B) is the correct answer.
10. The points (0, -1, -1), (-4, 4, 4), (4, 5, 1) and (3, 9, 4) are
(A) collinear (B) coplanar
(C) forming a square (D) none of these
hb : Equation of the plane passing through the points (0, -1, -1), (-4, 4, 4) and (4, 5, 1) is = 0 …. (1)
The point (3, 9, 4) satisfies the equation (1).
Hence (B) is the correct answer.
11. A variable plane passes through a fixed point (a, b, c) and meets the coordinate axes in A, B, C. The locus of the point common to plane through A, B, C parallel to coordinate planes is
(A) ayz + bzx + cxy = xyz (B) axy + byz + czx = xyz
(C) axy + byz + czx = abc (D) bcx + acy + abz = abc
hb : Let the equation to the plane be
( the plane passes through a, b, c)
Now the points of intersection of the plane with the coordinate axes are A(, 0, 0), B(0, , 0) & C(0, 0, )
Equation to planes parallel to the coordinate planes and passing through A, B & C are x = , y = and z = .
The locus of the common point is
(by eliminating , , from above equation)
Hence (A) is the correct answer.
12. Consider the following statements:
Assertion (A): the plane y + z + 1 = 0 is parallel to x-axis.
Reason (R): normal to the plane is parallel to x-axis.
Of these statements:
(A) both A and R are true and R is the correct explanation of A
(B) both A and R are true and R is not a correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
hb : Given plane y + z + 1 = 0 is parallel to x-axis as 0.1 + 1.0 + 1.0 = 0
but normal to the plane will be perpendicular to x-axis.
Hence (C) is the correct answer.
13. The equation of the plane containing the line is
a(x – ) + b(y – ) + c(z – ) = 0, where al + bm + cn is equal to
(A) 1 (B) –1
(C) 2 (D) 0
hb : Since straight line lies in the plane so it will be perpendicular to the normal at the given plane. Since direction cosines of straight line are l, m, n and direction ratios of normal to the plane are a, b, c. So, al + bm + cn = 0.
Hence (D) is the correct answer.
14. The shortest distance between the two straight lines and is
(A) (B) 3
(C) 0 (D) 6
hb : Since these two lines are intersecting so shortest distance between the lines will be 0.
Hence (C) is the correct answer.
15. A straight line passes through the point (2, –1, –1). It is parallel to the plane
4x + y + z + 2 = 0 and is perpendicular to the line . The equations of the straight line are
(A) (B)
(C) (D)
hb : Let direction cosines of straight line be l, m, n
4l + m + n = 0
l – 2m + n = 0
Equation of straight line is .
Hence (C) is the correct choice.
16. If centre of a sphere is (1, 4, –3) and radius is 3 units, then the equation of the sphere is
(A) x2 + y2 + z2 – 2x – 8y + 6z + 17 = 0 (B) 2(x2 + y2 + z2) – 2x – 8y + 6z + 17 = 0
(C) x2 + y2 + z2 – 4x + 16y + 12z + 17 = 0 (D) x2 + y2 + z2 + 2x + 8y – 6z – 17 = 0
hb : Equation of sphere will be (x – 1)2 + (y – 4)2 + (z + 3)2 = 9
Hence (A) is the correct answer.
17. If equation of a sphere is 2(x2 + y2 + z2) – 4x – 8y + 12z – 7 = 0 and one extremity of its diameter is (2, –1, 1), then the other extremity of diameter of the sphere will be
(A) (2, 9, –13) (B) (0, 9, 7)
(C) (0, 5, 7) (D) (2, 5, –13)
hb : The centre of the sphere is (1, 2, –3) so if other extremity of diameter is (x1, y1, z1), then
= 1, = 2, = –3
Required point is (0, 5, 7).
Hence (C) is the correct answer.
18. The direction cosines of the line which is perpendicular to the lines with direction cosines proportional to (1, – 2, – 2), (0, 2, 1) is
(A) (B)
(C) (D)
hb : Let direction ratios of the required line be
Therefore a 2 b 2 c = 0
And 2 b + c = 0
c = 2 b
a 2 b + 4b = 0 a = 2 b
Therefore direction ratios of the required line are < 2b, b, 2b> = <2, 1, 2>
direction cosines of the required line
=
19. The points (4, 7, 8), (2, 3, 4), (–1, –2, 1) and (1, 2, 5) are :
(A) the vertices of a parallelogram (B) collinear
(C) the vertices of a trapezium (D) concyclic
hb : Let A (4, 7, 8), B (2, 3, 4), C ( 1, 2, 1), D (1, 2, 5)
Direction cosines of AB =
Direction cosines of CD
=
So, AB parallel to CD
Direction cosines of AD
Direction cosines of BC
=
so, AD is parallel to BC.
Therefore ABCD is a parallelogram.
20. The equation of the plane parallel to the plane 4x – 3y + 2z + 1 = 0 and passing through the point (5, 1, – 6) is :
(A) 4x 3y + 2z 5 = 0 (B) 3x 4y + 2z 5 = 0
(C) 4x 3y + 2z + 5 = 0 (D) 3x 4y + 2z + 5 = 0
hb : Equation of the plane parallel to the plane 4x 3y + 2z + 1 = 0 is of the form of
4x – 3y + 2z + k = 0 again it passes through (5, 1, 6)
so, 20 3 12 + k = 0 k = 5
Therefore required equation is 4x 3y + 2z 5 = 0.
21. A plane is passed through the middle point of the segment A (–2, 5, 1), B (6, 1, 5) and is perpendicular to this line. Its equation is :
(A) 2x y + z = 4 (B) 2x y + z = 4
(C) x 3y + z = 5 (D) x 4y + 2z = 5
hb : Midpoint of A ( 2, 5, 1) and B (6, 1, 5) is (2, 3, 3)
direction ratios of the line joining A and B is <2, 1, 1>
Therefore equation of the line perpendicular to AB and passing through (2, 3, 3) is
2(x – 2) – 1(y – 3) + 1(z – 3) = 0 2x y + z = 4
22. A plane meets the co-ordinates axes in A, B, C such that the centroid of triangle ABC is (a, b, c). The equation of the plane is :
(A) (B)
(C) (D) None of these
hb : The plane meets the coordinate axes at A, B, C such that centroid of the triangle ABC is (a, b, c)
so, the plane cuts Xaxis at (3 a, 0, 0)
So, Xintercept = 3 a
The plane cuts Yaxis at (0, 3 b, 0)
Yintercept = 3 b
the plane cuts Zaxis at (0, 0, 3 c)
Zintercept = 3 c
Therefore required equation is = 1
= 3.
23. The radius of the sphere (x + 1)(x + 3) + (y – 2)(y – 4) + (z + 1)(z + 3) = 0 is:
(A) (B) 2
(C) (D) 3
hb : (x + 1) (x + 3) + (y 2) (y 4) + (z + 1)(z + 3) = 0 is the given equation of sphere.
So, end points of the diameter are
( 1, 2, 1) and ( 3, 4, 3)
radius = =
24. The sum of the direction cosines of a straight line is
(A) zero (B) one
(C) constant (D) none of these
hb : cos = l, cos = m, cos = n
sum of direction cosines cos + cos + cos
= l + m + n
which is constant.
25. The angle between straight lines whose direction cosines are and is
(A) cos–1 (B) cos–1
(C) cos–1 (D) none of these
hb : = cos1 = cos1
26. Which one of the following is best condition for the plane ax + by + cz + d = 0 to intersect the x and y axes at equal angle
(A) |a| = |b| (B) a = –b
(C) a = b (D) a2 + b2 = 1
hb : The plane a x + b y + c z + d = 0 intersects x and y axes at equal angles
Therefore |cos | = |cos |
|l| = |m|
|a| = |b|.
27. The equation of a straight line parallel to the x-axis is given by
(A) (B)
(C) (D)
hb : Equation of straight line parallel to xaxis is
because l = cos = 1, m = cos = cos = 0, n = cos = 0.
28. If P (2, 3, –6) and Q (3, –4, 5) are two points, the direction cosines of line PQ are
(A) – , – , – (B) , – ,
(C) , , – (D) – , – ,
hb : P (2, 3, 6), Q (3, 4, 5)
direction ratios = <1, 7, 11>
direction cosines =
=
29. The ratio in which yz–plane divide the line joining the points A(3, 1, –5) and B(1, 4, –6) is
(A) –3 : 1 (B) 3 : 1
(C) –1 : 3 (D) 1 : 3
hb : A (3, 1, 5), B (1, 4, 6)
Therefore = 0 = 3
Therefore required ratio is 3 : 1
30. A straight line is inclined to the axes of x and z at angels 450 and 600 respectively, then the inclination of the line to the y-axis is
(A) 300 (B) 450
(C) 600 (D) 900
hb : A straight line is inclined to the axes of x and z at an angle 450 and 600
l2 + m2 + n2 = 1
m2 =
m =
angle made by 600
31. The angle between two diagonals of a cube is
(A) cos = (B) cos =
(C) cos = 1/3 (D) none of these
hb : cos = of side is a
= .
32. Given that A (3, 2, –4), B (5, 4, –6) and C (9, 8, –10) are collinear. The ratio in which B divides AC
(A) 1 : 2 (B) 2 : 1
(C) –1 : 2 (D) –2 : 1
hb : = 5 9 5 = 2
= .
33. If P1P2 is perpendicular to P2P3, then the value of k is, where P1(k, 1, –1), P2(2k, 0, 2) and P3(2 + 2k, k , 1)
(A) 3 (B) –3
(C) 2 (D) –2
hb : Direction ratios of P1 P2 =
direction ratios of P2 P3 = <2, k, 1>
Therefore 2 k k 3 = 0
k = 3.
34. A is the point (3, 7, 5) and B is the point (–3, 2, 6). The projection of AB on the line which joins the points (7, 9, 4) and (4, 5, –8) is
(A) 26 (B) 2
(C) 13 (D) 4
hb : Distances of the line joining (7, 9, 4) and (4, 5, 8) is
Therefore required projection is = 2 (B)
Exercise # 1
35. The shortest distance of the point from the x-axis is equal to(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.1)
(A) (B)
(C) (D) None of these
Ans. (C)
hb . Foot of perpendicular drawn from P to x-axis will have its coordinates as (x, 0, 0).
Required distance =
36. The point of intersection of the xy plane and the line passing through the points and is:(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.4)
(A) (B)
(C) (D)
Ans. (B)
hb . Direction ratios of AB are 2, -3, 5.
Thus equation of AB is,
For the point of intersection of this line with xy–plane, we have
Z 0
Hence, the required point is
37. The projections of the line segment AB on the coordinate axes are –9, 12, -8 respectively. The direction cosines of the line segment AB are:(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.5)
(A) (B)
(C) (D) None of these
Ans. (A)
Length of segment
Thus direction cosines of AB are
.
38. The direction cosines of two mutually perpendicular lines are and . The direction cosines of the line perpendicular to both the given lines will be:(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.6)
(A) (B)
(C) (D)
Ans. (D)
Let the direction cosine of the required line be , m and n.
We must have,
We have
Thus
39. A directed line segment angles with the coordinate axes. The value of is always equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.7)
(A) -1 (B) 1
(C) -2 (D) 2
Ans. (A)
–1
40. The locus represented by xy + yz = 0 is:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.9)
(A) A pair of perpendicular lines (B) A pair of parallel lines
(C) A pair of parallel planes (D) A pair of perpendicular planes
Ans. (D)
Thus it represents a pair of planes
X 0, y z 0
that are clearly mutually perpendicular.
41. The plane 2x – 3y + 6z – 11 = 0 makes an angle (a) with x-axis. The value of ‘a’ is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.10)
(A) (B)
(C) (D)
Ans. (C)
If ‘’ be the angle between the plane and x–axis, then
42. The planes x + y = 0, y + z = 0 and x + z = 0:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.12)
(A) meet in a unique point (B) meet in a unique line
(C) are mutually perpendicular (D) none of these
Ans. (B)
Clearly, given planes have a common line of intersection namely the z-axis.
43. The equation of a plane passing through (1, 2, -3), (0, 0, 0) and perpendicular to the plane 3x – 5y + 2z = 11, is:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.13)
(A) (B) 4x + y + 2z = 0
(C) (D) x + y + z = 0
Ans. (D)
Let the required plane be
Ax by cz 0.
We have 3a – 5b 2c 0, a 2b – 3c 0
Thus plan is x y z 0
44. The direction ration of a normal to the plane passing through (1, 0, 0), (0, 1, 0) and making an angle with the plane x + y = 3 are:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.15)
(A) (B)
(C) (D)
Ans. (B)
Let the plane be
.
Also,
Thus direction rations are
or
45. The equation of a plane passing through the line of intersection of the planes x + y + z = 5, 2x – y + 3z = 1 and parallel to the line y = z = 0 is:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.16)
(A) 3x – z = 9 (B) 3y – z = 9
(C) x – 3z = 9 (D) y – 3z = 9
Ans. (B)
Plane will be in the form
(x y z – 5) a(2x – y 3z) –1) 0 i.e., x(1 2a) y(1 – a) z(1 3a) 5 a
It is parallel to the line y z 0.
Since, (1 2a) 0
Thus required plane is
i.e., 3y – z 9
46. The angle between lines whose direction cosines are given by , , is:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.17)
(A) (B)
(C) (D) None of these
Ans. (D)
.
We also have
Also,
Hence, direction cosines are lines are
,
Angle between these lines in both cases is zero.
47. Centroid of the tetrahedron OABC, where , , and O is the origin is (1, 2, 3). The value of is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.3Q.19)
(A) 75 (B) 80
(C) 121 (D) None of these
Ans. (A)
We have
4 a 1 2 0,
8 2 b 1 0
12 3 2 c 0.
48. The equation of the plane passing through the points (2, -1, 0), (3, -4, 5) and parallel to the line 2x = 3y = 4z is:(Ref.P.K.Sharma_Three Dimen._P.C6.3Q.20)
(A) 125x – 90y – 79z = 340 (B) 32x – 21y – 36z = 85
(C) 73x + 61y – 22z = 85 (D) 29x – 27y – 22z = 85
Ans. (D)
Give line is 2x 3y 4z
i.e.,
Let the plane be
Ax by cz 1.
We have
6a 4b 3c 0
2a – b 1
3a – 4b 5c 1.
Thus equation of plane is
29x – 27y – 22z 85.
Exercise # 2
49. A plane passes through the point . If the distance of this plane from the origin is maximum, then it’s equation is:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.1)
(A) x + 2y – 3z + 4 = 0 (B) x + 2y + 3z = 0
(C) 2y - x + 3z = 0 (D) x - 2y + 3z = 0
Ans. (B)
Clearly in this case OA will be a normal to the plane.
Direction cosine of segment OA are
and .
Thus the equation of plane is
x 2y 3z 14.
50. The shortest distance of the plane 12 + 4y + 3z = 327, from the sphere , is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.3)
(A) 39 units (B) 26 sq. units
(C) 13 units (D) None of these
Ans. (C)
Center and radius of given sphere are (–2, 1, 3) and 13 unit respectively.
Now, distance of center of the sphere from the given plane
Shortest distance (26 – 13) 13 units.
51. The lines x = ay + b, z = cy + d and x = a’y + b’, z = c’y + d’ will be mutually perpendicular provided:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.5)
(A) (a + a’)(b + b’)(c + c’) (B) aa’ + cc’ + 1 = 0
(C) aa’ + bb’ + cc’ + 1 = 0 (D) (a + a’) (b + b’) (c + c’) + 1 = 0
Ans. (B)
Give lines are
These lines will be mutually perpendicular, provided
.
52. The straight lines and , will intersect provided:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.6)
(A) k = {3, -3} (B) k = {0, -1}
(C) k = {-1, 1} (D) k = {0, -3}
Ans. (D)
Any point on the first line can be takes as
.
These lines will intersect if for some r1 and r2 we have
.
putting these values in the last condition, we get
k2 3k 0
.
53. The plane ax + by + cz = d, meets the coordinate axes at the points, A, B and C respectively. Area of triangle ABC is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.7)
(A) (B)
(C) (D) None of these
Ans. (B)
Area of triangle
Area of triangle
Area of triangle
If area of triangle ABC be , then
54. Equation of the plane passing through (-1, 1, 4) and containing the line , is:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.9)
(A) 9x – 22y + 2z + 23 = 0 (B) x + 22y + z = 25
(C) 9x + 22y - 3z = 1 (D) 22y – 9x + z = 35
Ans. (D)
Equation of any plane containing the line will be
where, 3a b 4c 0 …. (i)
It is given that plane passes through (–1, 1, 4).
–2a – b 4c 0 …. (ii)
From (i) and (ii), we get
.
Thus the equation of required plane is,
–9(x – 1) 22(y – 2) z 0
i.e., 22y – 9x z 35
55. Equation of the plane containing the lines and is,:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.10)
(A) x + y – 4z = 6 (B) x - y + 4z = 6
(C) x + y + 4z = 6 (D) None of these
Ans. (D)
Equation of any plane containing the line is
where a 3b – c 0
This plane will also contain the second line if
2a – 3b c 0
and 4a b(0 – 2) c(0 4) 0
Solving these equation, we get
a 0, b 0, c 0.
That means the given lines are non–coplanar.
56. Equation of the plane such that foot of altitude drawn from (-1, 1, 1) to the plane has the coordinate (3, -2, -1), is:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.11)
(A) x + y + z = 0 (B) 4x - 3y – 2z = 20
(C) 3x + y – z = 8 (D) 4x + 3y – z = 7
Ans. (B)
Clearly, the direction ratio of the plane are
4, –3, –2
Thus equation of plane will be
4x – 3y – 2z d
It will necessarily pass through (3, –2, –1)
i.e,, d = 12 6 2 20
Thus the equation of plane is
4x – 3y – 2z 20.
57. The distance of the point (-1, 2, 6) from the line , is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.13)
(A) 7 units (B) 9 units
(C) 10 units (D) 12 units
Ans. (A)
Any point on the line is
P (6r1 2, 3r1 3, –4r1 –4).
Direction ration of the line segment PQ, where Q (–1, 2, 6), are
6r1 3, 3r1 1, – 4r, – 10.
If ‘P’ be the foot of altitude drawn from Q to the given line, then
6(6r1 3) 3(3r1 1) 4(4r1 10) 0.
r1 = –1.
Thus, P (–4, 0, 0)
Required distance
7 units.
58. The point of intersection of the lines and is:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.14)
(A) (B)
(C) (D)
Ans. (B)
Any point on the first line is
(3r1 – 1, 5r1 – 3, 7r1 – 5)
and any point on the second line is
(r2 2, 3r2 4, 5r2 6).
At the point of intersection, we must have
3r1 – 1 r2 2,
5r1 – 3 3r2 4.
7r1 – 5 5r1 6.
Thus,
Hence required point is i.e.
59. The shortest distance between the line x + y + 2z – 3 = 2x + 3y + 4z – 4 = 0 and the z-axis is:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.15)
(A) 1 unit (B) 2 units
(C) 3 units (D) 4 units
Ans. (B)
We have,
Solving these equations, we get
Y –2.
Thus required shortest distance is 2 units.
60. The length of projection, of the line segment joining the points (1, -1, 0) and (-1, 0, 1), to the plane 2x + y + 6z = 1, is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.17)
(A) (B)
(C) (D)
Ans. (B)
Let A (1, –1, 0), B (–1, 0, 1).
Direction rations of segment AB are
2, –1, –1.
If ‘’ be the acute angle between segment AB and normal to plane,
.
Length of projection
(AB) sin
units.
61. Reflection of the line in the plane x + y + z = 7 is:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.18)
(A) (B)
(C) (D)
Ans. (C)
Given line passes through (1, 2, 4) and this point also lies on the given plane.
Thus required line will be in the form of .
Any point on the given line is
(–r1 1, 3r1 2, r1 4).
If r1 1, this point becomes P (0, 5, 5).
Let Q (a, b, c) be the reflection of ‘P’ in the given plane, then
i.e, a b c 4,
and
Thus, Q (–2, 3, 3)
Hence direction rations of reflected line are
–3, 1, –1
Thus it’s equation is
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