3D Geometry - 5

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 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.]

    

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 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(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


Sphere:

Equation of sphere whose centre is & radius is a:


In OPC,  
    
| - | = CP = a (radius)
| - |² = a²   ( - ).( - ) = A²
R² - 2. + c² = a²   r² - 2. + (c² - a²) = 0


Cartesian Form:


From Fig.   CP = R
                CP² = R²
By distance formula.
    (x - a)² + (y - b)² + (z - c)² = R²


GEneral eq. of sphere:

x² + y² + z² + 2ux + 2vy + 2wz + d = 0 is a sphere with centre (- u, - v, - w) &
radius =


Illustration: Find centre & radius of sphere 2x² + 2y² + 2z² - 2x - 4y + 2z + 3 = 0

Ans: 2x² + 2y² + 2z² - 2x - 4y + 2z + 3 = 0
or    x² + y² + z² - 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:

Vector Form: Suppose & be P.V of extremities A & B of diameter AB of sphere. Let be P.V. of any point P on sphere. Then,
     = -   & B  = -


Since, diameter of sphere subtends a right at any point of sphere.
  APB =   . = 0
( - ).( - ) = 0
||² - ( + ). + . = 0
or   AP² + BP² = AB²

| - |2 + | - | = | - |2

Cartesian Form: If A(x₁, y₁, z₁) & B(x₂, y₂, z₂) are extremities of diameter then, eq. of sphere is   (x - x₁)(x - x₂) + (y - y₁)(y - y₂) + (z - z₁)(z - z₂) = 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:

Vector Form: Plane . = d touches sphere | - | = R
if    = R

Dumb Question: Why = R ?

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


Cartesian Form: Plane ln + my + nz = p touches (ul + vm + wn + p)² = (l² + m² + n²)(u² + v² + w² -d)


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

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