3D Geometry - 6

Q.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: I^st line is   x = ay + b, z = cy + d
     = y, = y   = = ………………………….. (i)
and II^nd 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 to .




Q.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 = 0