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
= =
l^{2} + m^{2} + n^{2} cos^{2} + cos^{2} + cos^{2} = 1
3 cos^{2} = 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
l^{2} + m^{2} + n^{2} = 1
4R^{2} + R^{2} + 4R^{2} = 1 9R^{2} = 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:
= (x_{2} - x_{1}) + (y_{2} - y_{1}) + (z_{2} - z_{1})
Projection of on x-axis
= . = (x_{2} - x_{1}) = 4
Projection of on y-axis
= . = (y_{2} - y_{1}) = 5
Projection of on z-axis
= . = (z_{2} - z_{1}) = 13
Length of AB =
d.r’s = , &
Q.10. Find locus of mid point of chords of sphere r^{2} - 2. + k = 0 if chords being drawn || to vector .
Ans: r^{2} - 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 27k^{3} 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 (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}), (x_{3}, y_{3}, z_{3}) & (x_{4}, y_{4}, z_{4}) are coordinate of tetrahedron.]
Volume of tetrahedron = (Area of AOB).OC
27k^{3} = (ab)c =
27k^{3} =
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} + µ_{1} - µ_{2} = 0, 1 + _{1} - _{2} - µ_{2} = 0, µ_{1} - 1 + _{2} = 0