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A Surface is called a plane surface
or a plane if all the points of a straight line
joining any two points on the surface lie on it
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General equation of a PlaneGeneral equation of a Plane
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Let be a plane and P be a point
which is not in the plane . If M is the projection of P in the plane then the line
PM is called a normal or perpendicular
through P to the plane and PM is called perpendicular distance or length of
the perpendicular from P to the plane .
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The vector equation of the plane which
is at a distance of P from the origin along the unitvector n is r . n=p.
The locus of a point P in the
plane is The equation of the plane is
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1) Find the DCs of the normal plane 2x-y+2z+5=0 ?
Sol) the drs of the normal to the plane ax+by+cz+d=0
drs of the normal (a,b,c)= (2,-1,2)
DCs of the normal are
=
=lx+my+nz = p
2x-y+2z=9 => 2x-y+2z-9=0.
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The vector equation of a plane perpendicular to the vector m
is r.m=q where q is a real number.
The equation lx+my+nz=pwhere p is a positive real number .
The equation of a plane
through the origin is lx+my+nz=0
The equation of a plane is of
the form ax+by+cz+d=0
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The perpendicular distance ( length of theperpendicular)from (x,y,z) to the plane
ax+by+cz+d=0 is
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1) Find the perpendicular distance from a) (3,-4,1)to the plane x+y-z+7=0b)(0,0,0)to the plane 3x+4y+12z-26=0
Sol)
=>
=>
=>
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If (a,b,c)are the direction ratios of the normal toa plane then the equations of the plane is
ax+by+cz+d=0
The vector equation of the plane passing throughthe point A having position vector a and perpendicular
to the vector m is (r-a). m=0
The equation of a plane passing through ( )is
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1) Find the equation of the plane passing through (2,3,5)& whose
normal has the drs1,2,3
Sol) a(x-x1)+b(y-y1)+c(z-z1)=0
1(x-2)+2(y-3)+3(z-5)=0
x-2+2y-6+3z-15=0
x+2y+3z-23=0
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The first degree general equation ax+by+cz+d=0 in x,y,zrepresents a plane . Since this equation is of first degree atleast one of
a,b,c is non zero . If a0 then the equation becomes
Thus a plane must satisfy 3conditions . They are
1. The vector equation of the plane passing through the points A,B,C
having position vectors a,b,c is [r-a b-a c-a ] =0
2. the equation of the plane passing through ,
is =0
3.The equation of a plane passing through is
,
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1)Find the equation of the plane passing through the point (2,2,-1)(3,4,2)(7,0,6)
Sol) the equation of the plane through the point (2,2,-1) is
a(x-x1)+b(y-y1)+c(z-z1)=0a(x-2)+b(y-2)+c(z+1)=0
(3,4,2)lies in the plane => a(3-2)+b(4-2)+c(2+1)=0
=> a+2b+3c=0 --------(7,0,6)lies in the plane => a(7-2)+b(0-2)+c(6+1)=0
=> 5a-2b+7c=0 -------
From 1 & 2
a b c
2 3 1 2
-2 7 5 -2
=>
the equation of the required plane is 5(x-2)+2(y-2)-3(z+1)=0=> 5x-10+2y-4-3z-3=0
=> 5x+2y-3z-17=0
1
2
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If a plane cuts x-axis at A (a,0,0), y-axis at B (0,b,0)and z-
axis at C (0,0,c)then a is called x-intercept, b is called y-intercept
and c is called z-intercept of the plane
The equation of the plane having x-intercept a, y-intercept b, z-intercept c
is
NOTE:- The equation of a plane is referred as intercepts
formThe intercepts of the plane ax+by+cz+d=0 or d/a, -d/b, -d/c.
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1)Find the intercepts of the plane 2x+3y+6z-
12=0
Sol:-2x+3y+6z=12
Intercepts (6,4,2)
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The angle between the normal's of two planes is called the
angle between the planes
If is angle between the planesthen cos = _________ _________
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1)Find the angle between two planes x+2y+3z=5,3x+3y+z=9?
Sol) Given planes are x+2y+3z=5 1
3x+3y+z=9 2
Let be the angle between the 1&2 is
cos=
=
=
=cos ( )
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The planes are
i. Parallel:
ii. Perpendicular:
1)The equation of a plane parallel to ax+by+cz+d=0 is
ax+by+cz+k=0
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1)Find the distance between the parallel planes 2x-2y+z+5=0, 2x-2y+z-7=0
Sol) given planes are 2x-2y+z+5=0-------
2x-2y+z-7=0-------
distance between1 & 2 are
1
2
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The ratio in which the plane
=ax+by+cz+d=0 divides the line segment joining
A(x1,y1,z1), B(x 2,y2,z2) is 111:222
where111=(x1.y1,z1)=ax1+by1+cz1+d,222=
(x2,y2,z2)=ax2+by2+cz2+d
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1)Find wheather the points lie on the same side or opposite side of theplane 5x+2y-7z+9=0,A(1,-1,3)B(3,3,3)
Sol) Given plane is = 5x+2y-7z+9=0 ------
the given points are A=(1,-1,3)B(3,3,3)consider 111=5-2-21+9=-9
222=15+6-21+9=9
ratio of divides line joining A & B is 111 : 222
-(-9): 9 => 9: 9 =>1:1
the points lie on the opposite side of the plane .
1
1
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