# 3D Geometry - 5

__ Equation of planes Bisecting Angle b/w Two planes__:

Eq. of planes bisceting angle b/w planes, a

_{1}x + b

_{1}y + c

_{1}z + d

_{1}= 0 and a

_{2}x + b

_{2}y + c

_{2}z + d

_{2}= 0 is,

Proof: Let P(x, y, z) be point on plane bisecting angle b/w two planes 7 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.]

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

**Note**(ii)

__:__

**Bisector of acute & obtuse angles b/w**Let a

_{1}x + b

_{1}y + c

_{1}z + d

_{1}= 0

& a

_{2}x + b

_{2}y + c

_{2}z + d

_{2}= 0 where d

_{1}, d

_{2}> 0

(a) If a

_{1}a

_{2}+ b

_{1}b

_{2}+ c

_{1}c

_{2}> 0, origin lies in obtuse angle bisector & eq. of bisector of acute angle is

(b) If a

_{1}a

_{2}+ b

_{1}b

_{2}+ c

_{1}c

_{2}< 0, origin lies in atute angle bisector & eq. of acute angle bisector is

__: 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.__

**Illustration**Ans: 2x + y - 2z + 3 = 0 & 3x + 2y - 6z + 8 = 0 where d

_{1}, d

_{2}> 0

Now a

_{1}a

_{2}+ b

_{1}b

_{2}+ c

_{1}c

_{2}= 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 sign

5x + y + 4z - 3 = 0

__:__

**Intersection of line & plane**= r & plane ax + by + cz + d = 0

Let P be point of intersection coordinate of P(x

_{1}+ lr, y

_{1}+ mr, z

_{1}+ nr)

But it satisfy eq. of plane.

a(x

_{1}+ lr) + b(y

_{1}+ mr) + c(z

_{1}+ 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 & ax

_{1}+ by

_{1}+ cz

_{1}+ d = 0

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

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

sin = [Previously desired]

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

__:__

**Sphere**__:__

**Equation of sphere whose centre is & radius is a**In OPC,

| - | = CP = a (radius)

| - |

^{2}= a

^{2}( - ).( - ) = A

^{2}

R

^{2}- 2. + c

^{2}= a

^{2}r

^{2}- 2. + (c

^{2}- a

^{2}) = 0

__:__

**Cartesian Form**From Fig. CP = R

CP

^{2}= R

^{2}

By distance formula.

(x - a)

^{2}+ (y - b)

^{2}+ (z - c)

^{2}= R

^{2}

__:__

**GEneral eq. of sphere**x

^{2}+ y

^{2}+ z

^{2}+ 2ux + 2vy + 2wz + d = 0 is a sphere with centre (- u, - v, - w) &

radius =

__: Find centre & radius of sphere 2x__

**Illustration**^{2}+ 2y

^{2}+ 2z

^{2}- 2x - 4y + 2z + 3 = 0

Ans: 2x

^{2}+ 2y

^{2}+ 2z

^{2}- 2x - 4y + 2z + 3 = 0

or x

^{2}+ y

^{2}+ z

^{2}- x - 2y + z + = 0

centre is (- coeff. of x, - coeff. of y, - coeff. of z)

centre (, 1, - )

radius =

Thus sphere is point circle.

__:__

**Diameter form of Eq. of sphere**__: Suppose & be P.V of extremities A & B of diameter AB of sphere. Let be P.V. of any point P on sphere. Then,__

**Vector Form**= - & B = -

Since, diameter of sphere subtends a right at any point of sphere.

APB = . = 0

( - ).( - ) = 0

||

^{2}- ( + ). + . = 0

or AP

^{2}+ BP

^{2}= AB

^{2}

| - |^{2} + | - | = | - |^{2} |

__: If A(x__

**Cartesian Form**_{1}, y

_{1}, z

_{1}) & B(x

_{2}, y

_{2}, z

_{2}) are extremities of diameter then, eq. of sphere is (x - x

_{1})(x - x

_{2}) + (y - y

_{1})(y - y

_{2}) + (z - z

_{1})(z - z

_{2}) = 0

__:__

**Section of sphere by plane**If sphere is intersection by a plane. We get circle by intersection.

PM is radius of circle

PM =

__:__

**Condition of Tangency of plane to a sphere**__: Plane . = d touches sphere | - | = R__

**Vector Form**if = R

__: Why = R ?__

**Dumb Question**Ans: A plane touches to sphere if distaqnce from centre of sphere is equal to radius.

__: Plane ln + my + nz = p touches (ul + vm + wn + p)__

**Cartesian Form**^{2}= (l

^{2}+ m

^{2}+ n

^{2})(u

^{2}+ v

^{2}+ w

^{2}-d)

__: Find eq. of sphere whose centre has P.V. & which touches plane . = 10__

**Illustration**Ans: Center =

It touches plane . = 10

eq. of sphere | - | =

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 11:32