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Three Dimensional Geometry Test - 5

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Three Dimensional Geometry Test - 5
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Weekly Quiz Competition
  • Question 1
    1 / -0
    The locus of a point which moves so that the difference of the squares of its distances from two given points is constant, is a:
  • Question 2
    1 / -0

    The radius of the sphere through the points ( 4 ,3 , 0 ) , ( 0 , 4 , 3 ) ,( 0 , 5 , 0 ) and ( 4 , 0 , 3 ) is

    Solution

    we know that general equation of sphere is  x2+y2+z2+2gx+2fy+2hz+c=0


    since sphere passing through the points  ( 4 ,3 , 0 ) , ( 0 , 4 , 3 ) ,( 0 , 5 , 0 ) and ( 4 , 0 , 3 ) so putting these values one by one in given equation , we get

    16+ 9+ 8g + 6f + c=0    =>8g +6f +c = -25  -------------1)

    16 +9 +8f +6h +c=0    =>8f + 6h +c= -25   -----------2)

    25 + 10f +c  = 0       -----------------------3)

    16+ 9+ 8g + 6h +c=0    =>8g +6h +c= -25  -------------4)

    by 2) - 4 ) we have  f = g

    using this in 1) we have 16f +c = -25 -------5)

    Now solving 3) and 5) we get f=0 

    using value of f  we have g= 0 , h= 0 , c= -25

    Now radius =√(−g)2+(−f)+(−h)2−c = √25 =5

  • Question 3
    1 / -0
    The direction ratios of normal to the plane through (1, 0, 0) and (0, 1, 0), which makes an angle  with the plane x + y = 3, are
  • Question 4
    1 / -0

    If the direction cosines of a straight line are < k , k , k > , then what will be the value of k

    Solution

    We know that if  l , m , n are direction cosines of a lines then l2 +m2 +n2= 1

     so,  k2 + k2 +k2 = 1

           3 k2 = 1

           k2 = 1/3 


    taking squareroot on both sides , we have  k=1/√3ork=−1/√3

  • Question 5
    1 / -0
    The angle between the lines x = 1, y = 2 and y = 1, z = 0 is
  • Question 6
    1 / -0

    the numbers 3, 4 , 5 can be

    Solution

    the numbers 3, 4 , 5 can be direction ratio of any line these not satisfying any other option

  • Question 7
    1 / -0
    The equation of the line passing through (1, 2, 3) and parallel to the planes x - y + 2z = 5 and 3x + y + z = 6, is:
  • Question 8
    1 / -0
    If a and b are two non-zero vectors, then the component of b along a is:
  • Question 9
    1 / -0
    The cosine of the angle between any two diagonals of a cube
  • Question 10
    1 / -0
    The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x2 + y2 + z2 + 4x – 2y – 6z = 155 is:
  • Question 11
    1 / -0
    The angle between the two diagonals of a cube is
  • Question 12
    1 / -0
    In a triangle ABC, if D is the mid-point of side [BC], then find .
  • Question 13
    1 / -0
    If ABCDEF is a regular hexagon and , then is equal to
  • Question 14
    1 / -0
    The locus of a point which moves so that the difference of the squares of its distances from two given points is constant, is a:
  • Question 15
    1 / -0
    The equation x2 + 3y2 - 9x + 2y + 1 = 0, represents
  • Question 16
    1 / -0
    The point of intersection of the line  and plane 2x – y + 3z – 1 = 0 is
  • Question 17
    1 / -0
    The distance between the planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is
  • Question 18
    1 / -0
    The equation of XOY plane is
  • Question 19
    1 / -0

    The lines, which do not lie in the same plane, are called ?

    Solution

    Lines which lie in same plane can be parallel, coincident or intersecting.

    Skew lines are those lines which lie in different planes and never meet to each other 

  • Question 20
    1 / -0
    If = 1 – cos  + i sin, then amp z =
  • Question 21
    1 / -0

    The points ( 4,7 ,8) ,( 2, 3,4 ),( - 1 , -2 , 1 ) and (1 , 2, 5 ) are the vertices of a

    Solution

    The vertices of a quadrilateral are A( 4,7 ,8) , B( 2, 3,4 ) ,C (- 1 , -2 , 1 ) and D(1 , 2, 5 ) 

    AB =√(4−2)2+(7−3)2+(8−4)2=√4+16+16√=√36=6

    BC=√(2+1)2+(3+2)2+(4−1)2=√9+25+9=√43

    CD=√(1+1)2+(2+2)2+(5−1)2=√4+16+16=√36=6


    DA= √(4−1)2+(7−2)2+(8−5)2=√9+25+9=√43


    Here opposite sides of quadrilateral ABCD are equal , so  it may be parallelogram or rectangle

    Now , diagonal AC=√(4+1)2+(7+2)2+(8−1)2=√25+81+49=√155

    BD =√ (2−1)2+(3−2)2+(4−5)2=√1+1+1=√3

    Here length of diagonals AC and BD are different so ABCD is a parrallelogram

  • Question 22
    1 / -0
    The medians of a triangle are concurrent at the point called
  • Question 23
    1 / -0
    The area of the triangle formed by the points (a, b + c), (b, c + a) and (c, a + b) is equal to
  • Question 24
    1 / -0
    If I is the incentre of triangle ABC, thenis equal to
  • Question 25
    1 / -0
    The equation xy = 0 in three dimensional space represents
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