3D Geometry - 3

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


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 palne.


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.