3D Geometry - 3
Introduction
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.

Since L lies on line AB
P.V. of L = P.V. of line AB
=



dr's of

Since









Reflection of point in straight line:
Castesian Form:

From above, we get coordinate of L(foot of

But L is mid point of PQ

















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




DR's of PL are
= (3



= (3



DR's of AB are (3, 2, 4)
Since PL

3(3



9



29



Since L is mid point of PQ
So,


















where


Skew lines: Those lines which do not lies in same plane.
Shortest distence b/w two skew lines:

The line which is

Vector Form:
Let l1 & l2 are:


Since








Let












Dumb Question: How PQ = Projection


Ans: Form fig., it is clear that PQ is projection of



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


& BM is shortest distance b/w l1 & l2
sin






|






= |






Dumb Question: Why we have taken sin(


Ans:

Since direction of vector







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

Proof: General eq. of plane is ax + by + cz + d = 0 ....................................... (i)
It is passes through (x1, y1, z1)

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

A(a, 0, 0), B(0, b, 0), c(0, 0, c)





x =



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.





Dumb Question: What is normal to vector ?
Ans: Plane normal to vector means the every line in plane is

Proof: Let plane passes through A(




Since

















eq. of plane (



Eq. of plane in normal form:
Vector eq. of plane normal to unit vector



Proof:

ON is




Since






















eq. of plane is


Cartesian Form: Let l, m, n be d.r.'s of normal to given plane & P is length of
