# 3D Geometry - 4

Introduction

__: How this lx + my + nz = P is eq. of plane ?__

**Dumb Question**Ans: Since . = P = d is eq. of plane

&

lx + my + nz = P

__: Find vector eq. of plane which is at a distance of 8 units from origin & which is normal to vector .__

**Illustration**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 )

[] + [] + [] = []

__: If P is length of from origin on this plane,__

**Note**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 )

[] = []

__: Eq. of plane passing through a point (x__

**Cartesian form**_{1}, y

_{1}, z

_{1}) & || to two lines having direction ratios (

_{1},

_{1},

_{1}) & (

_{2},

_{2},

_{2}) is:

__: Find eq. of plane passing through points__

**Illustration**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**. = d

_{1}. = d

_{2}are two planers, then

cos =

__: Angle b/w planes is defined as angle b/w their normals.__

**Note**__: Let planes are a__

**Cartesion Form**_{1}x + b

_{1}y + c

_{1}z + d

_{1}= 0 & a

_{2}x + b

_{2}y + c

_{2}z + d

_{2}= 0

cos =

__:__

**Condition for**. = 0

or a

_{1}a

_{2}+ b

_{1}b

_{2}+ c

_{1}c

_{2}= 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(90

^{0}- ) =

__: If is angle b/w line & plane . = d__

**Vector form**sin =

__: Find angle b/w line & 3x + 2y - 2z + 3 = 0.__

**Illustration**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

a

_{1}x + b

_{1}y + c

_{1}z + d

_{1}= 0 & a

_{2}x + b

_{2}y + c

_{2}z + d

_{2}= 0 is

(a

_{1}x + b

_{1}y + c

_{1}z + d

_{1}) + k(a

_{2}x + b

_{2}y + c

_{2}z + d

_{2}) = 0

__: How this eq. is that of required plane ?__

**Dumb Question**Ans: Let P(, , ) be point on line of intersection of two planes. So, it lies on both the planes.

i.e. a

_{1}+ b

_{1}+ c

_{1}+ d

_{1}= 0

& a

_{2}+ b

_{2}+ c

_{2}+ d

_{2}= 0

So, point P(, , ) should lie on required plane.

i.e. (a

_{1}+ b

_{1}+ c

_{1}+ d

_{1}) + k(a

_{2}+ b

_{2}+ c

_{2}+ d

_{2}) = 0

P(, , ) lies on. plane (a

_{1}x + b

_{1}y + c

_{1}z + d

_{1}) + k(a

_{2}x + b

_{2}y + c

_{2}z + d

_{2}) = 0

__:__

**Vector Form**. = d

_{1}& . = d

_{2}

So, eq. of required plane is (. - d

_{1}) + k(. - d

_{2}) = 0

__: 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).__

**Illustration**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 (x

_{1}, y

_{1}, z

_{1}) & (x

_{2}, y

_{2}, z

_{2}) 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 = + -

__: How P.V. of M is +__

**Dumb Question**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 = || =

__: Length of from point P(x__

**Cartesian Form**_{1}, y

_{1}, z

_{1}) 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 (x

_{1}, y

_{1}, z

_{1}) & || to normal of plane so, we get this eq.]

Coordinates of any point on PM are

(x

_{1}+ ar, y

_{1}+ br, z

_{1}+ cr)

But this also coordinate of M & M also lies on plane

a(x

_{1}+ ar) + b(y

_{1}+ br) + c(z

_{1}+ 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 + d

_{1}= 0 & ax + by + cz + d

_{2}= 0

D =

__: . = d__

**Vector Form**_{1}& . =

_{2}

__: Find distance b/w parallel planes__

**Illustion**x + 2y + 2z + 2 = 0 & 2x + 4y + 4z + 3 = 0

Ans: Distance b/w ax + by + cz + d

_{1}= 0 & ax + by + cz + d

_{2}= 0 is

x + 2y + 2z + 2 = 0 & x + 2y + 2z + = 0