3D Geometry - 4 · KnowledgeBin.org

Introduction

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

Ans: Since

= P = d is eq. of plane

lx + my + nz = P

Illustration: 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,

) = 0
(

).{(

) x (

)} = 0

(

).(

) = 0

(

).(

) = 0

) =

) +

)

[

] + [

] = [

]

Note: If P is length of

from origin on this plane,
then P =

n = |

|

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

are || to plane.
So, vector

are coplanar

(

).(

) = 0

) =

)

[

] = [

]

Cartesian form: Eq. of plane passing through a point (x₁, y₁, z₁) & || to two lines having direction ratios (

₁,

₁) & (

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:

Angle b/w line & a plane:

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

) lies on. plane (a₁x + b₁y + c₁z + d₁) + k(a₂x + b₂y + c₂z + d₂) = 0

Vector Form:

= d₁ &

= d₂
So, 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

....................................................... (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

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