3D Geometry - 4
Introduction
Dumb Question: How this lx + my + nz = P is eq. of plane ? Ans: Since
.
= P = d is eq. of plane
& 
lx + my + nz = PIllustration: Find vector eq. of plane which is at a distance of 8 units from origin & which is normal to vector
. Ans: d = 8 &
= 
eq. of plane is,
. = d 
Eq. of plane passing through 3 given points:
Eq. of plane passing through three points A, B, C having P.V.'s
, 
, 
respectivelt. Let
be P>V> of any point P is the plane. So, vectors,
= - , 
= - , 
= - are coplanar Hence,
.( x ) = 0 (
- ).{( - ) x ( - )} = 0 ( - ).( x - x - x - x ) = 0 ( - ).( x + x + x ) = 0 .( x + x + x ) = .( x ) + .( x ) + .( x ) [] + [] + [] = []Note: If P is length of
from origin on this plane,then P =
n = | x + x + x | Eq. of plane that passes through a point A with position vector
& is || to given vectors & : Derivation:
Let
be P.V. of any point P in plane Then,
= 
- 
= - Sincen
x are || to plane.So, vector
- , & are coplanar
( - ).( x ) = 0 .( x ) = .( x ) [] = [] Cartesian form: Eq. of plane passing through a point (x1, y1, z1) & || to two lines having direction ratios (
1, 
1, 
1) & (2, 2, 2) is: 
Illustration: Find eq. of plane passing through points
P(1, 1, 1), Q(3, -1, 2), R(-3, 5, -4).
Ans: Let eq. of plane is ax + by + cz + d = 0
It passes through P(1, 1, 1). So,
eq. of plane
a(x - 1) + b(y - 1) + c(z - 1) = 0 .......................................... (i)
But it also passes through Q & R
2a - 2b + c = 0] x (- 2)
- 4a + 4b - 5c = 0
Similarly a = 6, b = 6
Putting these in eq. (i)
We get, 6(x - 1) + 6(y - 1) = 0
x + y = 2 Angle b/w two planes:
.
= d1 .
= d2 are two planers, then cos
= 
Note: Angle b/w planes is defined as angle b/w their normals.
Cartesion Form: Let planes are a1x + b1y + c1z + d1 = 0 & a2x + b2y + c2z + d2 = 0
cos = 
Condition for
: . = 0 or a1a2 + b1b2 + c1c2 = 0
Condition for parallelism:
or 
Angle b/w line & a plane:
If
, , be direction ratios of line & ax + by + cz + d = 0 be eq. of plane in which normal has d.r's a, b, c. Q is angle b/w line & plane. cos(900 - ) = 
Vector form: If
is angle b/w line 
& plane . = d sin = 
Illustration: Find angle b/w line
& 3x + 2y - 2z + 3 = 0. Ans: D.r's of line are 2, 3, 2 & d.r's of normal to plane are 3, 2, - 2
sin
= 
= sin-1
Eq. of plane passing through the Line of Intersection of planes
a1x + b1y + c1z + d1 = 0 & a2x + b2y + c2z + d2 = 0 is
(a1x + b1y + c1z + d1) + k(a2x + b2y + c2z + d2) = 0
Dumb Question: How this eq. is that of required plane ?
Ans: Let P(
, , ) be point on line of intersection of two planes. So, it lies on both the planes. i.e. a1
+ b1 + c1 + d1 = 0 & a2
+ b2 + c2 + d2 = 0 So, point P(
, , ) should lie on required plane. i.e. (a1
+ b1 + c1 + d1) + k(a2 + b2 + c2 + d2 ) = 0 P(, , ) lies on. plane (a1x + b1y + c1z + d1) + k(a2x + b2y + c2z + d2) = 0 Vector Form:
. = d1 & . = d2So, eq. of required plane is (
. - d1) + k(. - d2) = 0 Illustration: Find eq. of plane constaining line of intersection of plane x - y + z + 7 = 0 and x + 3y + 2z + 5 = 0 & passing through (1, 2, 2).
Ans: Eq. of plane through line of intersection of given planes is,
(x - y + z + 7) +
(x + 3y + 2z + 5) = 0 ....................................... (i) It passes through (1, 2, 3)
(1 - 2 + 2 + 7) +
(1 + 3 x 2 + 2 x 2 + 5) = 0 8 +
(7 + 4 + 5) = 
Putting
= - 
in eq. (i) We get,
(x - y + z + 7) + (-
)(x + 3y + 2z + 5) = 0 2x - 2y + 2z + 14 - x - 3y - 2z - 5 = 0
x - 5y + 9 = 0
Two sides of plane:
If ax + by + cz + d = 0 be a plane then points (x1, y1, z1) & (x2, y2, z2) are points lies on
Same side if
> 0 & opposite side if < 0 Distance of point from a plne:
Length of
from a point having P.V. to plane . = d is given by P =
Proof: PM is length of
from P to plane. Since line PM passes through P() & || to vector which is normal to plane. eq. of line = x ....................................................... (i) Since point M is intersection of line & plane. So, it lies on line as well as plane
Putting
in eq. (i) = + 
= P.V. of M - P.V. of P = + - Dumb Question: How P.V. of M is
+ 
Ans: M lies on line as well as plane. On solving value of
for line eq. We get P.V. of M. So, this is P.V. of M
PM = |
| = 
Cartesian Form: Length of
from point P(x1, y1, z1) to plane ax + by + cz + d = 0, Then eq. of PM is 
= r [Dumb Question: How this eq. of PM comes ?
Ans: It is passes through point (x1, y1, z1) & || to normal of plane so, we get this eq.]
Coordinates of any point on PM are
(x1 + ar, y1 + br, z1 + cr)
But this also coordinate of M & M also lies on plane
a(x1 + ar) + b(y1 + br) + c(z1 + cr) + d = 0 i.e. r = -
PM =
Distance b/w parallel planes:
Distance b/w || planes is difference of length of
from origin to two planes. Let ax + by + cz + d1 = 0 & ax + by + cz + d2 = 0
D =
Vector Form:
. = d1 & . = 2 
Illustion: Find distance b/w parallel planes
x + 2y + 2z + 2 = 0 & 2x + 4y + 4z + 3 = 0
Ans: Distance b/w ax + by + cz + d1 = 0 & ax + by + cz + d2 = 0 is
x + 2y + 2z + 2 = 0 & x + 2y + 2z +
= 0 