3D Geometry - 4

Dumb Question: How this   lx + my + nz = P is eq. of plane ?

Ans:   Since   . = P = d is eq. of plane
            &  
         
      lx + my + nz = P


Illustration: Find vector eq. of plane which is at a distance of 8 units from origin & which is normal to vector   .

Ans:   d = 8 & =
        
eq. of plane is,   . = d
        


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 plane
Then,   = - = -
Sincen 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 points
P(1, 1, 1), Q(3, -1, 2), R(-3, 5, -4).

Ans: Let eq. of plane is ax + by + cz + d = 0
It 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. Q 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 planes
a₁x + b₁y + c₁z + d₁ = 0   &   a₂x + b₂y + c₂z + d₂ = 0 is
(a₁x + b₁y + c₁z + d₁) + k(a₂x + b₂y + c₂z + d₂) = 0


Dumb 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) = 0

Vector Form:

     . = d₁   &   . = d₂
So, eq. of required plane is (. - d₁) + k(. - d₂) = 0


Illustration: 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) = 0
     2x - 2y + 2z + 14 - x - 3y - 2z - 5 = 0
     x - 5y + 9 = 0


Two 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 plne:


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   &   . = ₂
                    


Illustion: Find distance b/w parallel planes
x + 2y + 2z + 2 = 0   &   2x + 4y + 4z + 3 = 0

Ans: Distance b/w ax + by + cz + d₁ = 0 & ax + by + cz + d₂ = 0 is
      
       x + 2y + 2z + 2 = 0   &   x + 2y + 2z + = 0