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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

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    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 .

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