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Three Dimensional Geometry (part – 2) , The Planes | study material for IIT JEE | concept booster, chapter highlights/

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THREE DIMENSIONAL GEOMETRY ( PART – 2), THE PLANES

THE PLANES

Plane is the surface such that if any two points are taken on it, then the line joining the two points lies on it.
General equation of plane is given as –
ax + by + cz + d = 0 , where a ,b and c is not equal to zero.

           POINTS TO BE REMEMBERED
 1). a, b and c are the directions ratios of the normal to the plane ax + by + cz + d = 0.

2). Equation of the yz – plane is x = 0

3). Equation of the zx – plane is y = 0

4). Equation of the xy – plane is z = 0

5). Equation of the any plane parallel to the xy – plane is z = c. Similarly for Planes parallel to yz and zx is x = c and y = c

EQUATION OF THE PLANE IN NORMAL FORM

VECTOR FORM

If n̂ be a unit vector normal to a given plane and d be the length of the perpendicular from the origin to the plane, then the equation of the plane is given by –
                          r.n̂ = d

CARTESIAN FORM

If l, m, n are the directions cosines of the normal to the plane and d is the perpendicular distance from the origin to the plane , then the equation of the plane is –
                     lx + my + nz = d

GENERAL FORM TO NORMAL FORM

To reduce the general equation of the plane
ax+by+cz+d = 0 to the normal form
Follow the following working rule
1). Write all the variable terms on the left hand side and constant term on the right hand side.
2). If constant term on the right hand side is not positive then make it positive by multiplying both side by -1.
3). Divide each term by √(a²+b²+c²) , then we get
ax/±√(a²+b²+c²)+by/±√(a²+b²+c²)+cz/±√(a²+b²+c²)
= d/±√(a²+b²+c²)
+ Sign is taken when d>0 and – sign is taken when d

EQUATION OF THE PLANE PASSING THROUGH A GIVEN POINT

VECTOR FORM

The vector equation of the plane passing through a given point r1 and perpendicular to n̂
is = (r – r1).n̂ = d

CARTESIAN FORM

Equation of the plane passing through the point A(x1, y1, z1) and (a, b, c) the direction ratios of the normal vector to the plane is
 
a(x-x1)+b(y-y1)+c(z-z1) = 0
 

EQUATION OF THE PLANE IN INTERCEPT FORM

If the plane makes intercepts of length a , b and c with x axis , y axis and z axis then the equation of the plane is 
 
x/a + y/b + c/z = 1
 

EQUATION OF THE PLANE PASSING THROUGH THREE POINTS

VECTOR FORM

Equation of the plane passing through the three points whose position vector is a , b and c is

(r-a).[(b-a)×(c-a)] = 0

CARTESIAN FORM

The equation of the plane passing through the three non – collinear points A(x1,y1,z1) ,B(X2,y2,z2) and C(x3,y3,z3) is 

 

EQUATION OF THE PLANE PASSING THROUGH TWO GIVEN POINTS AND PARALLEL TO THE GIVEN VECTOR

VECTOR FORM

Equation of the plane Passing through the two points having position vector a and b and parallel to a given vector c is –
(r – a).[(b – c)×c] = 0

CARTESIAN FORM

The equation of the plane passing through the two points A(x1,y1,z1) and B(x2,y2,z2) and parallel to the given vector whose direction ratios is a, b and c is 
 

 

EQUATION OF THE PLANE PASSING THROUGH A GIVEN POINTS AND PARALLEL TO TWO GIVEN VECTORS

VECTOR FORM

Equation of the plane passing through a given point whose position vector is a and parallel to the two given vectors b and c is
r = a + λb + μc , where λ and μ are scalers
Or (r – a).(b × c) = 0

CARTESIAN FORM

The equation of the plane passing through a point (x1,y1,z1) and parallel to the two lines having direction ratios a1, b1,c1 and a2, b2, c2 is

PLANES PARALLEL TO A GIVEN PLANE

VECTOR FORM

Equation of the plane parallel to the plane
r.n̂ = d1 is r.n̂ = d2,  where d2 is a constant and is to be determined by the given condition.

CARTESIAN FORM

Equation of the plane parallel to the plane
ax + by + cz + d = 0 is ax + by + cz + k = 0 , where is any constant and can be determined by the given condition.

ANGLE BETWEEN THE TWO PLANE

Angles between the two Planes means angle between their normals.

VECTOR FORM

The angles between the two Planes r.n̂ = d1 and 
r.n̂ = d2 is given as –
 
Cosθ  = n1.n2/|n1||n2| 

CARTESIAN FORM

The angles between the two Planes
a1x +b1y + c1z + d1 =0
a2x + b2y + c2z + d2 = 0 is

                  POINTS TO BE REMEMBERED
 1). If a1a2 + b1b2 + c1c2 = 0 then the planes is perpendicular to each other.

2). If a1/a2 = b1/b2 = c1/c2 = λ , then the plane is parallel to each other.

ANGLE BETWEEN A LINE AND A PLANE

The angles between the line and a plane is angle between the normal and the line

VECTOR FORM

If θ is the angle between line r = a + λb and plane r.n = d , then
Sinθ = b.n/|b|.|n|

CARTESIAN FORM

If θ is the angle between the line
(x-x1)/a1 = (y-y1)/b1 = (z-z1)/c1 and plane
a2x + b2y + c2z + d = 0 , then

              POINTS TO BE REMEMBERED
 1). If the line x-x1/a1 = y-y1/b1 = z-z1/c1 is parallel to the plane a2x+b2y+c2z+d = 0 , then
a1a2 + b1b2 + c1c2 = 0

POINTS OF INTERSECTION OF A LINE AND A PLANE

The given working rule can be used to find the points of intersection of a line and a plane
Step 1). Write the coordinates of any point on the line in terms of some parameters λ (say)
Step 2). Substitute these coordinates in the equation of the plane to obtain the value of λ.
Step 3). And at last put the value of λ in the coordinates of the point in step 1.

               POINTS TO BE REMEMBERED
 1). The ratios in which the line segment PQ, joining the points P(x1,y1,z1) and Q(X2,y2,z2), is divided by the plane ax+by+cz+d =0 is given as –

-(ax1+by1+cz1+d/ax2+by2+cz2+d)

PLANES BISECTING THE ANGLES BETWEEN THE TWO PLANES

CARTESIAN FORM

The equation of the planes bisecting the  angles between the two Planes a1x+b1y+c1z+d1 =0 and
a2x+b2y+c2z+d2=0 ares :
(a1x+b1y+c1z+d1)/√(a1²+b1²+c1²) =
(a2x+b2y+c2z+d2)/√(a2²+b2²+c2²)

VECTOR FORM

The equation of the planes bisecting the angles between the two Planes r.n1 = d1 and r.n2 = d2 is –
|r.n1-d1|/|n1| = |r.n2-d2|/|n2|
Or , r.(n1±n2) = d1/|n1| = d2/|n2|

DISTANCE OF A POINT FROM A PLANE

VECTOR FORM

The lenght of the perpendicular drawn from the point having position vector a to the plane
r.n = d is given by
                         d = |a.n-d|/|n|

CARTESIAN FORM

The length of the perpendicular drawn from the point P(x1,y1,z1) to the plane ax+by+cz+d = 0
is given as –
p = |ax1+by1+cz1+d |/√(a²+b²+c²)

DISTANCE BETWEEN THE TWO PARALLEL PLANES

VECTOR FORM

The distance between the two parallel planes
r.n=d1 and r.n= d2 is –
p = | d1-d2 |/|n|

CARTESIAN FORM

The distance between the two parallel planes
a1x+b1y+c1z+d1= 0 and a1x+b1y+c1z+d2 =0
is given as –
p = |d1-d2|/√(a1²+b1²+c1²)

PLANES PASSING THROUGH THE INTERSECTION OF TWO PLANES

VECTOR FORM

The vector equation of the plane passing through the intersection of the two Planes
r n1 = d1 and r n2= d2 is given as –
(r.n1-d1)+λ(r.n2-d2) = 0
Or , r.(n1+λn2) = d1+λd2
Where λ is any arbitrary constant.

CARTESIAN FORM

The equation of the plane passing through the intersection of the two Planes
a1x+b1y+c1z+d1= 0 and a2x+b2y+c2z+d2=0 is
(a1x+b1y+c1z+d1)+λ(a2x+b2y+c2z+d2) = 0 , where λ is any arbitrary constant.

CONDITION FOR THE TWO LINES TO BE INTERSECTING (COPLANER) AND THE EQUATION OF THE PLANE CONTAINING THEM

VECTOR FORM

If the lines r = a1+λb1 and r = a2+μb2 are intersecting (coplaner) then –
            [a1 b1 b2 ] = [ a2 b1 b2 ]
And the equation of the plane containing them is –
          [ r b1 b2 ] = [ a1 b1 b2 ]
          [ r b1 b2 ] = [ a2 b1 b2 ]

CARTESIAN FORM

If the lines x-x1/l1 = y-y1/m1 = z-z1/n1 and
x-x2/l2 = y-y2/m2 = z-z2/n2 are interesting (coplaner) then
And the equation of the plane is –

IMAGE OF A POINT IN A PLANE

Let’s take P and Q two points and π be the plane such that
1). Line PQ is perpendicular to the plane π.
2). Mid point of line PQ lies on the plane π.
If this happens then the either of the point is the image of the other point in the plane π.
To find the image of the point in a given plane , follow the following working rule.
Step 1). Write the equation of the line passing through P and normal the given plane as
x-x1/a = y-y1/b = z-z1/c
Step 2). Write the coordinates of the image Q as
    (x1+aλ , y1+bλ , z1+cλ)
Step 3). Find tthe coordinates of the mid point M of PQ.
Step 4). Obtain the value of λ by putting the coordinates of M in the equation of the plane.
Step 5). Put the value of λ in the coordinates of Q.

IMAGE OF A LINE ABOUT A PLANE

Let consider a line x-x1/a1 = y-y1/b1 = z-z1/c1
and the plane a2x+b2y+c2z+d = 0.
Find point of intersection (say)P of the line and the plane. Then find image (say)Q of the point (x1,y1,z1) about a plane, and the line PQ is the reflected line.

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