THREE–DIMENSIONAL GEOMETRYIntroduction

Three dimensional geometry developed accordance to Einstein’s field equations. It is useful in several branches of science like it is useful in Electromagnetism. It is used in computer algorithms to construct 3D models that can be interactively experienced in virtual reality fashion. These models are used for single view metrology. 3-D Geometry as carrier of information about time by Einstein. 3-D Geometry is extensively used in quantum & black hole theory. In this chapter we present a vector–algebra approach to three–dimensional geometry. The aim is to present standard properties of lines and planes, with minimum use of complicated three–dimensional diagrams such as those involving similar triangles. We summarize the chapter: Points are defined as ordered triples of real numbers and the distance between points P1 = (x1, y1, z1) and P2 = (x2, y2, z2) is defined by theformulaP1P2 = √(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.Directed line segments

Points in space Points in space can be represented by vectors as this is equivalent to a displacement relative to the origin of the coordinate system.

For example the vector shown above can represent the point (5,8,3)

Section Formula:

(1)Integral division: If R(x, y, z) is point dividing join of P(x1, y1, z1) & Q(x2, y2, z3) in ratio of m : n.

Then,   x = , y = , z =(2)External division: Coordinates of point R which divides join of P(x1, y1, z1) & Q(x2, y2, z2) externally in ratio m : n are  

Illustration: Show that plane ax + by + cz + d = 0 divides line joining (x1, y1, z1) & (x2, y2, z) in ratio of

Ans: Let plane ax + by + cz + d = 0 divides line joining (x1, y1, z1) & (x2, y2, z2) in ratio K : 1 Coordinates of P

must satisfy eq. of plane.ax + by + cz + d = 0

Direction Cosines:

Let

is a vector , ,inclination with x, y & z-axis respectively. Then cos

, cos & cos

are direction cosines of

. They denoted by    

direction angles.& lies   0 , , Note: (i) Direction cosines of x-axis are (1, 0, 0)Direction cosines of y-axis are (0, 1, 0)Direction cosines of z-axis are (0, 0, 1)

(ii) Suppose OP be any line through origin O which has direction l, m, n

(r cos , r cos , r cos ) where OP = r

coordinates of P are (r cos , r cos , r cos )or   x = lr, y

= mr, z = nr

(iii) l2+ m2+ n2= 1 Proof:   | | = | | =

| |2= x2+ y2+ z2= l2| |2+ m2| |2+ n2| |2

l2+ m2+ n2= 1

(iv)

Direction ratios: Suppose l, m, & n are direction cosines of vector

& a, b, c are no.s such that a, b, c are proportional to l, m, n. These a, b, c are direction ratios.

= kSuppose a, b, c are direction ratios of vector

having direction cosines l, m, n. Then,

= l = a, m = b, n = c

  l2+ m2+ n2= 1  ⇨     a2 2+ b2 2+ c2 2= 1  =

l = ± , m = ± , n = ±

Note: (i) If

having direction cosines l, m, n. Then, l =

, m = , n =

(ii) Direction ratios of line joining two given points(x1, y1, z1) & (x2, y2, z2) is (x2- x1, y2- y1, z2- z1)

(iii) If direction ratio's of

are a, b, c

=(iv) Projection of segment joining points P(x1, y1, z1) and Q(x2, y2, z2) on a line with direction cosines l, m, n, is:(x2- x1)l + (y2- y1)m + (z2- z1)n

Illustration

: If a line makes angles

, &with coordinate axes, prove that

sin2 +sin2+ sin2 =2

Ans: Line is making , &

with coordinate axes.Then, direction cosines are l = cos

, m = cos & n = cos . But   l2+ m2+ n2= 1      

cos2 + cos2 + cos2 = 1,      

1-sin2 +1-sin2 + 1 - sin2 = 1   sin2 + sin2 + sin2 = 2Angle b/w two vectors in terms of direction cosines &

Cos =

   cos = l1l2 + m1m2 + n1n2

(ii) IF a1, b1, c1 and a2, b2, c2 are d.r.s of & . Then

&  

cos =

cos =

Note: (i) If two lines are then,             cos = 0   or  

l1l2+ m1m2+ c1c2=0            

or a1a2+ b1b2+ c1c2=0

(ii)If two lines are || then

cos =1 or

or  

Illustration: Find angle b/w lines whose direction cosines are

& .

Ans: Let be angle

cos =l1l2+m1m2+n1n2

cos =- =1200

Straight line Vectorial eq. of line:

Let a line passing through a point of P.V. & || to given line EF( ).

Then, eq. of line

Proof: Let P be any point on line AP & its P.V. is

Then, = = (By law)

Note: Eq. of line through origin & || to is = Cartesion eq. of straight line:

Passing through point & parallel to given vector .

Let line is passing through A(x1, y1, z1) & || to EF whose d.r.'s are a, b, c.

d.r.'s of AP = (x - x1, y - y1, z - z1) d.r.'s of EF = (a, b, c) Since EF || AP, then,

Eq. of line in parametric form.Illustration

: Find eq. of line || to & passing through pt.(1, 2, 3) ?

Ans: A(1, 2, 3) =

eq. of line passing through & || to

= ( ) + ( )

Vector eq. of line passing through two given points :Vector eq. of line passing through two points whose P.V. S

& is: = + ( - )

Proof: is collinear with

=

Cartesian form :Eq. of line passing through (x1, y1, z1) & (x2, y2, z2)

D.R.'s of AB = (x2- x1, y2- y1, z2- z1)D.R.'s of AP = (x - x1, y - y1, z - z1)Since AB || AP

Illustration: Find cartesian eq. of line are 6x + 2 = 3y - 1 = 2z + 2. Find its direction ratios.

Ans: is cartesion eq. of line 6x + 2 = 3y - 1 = 2z + 2

6(x + ) = 3(y - ) = 2(z + 1)

on comparing a = , b = , c =

Angle b/w Two lines:

=

Angle b/w two lines cos = If

=

Condition for perpendicularity:

. = 0 a1a2+ b1b2+ c1c2 =0

Condition for parallelism :Dumb Question: Why angle between pair of lines

Ans:

&

Perpendicular distance of point from a line :(a)Castesian form :

Suppose 'L' is foot of line of . Since L lies on line AB so,

coordinate of L is =

i.e. L(x1+ a, y1+ b, z1+ c) direction ratios of PL are: (x1+ a ,

y1+ b - , z1+ c - )also direction ratios of AB are (a, b, c) Since PL AB

a(x1+ a - ) + b( b - ) + c(z1+ c - ) = 0

Vector Form:

Since L lies on line AB , P.V. of L = P.V. of line AB = +

dr's of

Since

P.V. of L is +

Reflection of point in straight line: Castesian Form:

From above, we get coordinate of L(foot of ) But L is mid point of PQ

= x1+ a , = y1 + b , = z1 + c

' = 2(x1 + a ) - , ' = 2(y1 + b ) - , ' = 2(z1 + c ) -

Vector Form:

From above, we get , P.V. of L, +

Let P.V. of Q is

Since L is mid point of PQ

Illustration: Find reflection of point P(2, 3, 1) in line

Ans:

Since L lies on line AB coordinate of L (3 + 2, 2 + 1, 4 - 3)

DR's of PL are = (3 + 2 - 2, 2 + 1 - 3, 4 - 3 - 1)

= (3 , 2 - 2, 4 - 4) , DR's of AB are (3, 2, 4) Since PL AB

3(3 ) + 2(2 - 2) + 4(4 - 4) = 0

9 + 4 - 4 + 16 - 16 = 0

29 = 20 = Since L is mid point of PQ

So, = 3 + 2, = 2 + 1, = 4 - 3

= (6 + 4 - 2), = (4 + 2 - 3), = 8 - 6 - 1

= 6 + 2, = 4 - 1, = 8 - 7

where =

Skew lines: Those lines which do not lies in same plane.

Shortest distence b/w two skew lines:

The line which is to both line l1 & l2 are c/d line of shortest distance.

Vector Form: Let l1 & l2 are:

& respectively.

Since is to both l1 & l2 which are parallel to &

is || to x

Let be unit vector along , then = ±

PQ = Projection of on

= .

Dumb Question: How PQ = Projection on ? Ans: Form fig., it is clear that PQ is projection of on .

PQ = Cartesian form: Two skew lines

& Shortest distance :

Condition for lines to intersect: Two lines intersects if shortest distance = 0

or

Shortest distance b/w parallel line:

Let l1 & l2 are respectively & BM is shortest distance b/w l1 & l2

sin = BM = AB sin = | | sin

| x | = | | | | sin( - )

= | | | | sin = (| | sin ) | | =

Dumb Question: Why we have taken sin( - ) ? Ans:

Since direction of vector is opposite to || lines. So, we have taken

( - ) instead of

BM =

Illustration: Find shortest distance b/w lines:

Ans:

On comparing

Plane: (i) Eq. of plane passing through a given point (x1, y1, z1) is:

a(x - x1) + b(y - y1) + c(z - z1) = 0 where a, b, c constants. Proof: General eq. of plane is ax + by + cz + d = 0....................(i)

It is passes through (x1, y1, z1) ax1 + by1 + cz1 + d = 0 ............................ (ii)

By (i) - (ii), we get a(x - x1) + b(y - y1) + c(z - z1) = 0 Concurrence and Coplanarity

Four planes a x+b y+c z+d =0, a x+b y+c z+d =0, a x+b y+c z+d =0 and a x+b y+c z+d =0 are concurrent if and only if

Four points (x ,y ,z ), (x ,y ,z ), (x ,y ,z ) and (x ,y ,z ) are coplanar if and only if

(Both of these assertions remain true in oblique coordinates.)

Intercept form of a plane: eq. of plane of intercepting lengths a, b & c with x, y & z-axis respectively is,

Illustration: A variable plane moves in such a way that sum of reciprocals of its intercepts on 3 coordinate axes is constant. Show that plane passes through fixed point.

Ans: Let eq. of plane is . Then, intercepts of plane with axes are: A(a, 0, 0), B(0, b, 0), c(0, 0, c)

= constant (k) (given)

= 1 & comparing with fixed point

x = , y = , z =

This shows plane passes through fixed point ( , , )

Vector eq. of plane passing through a given point & normal to given vector:

Vector eq. of plane passing through u point of P.V. & normal to

vector is ( - ). = 0

Dumb Question: What is normal to vector ?

Ans: Plane normal to vector means the every line in plane is to that given vector.

Proof: Let plane passes through A( ) & normal to vector & be P.V. of every point 'P' on plane.

Since lies in plane & is normal to plane.

. = 0

( - ). = 0 ( = - )

eq. of plane ( - ). = 0

Eq. of plane in normal form: Vector eq. of plane normal to unit

vector & at O distance d from origin is . = d. Proof:

ON is to plane such that & =

Since . = 0

( - ). = 0

( - d ).d = 0 .d - d2 . = 0

d(⇨ . ) - d2 = 0 . = d

eq. of plane is . = d

Cartesian Form: Let l, m, n be d.r.'s of normal to given plane & P is length of from origin to plane, then eq. of plane is lx + my + nz = P.

Eq. of plane passing through 3 given points:

Eq. of plane passing through three points A, B, C having P.V.'s ,

, respectivelt. Let be P.V. of any point P is the plane.

So, vectors, = - , = - , = - are coplanar

Hence, .( x ) = 0

( - ).{( - ) x ( - )} = 0

( - ).( x - x - x - x ) = 0

( - ).( x + x + x ) = 0

.( x + x + x ) = .( x ) + .( x )

+ .( x )

[ ] + [ ] + [ ] = [ ]Note: If P is length of from origin on this plane,

then P = n = | x + x + x |

Eq. of plane that passes through a point A with position vector &

is || to given vectors & : Derivation:

Let be P.V. of any point P in planeThen, = - = -

Since x are || to plane.So, vector - , & are coplanar

( - ).( x ) = 0 .( x ) = .( x ) [ ] = [ ]Cartesian form: Eq. of plane passing through a point (x1, y1, z1) & || to two lines

having direction ratios ( 1, 1, 1) & ( 2, 2, 2) is:

Illustration : Find eq. of plane passing through pointsP(1, 1, 1), Q(3, -1, 2), R(-3, 5, -4).Ans: Let eq. of plane is ax + by + cz + d = 0It passes through P(1, 1, 1). So,eq. of plane a(x - 1) + b(y - 1) + c(z - 1) = 0 .............................. (i)But it also passes through Q & R 2a - 2b + c = 0] x (- 2) - 4a + 4b - 5c = 0

Similarly a = 6, b = 6, Putting these in eq. (i)We get, 6(x - 1) + 6(y - 1) = 0

x + y = 2

Angle b/w two planes : . = d1

. = d2 are two planers, then cos = Note: Angle b/w planes is defined as angle b/w their normals.Cartesion Form : Let planes are a1x + b1y + c1z + d1= 0 & a2x + b2y +

c2z + d2= 0 cos =

Condition for : . = 0 or a1a2+ b1b2+ c1c2 = 0

Condition for parallelism:

or

Angle b/w line & a plane:

If , , be direction ratios of line & ax + by + cz + d = 0 be eq. of

plane in which normal has d.r's a, b, c. Ѳ is angle b/w line & plane.

cos(900- ) =

Vector form: If is angle b/w line & plane . = d sin

=

Illustration : Find angle b/w line & 3x + 2y - 2z + 3 = 0.Ans: D.r's of line are 2, 3, 2 & d.r's of normal to plane are 3, 2, - 2

sin =

= sin -1

Eq. of plane passing through the Line of Intersection of planesa1x + b1y + c1z + d1= 0 & a2x + b2y + c2z + d2= 0 is

(a1x + b1y + c1z + d1 ) + k(a2x + b2y + c2z + d2) = 0Dumb Question : How this eq. is that of required plane ?

Ans: Let P( , , ) be point on line of intersection of two planes.

So,it lies on both the planes.i.e. a1 + b1 + c1 + d1= 0& a2 + b2 + c2 + d2= 0

So, point P( , , ) should lie on required plane.i.e. (a1 + b1 +

c1 + d1) + k(a2 + b2 + c2 + d2) = 0 P( , , ) lies on. Plane (a1x + b1y + c1z + d1) + k(a2x + b2y + c2z + d2) = 0Vector Form:

. = d1 & . = d2

So, eq. of required plane is ( . - d1) + k( . - d2) = 0Illustration: Find eq. of plane constaining line of intersection of plane x - y + z + 7 = 0 and x + 3y + 2z + 5 = 0 &

passing through (1, 2, 2).Ans: Eq. of plane through line of intersection of given planes is,

(x - y + z + 7) + (x + 3y + 2z + 5) = 0 .............. (i)It passes through (1, 2, 3)

(1 - 2 + 2 + 7) + (1 + 3 x 2 + 2 x 2 + 5) = 0

8 + (7 + 4 + 5) =

Putting = - in eq. (i)We get, (x - y + z + 7) + (- )(x + 3y + 2z + 5) = 02x - 2y + 2z + 14 - x - 3y - 2z - 5 = 0 ⇨ x - 5y + 9 = 0Two sides of plane :If ax + by + cz + d = 0 be a plane then points (x1, y1, z1) & (x2, y2, z2)

are points lies on Same side if > 0 & opposite side if

< 0**Distance of point from a plane:

Length of from a point having P.V. to plane . = d is

given by P =

Proof: PM is length of from P to plane. Since line PM passes

through P( ) & || to vector which is normal to plane. eq. of

line = x .................................... (i)Since point M is intersection of line & plane. So, it lies on line as well as plane

Putting in eq. (i) = +

= P.V. of M - P.V. of P = + -

Dumb Question: How P.V. of M is +

Ans: M lies on line as well as plane. On solving value of for line eq. We get P.V. of M.So, this is P.V. of M

PM = | | =

Cartesian Form : Length of from point P(x1, y1, z1) to plane ax + by

+ cz + d = 0, Then eq. of PM is = r

Dumb Question: How this eq. of PM comes ? Ans: It is passes through point (x1, y1, z1) & || to normal of plane so, we get this eq.]Coordinates of any point on PM are (x1+ ar, y1+ br, z1+ cr)But this also coordinate of M & M also lies on plane

a(x1+ ar) + b(y1+ br) + c(z1+ cr) + d = 0

i.e. r = PM

Distance b/w parallel planes:

Distance b/w || planes is difference of length of from origin to two planes.Let ax + by + cz + d1= 0 & ax + by + cz + d2= 0

D =

Vector Form: . = d1 & . =d2

Illustration : Find distance b/w parallel planesx + 2y + 2z + 2 = 0 & 2x + 4y + 4z + 3 = 0Ans: Distance b/w ax + by + cz + d1= 0 & ax + by + cz + d2 = 0 is

x + 2y + 2z + 2 = 0 & x + 2y + 2z + = 0

Equation of planes Bisecting Angle b/w Two planes: Eq. of planes bisceting angle b/w planes, a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is,

Proof: Let P(x, y, z) be point on plane bisecting angle b/w two planes PL & PM is length of from P to planes.

By property of angle bisector.

PL = PM

Dumb Question: How PL = PM ?Ans: In fig 1. OPL & OPM are congruent by AAA symmetry. So, all sides are equal. So, PL = PM.]

Note: (i) Eq. of bisector of angle b/w two planes containing origin is

(ii) Bisector of acute & obtuse angles b/w:

Let a1x + b1y + c1z + d1 = 0 & a2x + b2y + c2z + d2 = 0 where d1, d2 > 0

(a) If a1a2 + b1b2 + c1c2 > 0, origin lies in obtuse angle bisector & eq. of bisector of acute angle is

(b) If a1a2 + b1b2 + c1c2 < 0, origin lies in atute angle bisector & eq. of acute angle bisector is

Illustration: Find eq. of bisector planes of angle b/w planes 2x + y - 2z + 3 = 0 & 3x + 2y - 6z + 8 = 0 specify obtuse & acute angle bisectors.

Ans: 2x + y - 2z + 3 = 0 & 3x + 2y - 6z + 8 = 0 where d1, d2 > 0 Now a1a2 + b1b2 + c1c2 = 2 x 3 + 1 x 2 + 2 x 6 > 0

............................ (i) (i) is obtuse angle bisector plane

& ............................ (ii)(ii) is acute angle bisector plane.

14x + 7y - 14z + 21 = ± 9x + 6y - 18z + 24 For obtuse angle bisector plane, Take -ve sign5x + y + 4z - 3 = 0

Intersection of line & plane:

= r & plane ax + by + cz + d = 0

Let P be point of intersection coordinate of P(x1 + lr, y1 + mr, z1 + nr) But it satisfy eq. of plane. a(x1 + lr) + b(y1 + mr) + c(z1 + nr)

Condition for line to be || to a plane:

be || to plane ax + by + cz + d = 0 if al + bm + cm = 0 or sin = 0

Condition for a line to lie in the plane:

line lie in plane ax + by + cz + d = 0

if al + bm + cn = 0 & ax1 + by1 + cz1 + d = 0

Dumb Question: How line be || to plane if al + bm + cn = 0 ?

Ans: If is angle b/w line & plane, then

sin = [Previously desired]

line is || to plane if sin = 0 or al + bm + cn = 0

Easy Type Q.1. Find ratio in which 2x + 3y + 5z = 1 divides line joining the points (1, 0, 2) & (1, 2, 5) Ans: Let the ratio be k = 1 at point P

Then, P = must satisfy 2x + 3y + 5z = 1

2(K + 1) + 6K + 25K + 10 = K + 1 K + 1 + 6K + 25K + 10 = 0

32K + 11 = 0 K = Thus line divides externally in rstio of 11:32Q.2. What are direction cosines of a line which is equally inclined to axes ?

Ans: If , , are angles, if line is equally inclined

= = l2 + m2 + n2 cos2 + cos2 + cos2 = 1

3 cos2 = 1 cos = ± = cos = cos

d.c's arw ( , , ) or (- , - , - )

Q.3. Find direction cosines of line which is to lines with d.r (1, -2, -2) & (0, 2, 1)

Ans: Let l, m, n be d.c's line 1 to given line, d.c's are proportional to d.r's

d.c'r of lines are ( , - 2 , - 2 ) & (0, 2µ, µ)

Since line is to given lines

l + (- 2m ) + (- 2n ) = 0 l - 2m - 2n = 0

& 0 + 2mµ + nµ = 0 2m + n = 0

By cross multiplication,

l = 2R, m = - R & n = 2R , l2 + m2 + n2 = 1

4R2 + R2 + 4R2 = 1 9R2 = 1

Q.4. Prove that line x = ay + b, z = cy + d & x = a'y + b', z = c'y + d' are

if aa' + cc' + 1 = 0 Ans: Ist line is x = ay + b, z = cy + d

= y, = y = = ...........(i)

and IInd line ................................. (ii)

These lines are if aa' + |x| + cc' = 0 aa' + cc' + 1 = 0

Dumb Question: For what value of k, lines &

intersect?

Ans: =

and = µ Since these lines have point of intersection in common. then

(2 + 1, 3 - 1, 4 + 1) = (µ + 3, 2µ + k, µ)

or 2 + 1 = µ + 3 ....... (i), 3 - 1 = 2µ + k ....... (ii) & 4 + 1 =

µ ....... (iii) on solving (i) & (iii), we get = - 3/2 & µ = - 5

Substituting in (ii) - - 1 = - 10 + k k =

Q.6. Show that points and are

equidistant from plane .( ) + 9 = 0

Ans: .( ) = - 9

length of from ( )

So, length of is equal.

Q.7. Find the point in which the plane; = ( - ) + m( + + )

+ n( + - ) is cut by line through point 2 + 3 & parallel

to .

Ans: eq. of line through point 2 + 3 & || to

= (2 + 3 ) +

Since line cuts plane For one point,

(2 + 3 ) + = - + m( + + ) + n( + - )

Equating coeff. of , & on both side, we get

m + n = 1, m - n = 4, - m + n =

m = , n = & = - 4

P.V. of required point is: 2 + 3 - 4

Q.8. Show that line of intersection of planes .( ) = 0

& .( ) = 0 is equally inclined to & . Ans: Note: Line of intersection of two planes will be to normals to the planes. Hence it is || to vector

Now,

&

So, line is equally inclined to &

Q.9. Projection of line segment on 3 axes 4, 5 & 13 respectively. Find length & direction cosines of line segment. Ans:

= (x2- x1) + (y2- y1) + (z2- z1)

Projection of on x-axis = . = (x2- x1) = 4

Projection of on y-axis = . = (y2- y1) = 5

Projection of on z-axis = . = (z2- z1) = 13

Length of AB =

d.r's = , &

Q.10. Find locus of mid point of chords of sphere r2- 2 . + k = 0 if

chords being drawn || to vector .

Ans: r2- 2 . + k = 0 is sphere of centre chord AB || .

locus of M is ( - ). = 0 . = . represents a plane.

Dumb Question: Why is same as ?

Ans: Since || . So, if is . So, it will also toMedium TypeQ.1. If a variable plane forms a tetrahedron of constant volume

27k3 with coordinate planes, find locus of centroid of tetrahedron.Ans: sLet variable plane cuts coordinate axes at A(a, 0, 0), B(0, b, 0), C(0, 0, c)

Then, eq. of plane will be

= 1

Let P( , , ) be centroid of terahedron OABC, then,

= , = , =

Dumb Question : How this centroid tetrahedron OABC comes

= , = , = ? Ans: Centroid of tetrahedrold

where (x1, y1, z1), (x2, y2, z2), (x3, y3, z3) & (x4, y4, z4) are coordinate of tetrahedron.]

Volume of tetrahedron = (Area of AOB).OC 27k3= (

ab)c =

27k3=

Required locus of P( , , ) is

Q.2. Find vector eq. of straight line passing through intersection of plane

, , are non coplanar vectors.Ans: At points of intersection of two planes.

Since , , are non coplanar, then - 1+ µ1 - µ2= 0, 1 + 1

- 2 - µ2= 0, µ1- 1 + 2= 0On solving, we get 1 = 0 & µ1 = µ2

= + 1( - ) + µ1( + )

Since 1 = 0

= + µ1( + ) is required eq. of straight line.

Q.3. Prove that three lines from O with direction cosines l1, m1, n1; l2, m2, n2; l3, m3, n3 are coplaner if

l1(m2n3 - n2m3) + m1(n2l3 - l2n3) + n1(l2m3 - l3m2) = 0 Ans: Note: Three given lines are coplanar if they have common

perpendicular. Let d.c's of common be l, m, n

ll1 + mm1 + nn1 = 0 ............................. (i)

ll2 + mm2 + nn2 = 0 ............................. (ii)

ll3 + mm3 + nn3 = 0 ............................. (iii)

Soplving (ii) & (iii) by cross multiplication ......

l = k(m2n3 - n2m3), m = k(n2l3 - n3l2), n = k(l2m3 - l3m2) Substituting in (i), we get

k(m2n3 - n2m3)l1 + k(n2l3 - n3l2)m1 + k(l2m3 - l3m2)n1 = 0 l1(m2n3 - n2m3) + m1(n2l3 - n3l2) + n1(l2m3 - l3m2) = = 0

Q.4. Solve the equation x + y + z =

Ans: x + y + z = .................................. (i)

Taking dot product by x , we get

x .( x ) + y .( x ) + z .( x ) = .( x )

x[ ] = [ ] x =

& z =

x + y + z =

[ ] + [ ] + [ ] = [ ] is required solution.

Q.5. If planes x - cy - bz = 0, cx - y + az = 0 & bx + ay - z = 0 pass through a straight line, the find value of a2 + b2 + c2 + 2abc. Ans: Givem planes are:

x - cy - bz = 0 ............................. (i)

cx - y + az = 0 ............................. (ii)

& bx + ay - z = 0 ............................. (iii)

Eq. of plane passing through line of intersection of plane (i) & (ii) is

(x - cy - bz) + (cx - y + az) = 0

x(1 + c) - (c + ) + z(- b + a ) = 0 ............................. (iv)

If plane (iv) & (iii) are same, then,

and

a - a3 + bc - a2bc = a2bc + ac2 + ab2 + bc

a(2abc + c2 + b2 + a2 - 1) = 0

a2 + b2 + c2 + 2abc = 1

ANOTHER METHODSThe parametric vector equation

Lines can be specified in a variety of ways. One way is described as follows. Select a point P0 on the line l, and a non-zero vector v parallel to the line. The line l is then the unique line passing through P0 and parallel to v.

Now P lies on l if and only if is parallel to v. As

this condition is equivalent to r - r0 = tv for some scalar t. The equation

is called a parametric vector equation of the line l. (It is not unique, as a different point P0 on the line could have been chosen, changing r0, and v can be replaced by any other non-zero vector parallel to l.)

Each value of the parameter t determines a unique point P, with position vector r = r0 + tv, on the line l. As t takes all possible values, P takes all possible positions on the line l.

The parametric scalar equations

Suppose that the point P0 has coordinates x0,y0,z0, and the non-zero vector v has components a,b,c.

Let x,y,z be the coordinates of an arbitrary point P on the line l.

Then

and the position vectors of P0 and P are given by

and

respectively. The parametric vector equation may be rewritten as

that is,

with t R. These equations are called parametric scalar equations of the line.

The Cartesian equations

For each value of the parameter t, there is a point P with coordinates

on the line.

If a,b,c are all non-zero, then we can eliminate the parameter t in the parametric (scalar) equations to obtain

We call

The vector equation of a plane

A plane can be described in many ways. The plane, for example, can be specified by three non-collinear points of the plane: there is a unique plane containing a given set of three non-collinear points in space.

An alternative way to specify a plane is given as follows.

Select a point P0 in the plane. There is a unique line through P0 perpendicular to the plane. This line is called the normal to to the plane at P0. A vector n 0 parallel to this normal is called a normal vector for the plane.

There is a unique plane which passes through P0 and has n as a normal vector.

Now P lies in the plane through P0 perpendicular to n if and only if and n are perpendicular.

As = r - r0, this condition is equivalent to

The Cartesian equation of a plane

Suppose that P0 has coordinates x0,y0,z0 and n has components a,b,c.

Let P, with coordinates x,y,z, be an arbitrary point on the plane.

The position vectors of P0 and P are

and

Substituting

into the vector equation, we obtain

which, when multiplied out, gives

This is called a Cartesian equation of the plane.

It simplifies to

where d is the constant ax0 + by0 + cz0.

An equation of the form

where a,b,c and d are constants and not all a,b,c are zero, can be taken to be an equation of a plane in space.

The coefficients a, b and c are the components of a normal vector

for the plane described by the equation.

The Cartesian equation of a plane

Suppose that P0 has coordinates x0,y0,z0 and n has components a,b,c.

Let P, with coordinates x,y,z, be an arbitrary point on the plane.

The position vectors of P0 and P are

and

Substituting

into the vector equation, we obtain

which, when multiplied out, gives

This is called a Cartesian equation of the plane.

It simplifies to

where d is the constant ax0 + by0 + cz0.

An equation of the form

where a,b,c and d are constants and not all a,b,c are zero, can be taken to be an equation of a plane in space.

The coefficients a, b and c are the components of a normal vector

for the plane described by the equation.

EXTRA QUESTIONS:

Class – XII Subject – Mathematics (Three Dimensional Geometry)

1. Find the d.c’s of X, Y and Z-axis.[example 1,0,0(x-axis)0,1,0(y-axis]

2. The equation of a line is given by (4 – x)/2 = (y + 3)/3 = (z + 2)/6. Write the direction cosines of a line parallel to the above line. [Ans: – 2/7, 3/7, 6/7]

3. The equation of a line is (2x – 5)/4 = (y + 4)/3 = (6 – z)/6. Find the direction cosines of a line parallel to this line. [ Ans : 2/7, 3/7, – 6/7]

4. Find coordinates of the foot of the per. drawn from the origin to the plane 2x – 3y + 4z -6 = 0. [example ans. 12/29, -18/29, 24/29]

5. Find the angle between the line (x + 1)/2 = y/3 = (z-3)/6 and the

plane 10x + 2y – 11z = 3.[example ans. sinφ = | b⃗ . n⃗

|b⃗|∨ n⃗∨¿¿ | = 8/21]

*6. Find the image of the point (1, 6, 3) in the line x = (y – 1)/2 = (z – 2)/3. [ Hint: it will be found that foot of per. from P to the line is N (1,3,5). If Q(α ,β , γ) is the image of P then N is the mid point of PQ

⇨ α+12

=1 , β+62

=3 , γ+32

=5 ⇨ Q(1, 0, 7)]

7. Find the foot of the perpendicular drawn from the point P(1, 6, 3) on the line x/1 = (y – 1)/2 = (z – 2)/3. Also find the distance from P. [ Ans : √13 units., eqn. Of line is (x-1)/0=(y-6)/-3=(z+1)/2]

8. Find the length and the foot of the perpendicular drawn from the point (2, – 1 , 5) to the line (x – 11)/10 = (y + 2)/– 4 = (z + 8)/ –11. [ Ans. = Point (1, 2, 3); distance = √14 units.]

*9. Show that the angles between the diagonals of a cube is cos-

1(1/3).

Ans. (same as example. 26 of misc.) Let the length of cube be a , in fig. Of example : A,B,C are on x-axis, y-axis, z-axis resp.& F,G,D are in yz, xz ,xy –axis , O is origin , d.r’s of OE (E (a,a,a)) & AF are a,a,a & -a,a,a resp. ∴ cosѲ =(-a2+a2+a2)/√3a.√3a = 1/3(angle b/w two lines)

10. Find the length of the perpendicular drawn from the point (2, 3, 7) to the plane 3x – y – z = 7 . Also find the coordinates of the foot of the perpendicular. [ same as Q. 8 ]

*11. Find the point on the line (x + 2)/3 = (y + 1)/2 = (z – 3)/2 at a distance 3√2 from the point (1, 2, 3).

Ans. (x + 2)/3 = (y + 1)/2 = (z – 3)/2 =k , any point on the line A(3k-2, 2k-1, 2k+3) , if its distance from B(1,2,3) is 3√2, then AB² = (3√2)²⇨ K= 0 OR 30/17 , ∴ point A(-2,-1,3) OR A(56/17,43/17,111/17) [by putting value of k in A]

12. Find the equation of the perpendicular drawn from the point (2, 4, – 1) to the line (x + 5)/1 = (y + 3)/4 = (z – 6)/–9 .

[ Ans : (x – 2)/6 = (y – 4)/3 = (z + 1)/2]

*13. Find the equation of the line passing through the point (-1, 3, -2) and perpendicular to the lines x = y/2 = z/3 and (x + 2)/(-3) = (y – 1)/2 = (z + 1)/5.

Ans. eqn. Of line (x+1)/a=(y-3)/b=(z+2)/c ........(i) ,where a,b,c are d.r’s , since (i) is per. to lines ∴ a+2b+3c=0 & -3a+2b+5c=0, after solving , we get a/4=b/-14=c/8 =k(say) , so eqn. Of line (by putting the values of a=4k, b=-14k,c=8k) is (x+1)/4 = (y-3)/-14 = (z+2)/8

14. Find the foot of the per. drawn from the point (0, 2, 3) on the line (x + 3)/5 = (y – 1)/2 = (z + 4)/3 Also, find the length of the perpendicular. [ans..(2,3,-1), use distance formula, length= √21]

*15. Find the equation of the plane passing through the line intersection of the planes x – 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios (1, 2, 1) Also, find the perpendicular distance of the point P(3, 1, 2) from this plane.

Ans. equation of the plane passing through the line intersection of the planes x – 2y + z -1 +k(2x + y + z – 8)=0 or (1+2k)x+(-2+k)y+(1+k)z-1-8k=0 ........(i) is parallel to line ∴ (1+2k).1+(-2+k).2+(1+k).1=0 ⇨ k= 2/5(normal to the plane must be per. to the line), put k in (i) ⇨ 9x-8y+7z-21=0, per. dis. From P Is 22/√(194)

*19. Find the shortest distance between the following lines : (x – 3)/1 = (y – 5)/–2 = (z – 7)/1 and (x + 1)/7 = (y + 1)/–6 = (z + 1)/1.

[ncert Ans : 2√29 , use formula ]

20. Find the equation of the plane passing through the points (1, 2, 3) and (0, – 1, 0) and parallel to the line (x – 1)/2 = (y + 2)/3 = z/–3. [Ans :same as Q. 26 , 6x – 3y + z = 3]

21. Find the shortest distance between the following pairs of lines whose cartesian equations are : (x -1)/2 = (y + 1)/3 = z and (x + 1)/3 = (y – 2) , z = 2. [ SAME as Q. 19 ]

*22. Find the distance of the point P(2, 3, 4) from the plane 3x + 2y +2 z + 5=0 measured parallel to the line (x+3)/3 = (y-2)/6 = z/2

Ans. Any point on line (x+3)/3 = (y-2)/6 = z/2=k is Q(3k+2,6k+3,2k+4) It lies on plane ⇨ k=-1 ∴ point Q( -1,-3,2),PQ=7

23. Find the equation of the plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4). [ Ans : 5x – 7y + 11z + 4 = 0]

*24. Find the equation of the plane passing through the point (–1, –1, 2) and perpendicular to each of the following planes : 2x + 3y – 3z = 2 and 5x – 4y + z = 6.

[ Ans : 9x + 17y + 23z – 20 = 0, same as example of misc. Eqn. Of plane a(x+1)+b(y+1)+c(z-2)=0........(i) By condition of perpendicularity to the plane (i) with the planes 2a+3b-3c=0 & 5a-4b+c =0 , after solving , we get a=9c/23 & b=17c/23, put in (i) ]

25. Find the equation of the plane passing through the point (1, 1, -1) and perpendicular to the planes x + 2y + 3z – 7=0 and 2x – 3y + 4z = 0. [ SAME as Q. 24]

26. Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line (x + 3)/2 = (y – 3)/7 = (z – 2)/5.

Ans : 8x – 13y + 15z + 13 = 0, eqn. Of plane a(x-3)+b(y-4)+c(z-1)=0.......(i), It passes through (0,1,0) ,then -3a-3b-c=0 // to line ∴ 2a+7b+5c=0, after solving a=8c/15, b=-13c/15, put in (i)

*27. Find the image of the point P(1,2,3) in the plane x+2y+4Z=38

[ Ans :Let M is foot of per. (mid point of PQ) , if Q (α ,β , γ ) , d.r’s of a normal to the plane are 1,2,4, eqn. Of line PM is (x-1)/1=(y-2)/2=(z-3)/4 = k ⇨ any point on a line Q(1+K,2+2K,3+4K) ∴ M{(2+k)/2,(4+2k)/2,(6+4k)/2} put in plane ⇨ k=2 ∴ Q(3,6,11) ]

28. Find the co-ordinates of the image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0. [Ans : (–3, 5, 2)]

*29. Find the distance between the point P(6, 5, 9) and the plane determined by the points A(3, – 1 , 2), B(5, 2, 4) and C(– 1, – 1 , 6).

Ans. [example28 of misc. Ans: 6/√(34),we can find eqn. Of plane through A,B,C and find distance of P from plane or can be found as PD = A⃗P . A⃗B ×⃗ AC ( proj. Of A⃗P on A⃗B ×⃗ AC, D is foot of per. from P) or find eqn. of plane passes through A,B,C , then find dist. ]

30. Find the co-ordinates of the point where the line (x + 1)/2 = (y + 2)/3 = (z + 3)/4 meets the plane x + y + 4z = 6. [Ans: P(1, 1, 1)]

*31. Find the distance of the point A(– 2, 3 – 4) from the line (x + 2)/3 = (2y + 3)/4 = (3z + 4)/5 measured parallel to the plane 4x + 12y – 3z + 1 = 0.

[Ans: |AP|= 17/2, eqn. Can be written as (x + 2)/3 = (y + 3/2)/2 = (z + 4/3)/5/3 = t ⇨ P{ -2+3t, (-3/2)+2t, (-4/3)+5t/3} be any point on line D.R’S of AP {3t, 2t-(9/2), (5t/3)+(8/3)} , since AP is parallel to plane 4.3t+ (12).[ 2t-(9/2)]+ (-3). (5t/3)+(8/3) = 0 ⇨ t=2 ∴ P (4,5/2,2) ]

Ques. 32 Find the distance of the point P(-2, 3, -4) from the line (x+2)/3 = (2y+3)/4 = (3z+4)/6 measured // to the plane 4x + 12y -3 z + 1=0.

Ans. same as Q. 22, point on lineQ(3k-2,2k-(3/2),(5k/3)-(4/3)), d.r’s of PQ are (3k,2k-(9/2),(5k/3)+(8/3)) , since PQ is // to plane ∴ PQ is

per. to normal to the plane ⇨ (3k).4+{2k-(9/2)}.12+{(5k/3)+(8/3)}.(-3)=0 ∴ k= 2 ⇨ Q(4,5/2,2), PQ = 17/2

Question 33 find whether the lines r⃗ = (i-j-k) +λ (2i+j) and r⃗ = (2i-j) +μ (i+j-k) intersect or not . if intersecting, find their point of intersection.

Ans. a⃗2 -a⃗1 = I +k , b⃗1× b⃗2 = i j k2 1 01 1 −1

= -i+2j+k , its magnitude is √6

Shortest distance = = 0/√6 = 0

Hence lines intersect. Point of intersection is given by

(i-j-k) +λ (2i+j) = (2i-j) +μ (i+j-k) ⇨ (1+2λ) = 2+μ , -1+λ - -1+μ and -1 = -μ ⇨ λ = 1 and μ =1 satisfy μ = 2λ -1 , put the values of λ and μ in given lines , we obtain the position vector of point of intersection of the two given lines as 3i- k i.e., the point of intersection is (3,o,-1).

Ques. 34 Show that the four points (0,-1,-1), (4,5,1), (3,9,4) and (-4,4,4) are coplanar. Also find the eqn. of the plane containing them.

Ans. Eqn. of plane passes through (0,-1,-1) is a(x-0)+b(y+1)+(z+1)=0

It passes through (4,5,1), (3,9,4), we get 4a+6b+2c=0, 3a+10b+5c=0⇨ eqn. of plane is 5x-7y+11z+4=0 and (-4,4,4) will satisff the eqn. of plane to be coplanar.

Q. 35 Show that lines (x+3)/-3 = (y-1)/1 = (z-5)/5 & (x+1)/-1=(y-2)/2 =(z-5)/5 are coplanar . find the eqn. of plane.

Ans.(as example 21), The given lines are coplanar if

(x 2−x1) ( y 2− y 1) (z 2−z1)l1 m1 n1l 2 m2 n2

= (−1+3) (2−1) 0

−3 1 5−1 2 5

= 0 , so lines are coplanar. Eqn of plane

= (x−x 1) ( y− y1) (z−z1)

l 1 m1 n1l 2 m2 n2

=(x+3) ( y−1) (z−5)−3 1 5−1 2 5

= x-2y+z=0.

Q.36 Find the shortest distance and eqn. Of shortest distance b/w the lines (x-6)/3 = (y-7)/-1 = (z-4)/1 & x/-3 = (y+9)/2 = (z-2)/4 .

S (3p+6,-p+7,p+4) ......(i)

D (-3q,2q-9,4q+2) .......(ii)

Ans. let SD be the shortest dist. , then co-ordinates of eqn.(i) are ( 3p+6, -p+7, p+4) & eqn.(ii) are (-3q, 2q-9, 4q+2)

d.c’s of SD are proportional to -3q-3p-6, 2q+p-16, 4q-p-2 and if SD is at right angles to both lines , we get 3.( -3q-3p-6)+(-1).( 2q+p-16)+ 1. (4q-p-2)=0 and -3.(-3q-3p-6)+2.( 2q+p-16)+4.( 4q-p-2) =0 ⇨ 7q+11p+4=0 & 29q+7p-22=0 ⇨ p=-1 , q=1 and SD=3√(30) , eqn. Will be (x-3)/-6 = (y-8)/-15 = (z-3)/3

Q.37 Find the shortest distance and eqn. b/w the lines r⃗ = (8+3λ)i-(9+16λ)j+(10+7λ)k & r⃗ = (15i+29j+5k)+μ(3i+8j-5k).

Ans. let SD be the shortest dist. b/w two lines , let position vector of S be (8i-9j+10k)+λ(3i-16j+7k) and position vector of D is (15i+29j+5k+μ(3i+8j-5k), S⃗D = D⃗− S⃗ = (3μ-3λ+7)i+(8μ+16λ+38)j+(-7λ-5μ-5)k since SD is perpendicular on line (i) & (ii) then

[(3μ-3λ+7)i+(8μ+16λ+38)j+(-7λ-5μ-5)k]. (3i-16j+7k)=0 ⇨ 157λ+77μ+311=0...(iii) & 77λ+49μ+175=0...(iv), by solving λ = -1,μ=-2 ∴ S⃗D = D⃗− S⃗ = 4i+6j+12k ⇨| S⃗D |= 14 & eqn. Is r⃗ = (5i+7j+3k)+t[(9i+13j+15k)-(5i+7j+3k)] = (5i+7j+3k)+t(4i+6j+12k) where S⃗ =5i+7j+3k, D⃗=9i+13j+15k & user⃗ =a⃗ +t(b⃗−a⃗).

EX.11.2(ncert)

QUESTION 5 Find the equation of the line in vector and in Cartesian form that passes through the point with position vector and is in the direction .Ans.: Eqn of line passing through a point =a⃗ , parallel to b⃗= is r⃗ =a⃗+ λ b⃗ (vector form)

Cartesian form of line is xi+yj+zk = ( ) +λ ( ) =

(2+λ)i+(-1+2λ)j+(4-λ)k ⇨ λ = (x-2)/1 =(y+1)/2 = (z-4)/-1

Ques. 7 The Cartesian equation of a line is . Write its vector form

Ans. vector form is r⃗ =a⃗+ λ b⃗ = (5i-4j+6k) +λ (3i+7j+2k)

Ques.9 Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

Ans. vector form of a line is r⃗ =a⃗+ λ b⃗ =(3i-2j-5k)+λ( 11k)

¿ =(3-3)i+(-2+2)j+(6+5)k)

Cartesian form of a line is (x-x1)/a = (y-y1)/b = (z-z1)/c

(x-3)/0 = (y+2)/0 = (z+5)/11

Question12 Find the values of p so the line and

are at right angles.

Ans: write above lines in standard form , the d,r’s of given lines are -3, 2p/7, 2 & -3p/7, 1, -5. are perpendicular to each other, hence their dot product should be equal to 0 ⇨ p=70/11.

Ques. 13 Show that the lines and are perpendicular to each other.

Ans.: diff. Method , let b⃗1, b⃗2 be two vwctors parallel to given lines

b⃗1= 7i-5j+k, b⃗2 = i+2j+3k⇨ cosφ =( b⃗1. b⃗2)/¿⃗b1∨¿¿.| b⃗2| = п/2

Ques.14 Find the shortest distance between the lines

Ans.: r⃗ = a⃗1 +λ b⃗1 & r⃗ = a⃗2 +μ b⃗2 , shortest distant (d) =

Nr. = (i-3j-2k).(-3i+3k) = -9 & Dr. = 3√2, so d= 3√2/2

Ques.16 Find the shortest distance between the lines whose vector equations are

ANS.: Same as above Nr. = (3i+3j+3k).(-9i+3j+9k) = 9, Dr. = 3√19.

EX 11.3 Ques.4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

(a) (b)

(c) (d) [ co-ordinates of foot of per.(ld, md, nd)]

Ans. (a) let foot of per.P (x1,y1,z1) & eqn. Of plane in normal form

Lx+my+nz=d ⇨2x/√29 + 3y/√29 + 4z/√29 = 12/√29, so, 2/√29 ,3/√29 , 4/√29 are d.c’s of OP (O is orgin, P is foot of per.) & D.R’S of

OP is (x1,y1,z1) , they are proportional to each other (d.c’s & d.r’s of

a line) ⇨ x12

√29 =

y13

√29 =

z14

√29 = k , put the values of x1,y1,z1 in plane ,

we get k = 12/√29 ⇨foot of per. Is ( 24/29, 36/29, 48/29)

(b) eqn. Of plane is 3y/5+ 4z/5 = 6/5, d.c’s of OP are 0, 3/5, 4/5 &

d.r’s of OP are x1,y1,z1 , they are prop. ⇨ x10

= y135

= z145

= k , we get k=

6/5 , foot of per. (0, 18/25, 24/25)

(c) k= 1/√3 , foot of per. (1/3, 1/3, 1/3) (d) k= -8/5 , foot of per. (0,-8/5,0)

Ques.5 Find the vector and Cartesian equation of the planes

(a) that passes through the point (1, 0, −2) and the normal to the plane is .

(b) that passes through the point (1, 4, 6) and the normal vector to the plane is .

Ans. (a) eqn. Of plane (vector ) (r⃗ - a⃗ ) .n⃗ =0 ⇨[r⃗ - (i-2k)].(i+j-k) =0⇨ r⃗ .( i+j-k) – (i-2k). (i+j-k)=0 ⇨ r⃗ . (i+j-k)-3 = 0

Cartesian form x+y-z=3 [ eqn. Of plane 1(x-1)+1(y-0)-1(z+2)=0

(b) ) eqn. Of plane (vector ) (r⃗ - a⃗ ) .n⃗ =0 ⇨[r⃗ - (i+4j+6k)].(i-2j+k) =0⇨ r⃗ .( i-2j+k) – (i+4j+6k. (i-2j+k) =0 ⇨ r⃗ . (i-2j+k)+1 = 0

Cartesian form is x-2y+z+1-0 [ eqn. Of plane 1(x-1)-2(y-4)+1(z-6)=0]

Ques.6 Find the equations of the planes that passes through three points.

(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3)

(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)

Ans. let the eqn. Of plane passes through (1,1,-1) be

a(x-1)+b(y-1)+c(z+1)=0......(i) , it passes through (6, 4, −5), (−4, −2, 3)⇨ 5a+3b-4c=0....(ii) & 5a+3b-4c=0....(iii) , from (ii) & (iii) ⇨no unique plane can be drawn.

(b) let the eqn. Of plane passes through (1,1,0) be

a(x-1)+b(y-1)+cz=0......(i) , it passes through (1,2,1), (-2,2,-1)⇨ b+c=0....(ii) & 3a-b+c=0....(iii) , from (ii) & (iii) a = -2c/3, put the

Values of a & b in terms of c in (i) ⇨ 2x+3y-3z=5.

Ques. 10 Find the vector eqn. Of the plane through the line of intersection of the planes r⃗ . (2i+2j-3k) = 7 & r⃗ . (2i+5j+3k) = 9 and through the plane (2, 1, 3).

Ans. Eqn. Of the plane through the line of intersection of the planes r⃗ . (2i+2j-3k) - 7 +λ[ r⃗ . (2i+5j+3k) – 9]=0⇨ r⃗ . [(2+2λ)i+(2+5λ)j+(-3+3λ)k]-7-9λ = 0 .........(i) It passes (2,1,3)⇨ (2i+j+3k). [(2+2λ)i+(2+5λ)j+(-3+3λ)k]-7-9λ = 0 ⇨ λ =10/9, put in (i)

We get, r⃗ . (38i+68j+3k) = 153.

Ques.11 Find the equation of the plane through the line of intersection of the planes and which is perpendicular to the plane

Ans. Eqn. of the plane through the line of intersection of the planes x+y+z-1+λ(2x+3y+4z-5) = 0 or (1+2λ)x+(1+3λ)y+(1+4λ)z-1-5λ=0.....(i)

D.R’s of normal of the plane (i) are (1+2λ), (1+3λ), (1+4λ)

Eqn. (i) is per. to , its D.R’s of normal are 1, -1, 1

⇨ 1. (1+2λ)+ (-1).(1+3λ)+ 1. (1+4λ) =0 ⇨ λ =-1/3 , put in (i) ⇨x-z+2=0

Ques.13 In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

(a)

(b)

(c)

(d)

(e)

Ans. (a) d.r’s of the normal of the planes are 7,5,6 & 3, -1,-10

We shall check a1.a2+b1.b2+c1.c2 =0 & a1/a2 = b1/b2 = c1/c2 for per. & parallel , neither per. nor parallel, now find angle b/w them

cosѲ = ⇨ 7.3+5. (−1 )+6.(−10)√110 .√110 = -2/5

(b) per. planes [(c)&(d)] parallel planes (e) cosѲ = 1/√2 ⇨Ѳ=п/4

Ques.14 In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(a) (0, 0, 0)

(b) (3, −2, 1)

(c) (2, 3, −5)

(d) (−6, 0, 0)

Ans. Use distance of point from plane

(a) per. distance =¿ 3.0−4.0+12.0−3√3²+ (−4 ) ²+(12) ² | = 3/13

(b) 13/3, (c) 3, (d) 2.

MISCELLANOUS **Ques.2 If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m1n2

− m2n1, n1l2 − n2l1, l1m2 − l2m1.

Ans Let l, m, n be the d.c’s of the line per. to each one of the given lines. Then ll1 +mm1+ nn1 = 0 & ll2 +mm2+ nn2 = 0 , after solving these two results, we get (by cross mult.)

l

m1n2−m2 . n1 =

mn1l2−n2 . l1

= n

l1m2−l2 .m1 =

√l ²+m ²+n ²

√∑ (m1 . n2−m2 . n1 ) ² = 1sinθ

Where Ѳ is the angle b/w the given lines. But Ѳ= п/2(given)∴ sinѲ=1 ⇨ l = m1n2 − m2n1, m = n1l2 − n2l1, n = l1m2 − l2m1.

Ques.7 Find the vector equation of the plane passing through (1, 2,

3) and perpendicular to the plane

Ans. d.r’s of the normal of plane are 1, 2, -5.

Eqn. Of line passes through (1,2,3) and having d.r’s 1,2,-5 is r⃗ = (i+2j+3k)+λ (i+2j-5k) [ ∵ eqn. Of line passing through a⃗ with d.r’s of b⃗ is r⃗ = a⃗ + λ b⃗ ]

Ques.10 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane

Ans. Eqn. Of line passes through (5, 1, 6) and (3, 4, 1)

(x-5)/2 = (y-1)/-3 = (z-6)/5 = λ , point on the line is (5+2λ, 1-3λ, 6+5λ)

Crosses the YZ-plane [ eqn. Of YZ-plane is X = 0] ⇨ λ = -5/2, Required point is (0,17/2,-13/2)

Ques.12 Find the coordinates of the point where the line through (3, −4, −5) and (2, − 3, 1) crosses the plane 2x + y + z = 7.

Ans. Eqn. Of line passes through (3, −4, −5) and (2, − 3, 1)

(x-3)/-1 = (y+4)/1 = (z+5)/6 = λ, point on the line is (3-λ,-4+λ,-5+6λ)

Lies on 2x + y + z = 7 ⇨ λ = 2 ,required point is (1,-2,7)

Ques.15 Find the equation of the plane passing through the line of

intersection of the planes and and parallel to x-axis.

Ans. Equation of the plane passing through the line of intersection of the planes r⃗(2i+3j-k)+4+ λ[ r⃗ . (i+j+k)-1] = 0

r⃗ .[(2+λ)i+(3+λ)j+(-1+λ)k]+4-λ = 0.......(i), X-axis is parallel to plane (i) ⇨ its normal is per. to x-axis , d.r’s of x-axis are 1,0,0 & d.r’s of normal are 2+λ, 3+λ, -1+λ and they are per. ⇨ λ = -2, put in (i) ,we get r⃗. (j-3k)+6 =0Ques. 16 If O be orgin and the coordinates of P be (1,2,-3), then find the eqn. Of the plane passing through P and per. to P.

Ans. Eqn. Of plane passes through (x1,y1,z1) is

A(x-x1)+b(y-y1)+c(z-z1) = 0, where a,b,c are d.r’s of normal

Eqn. Of plane is 1.(x-1)+2(y-2)-3(z+3) = 0⇨ x+2y-3z-14=0

Ques.17 Find the equation of the plane which contains the line of

intersection of the planes , and

which is perpendicular to the plane .

Ans. equation of the plane which contains the line of intersection of

the planes +λ( )......(i)

r⃗ .[(1+2λ)i+(2+λ)j+(3-λ)k]-4+5λ=0 is per. to

(1+2λ).5+(2+λ).3+(3-λ).(-6) =0 ⇨ λ = 7/19 , put in (i), we get

r⃗ . (33i+45j+50k) – 41 = 0

Ques.18 Find the distance of the point (−1, −5, −10) from the point

of intersection of the line and the plane

.

Ans. put the value of r⃗ from first given eqn. in second eqn. , we get

[2i-j+2k+λ(3i+4j+2k)].(i-j+k) = 5 ⇨ λ =0, then point of intersection of plane & line is 2i-j+2k (2,-1,2) & distance b/w two points is 13

Ques.20 Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines:

Ans. The vector equation of the line passing through the point (1, 2, − 4) is r⃗ = i+2j-4k+λ(b1i+b2j+b3k) or (x-1)/b1=(y-2)/b2=(z+4)/b3 ........(i) Where b1, b2, b3 are d.r’s of above line which is per. to given two lines whose d.r’s are 3, -16, 7 & 3, 8, -5 ⇨ 3b1+(-16) b2+7 b3 = 0..........(ii)

& 3b1+8 b2+(-5) b3 =0.......(iii) (by condition of perpendicularity)

By solving (ii) & (iii) ,we get b1/2 = b2/3 = b3/6 = p, put in (i)

Ques.21 Prove that if a plane has the intercepts a, b, c and is at a

distance of P units from the origin, then

Ans. use formula , we get

P = 1

√ 1a ² + 1b ² + 1c ² ⇨ 1/ P² = 1a ²

+ 1b ²

+ 1c ²

(For class xi)QUES. 1:.Show that the points (-2, 3 , 5 ) , ( 1, 2 , 3 ) and ( 7, 0 , -1 ) are collinear.QUESTION 2: Find the equation of the set of points which are equidistant from the points (1, 2 , 3) and ( 3 , 2 , -1)

3- DIM. GEOMETRY FOR XI class (ncert)WITH HINTS

Ex12.2 Question 4: Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).Ans.Let P(x,y,z) be a point which is equidistant ∴ PA =PB ⇨ PA² = PB² ⇨ x – 2z = 0 (by distance formula P1P2 = √(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.)Ex 12.3 Question 4: Using section formula, show that the points A

(2, –3, 4), B (–1, 2, 1) and are collinear.Ans. Integral division: If C(x, y, z) is point dividing join of A(x1, y1, z1) & B(x2, y2, z3) in ratio of k:1.

Then,   x = , y = , z =

, we get (−k+2k+1 ,

2k−3k+1 ,

k+4k+1 ) = (0, 1/3, 2) ⇨ k=2

Question 5: Find the coordinates of the points which trisect the line segment joining the points P (4, 2, –6) and Q (10, –16, 6).

Ans. Let A, B be two points which trisect PQ ∴ PA = AB = BQ A divides PQ in the ratio 1:2 , ∴ A ( 6, -4, -2) ( by using section formula) , B divides PQ in the raio 2:1, ∴ B(8, -10, 2)Misc.Question 1: Three vertices of a parallelogram ABCD are A (3, –1, 2), B (1, 2, –4) andC (–1, 1, 2). Find the coordinates of the fourth vertex.Ans. let D(x,y,z) be the coordinates of the fourth vertex.by property of //gm. We know that diagonals of //gm. Are bisect each other , so mid point of AC & BD are same ⇨ mid point of AC is (1, 0, 2)= mid point of BD is ((x+1)/2, (y+1)/2, (z-4)/2) ⇨ x=1, y=-2, z= 4Question 2: Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and C(6, 0, 0).Ans. Take AD, BE& CF are medians (D,E,F are mid points on sides opposite to A,B,C resp.)by distance formula ⇨ AD=7,BE=√34& CF=7Question 4: If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find the values of a, b and c.Ans. co-ordinates of centriod= {(x1+x2+x3)/3, (y1+y2+y3)/3,(z1+z2+z3)/3 } ⇨ a=-2, b= -16/3, c= 2Question 5: Find the coordinates of a point on y-axis which are at a distance of from the point P (3, –2, 5).Ans. the coordinates of a point on y-axis is A(0,y,0) by distance formula ⇨ PA = ⇨ Y = -6, 2 Question 6: A point R with x-coordinate 4 lies on the line segment joining the pointsP (2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R.[Hint suppose R divides PQ in the ratio k: 1. The

coordinates of the point R are given by⇨ (8K+2)/(K+1) = 4 ⇨ K = ½ , SO R divides PQ in the ratio 1:2 , put the value of k , we get coordinates of R ( 4, -2, 6)Question 7: If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant.

Ans. Let P(x,y,z), use distance formula (x-3)2+(y-4)2+(z-5)2+(x+1)2+(y-3)2+(z+7)2= k2 ⇨ 2x2+2y2+2z2-4x-14y+4z = k2 -109