3D Geometry - 6
Introduction
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: Ist line is x = ay + b, z = cy + d
= y, 
= y = 
= ................................ (i) and IInd 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:
.() = - 9length 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 . Medium Type
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
= 1Let 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).OC27k3 =
(
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