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Introduction to Three Dimensional Geometry Test - 25

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Introduction to Three Dimensional Geometry Test - 25
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  • Question 1
    1 / -0
    Find the ratio in which (the plane) $$2x+3y+5z=1$$ divides the line joining the points $$(1,0,-3)$$ and $$(1,-5,7)$$.
    Solution
    Here, $$2x+3y+5z=1$$ divides $$(1,0,-3)$$ and $$(1,-5,7)$$ in the ratio of $$k:1$$ at point $$P.$$
    Then, $$\displaystyle P=\left( \frac { k+1 }{ k+1 } ,\frac { -5k }{ k+1 } ,\frac { 7k-3 }{ k+1 }  \right) $$ which must satisfy $$2x+3y+5x=1$$
    $$\displaystyle \Rightarrow 2\left( \frac { k+1 }{ k+1 }  \right) +3\left( \frac { -5k }{ k+1 }  \right) +5\left( \frac { 7k-3 }{ k+1 }  \right) =1$$
    $$\Rightarrow 2k+2-15k+35k-15=k+1\Rightarrow 21k=14$$
    $$\displaystyle \Rightarrow k=\frac { 2 }{ 3 } $$
    $$\therefore 2x+3y+5z=1$$
    It divides the joint of $$(1,0-3)$$ and $$(1,-5,7)$$ in the radio of $$2:3.$$
  • Question 2
    1 / -0
    A hall has dimensions $$24 m \times 8 m \times 6 m$$. The length of the longest pole which can be accommodated in the hall is
    Solution

    Given that,

    Dimensions of the hall x $$=24cm\times 8cm\times 6cm$$

    Now Leght of the longest pole which can be accommodated in the hall x $$=\sqrt{{{24}^{2}}+{{8}^{2}}+{{6}^{2}}}=26cm$$

  • Question 3
    1 / -0
    The coordinates of a point which divides the line joining the points $$P(2,3,1)$$ and $$Q(5,0,4)$$ in the ratio $$1:2$$ are
    Solution
    Using section formula,
    Coordinate of the point which divides $$P(2,3,1)$$ and $$Q(5,0,4)$$ in ratio $$1:2$$ is
    $$\left(\dfrac{2\cdot 2+5}{2+1}, \dfrac{2\cdot 3+0}{2+1},\dfrac{2\cdot 1+4}{2+1} \right)$$ $$=\left(\dfrac{9}{3}, \dfrac{6}{3},\dfrac{6}{3} \right)=(3,2,2)$$
  • Question 4
    1 / -0
    In $$\triangle ABC$$ the mid points of the sides $$AB, BC$$ and $$CA$$ are respectively $$\left( l,0,0 \right) ,\left( 0,m,0 \right) $$ and $$\left( 0,0,n \right) $$. Then, $$\dfrac { { AB }^{ 2 }+{ BC }^{ 2 }+{ CA }^{ 2 } }{ { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } } $$ is equal to
    Solution
    Let $$A=(x_1,y_1,z_1),B=(x_2,y_2,z_2),C=(x_3,y_3,z_3)$$

    From the figure,
    $${ x }_{ 1 }+{ x }_{ 2 }=2l, { y }_{ 1 }+{ y }_{ 2 }=0, { z }_{ 1 }+{ z }_{ 2 }=0$$,      [midpoint formula]

    $${ x }_{ 2 }+{ x }_{ 3 }=0, { y }_{ 2 }+{ y }_{ 3 }=2m, { z }_{ 2 }+{ z }_{ 3 }=0$$

    and $${ x }_{ 1 }+{ x }_{ 3 }=0, { y }_{ 1 }+{ y }_{ 3 }=0, { z }_{ 1 }+{ z }_{ 3 }=2n$$

    On solving, we get
    $${ x }_{ 1 }=l, { x }_{ 2 }=l, { x }_{ 3 }=-l$$,

    $${ y }_{ 1 }=-m, { y }_{ 2 }=m, { y }_{ 3 }=m$$

    and $${ z }_{ 1 }=n, { z }_{ 2 }=-n, { z }_{ 3 }=n$$

    $$\therefore $$ Coordinates are $$A\left( l,-m,n \right) ,B\left( l,m,-n \right) $$ and $$C\left( -l,m,n \right) $$

    $$\therefore \dfrac { { AB }^{ 2 }+{ BC }^{ 2 }+{ CA }^{ 2 } }{ { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } } $$

    $$=\dfrac { 4{ m }^{ 2 }+4{ n }^{ 2 }+4{ l }^{ 2 }+4{ n }^{ 2 }+\left( 4{ l }^{ 2 }+4{ m }^{ 2 } \right)  }{ { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 } } $$

    $$=8$$

  • Question 5
    1 / -0
    The distance of the point (1,−2,4) from the plane passing through the point (1,2,2) and perpendicular to the planes $$x−y+2z=3$$ and $$2x−2y+z+12=0$$ is :
    Solution

  • Question 6
    1 / -0
    The point $$(3, 0, -4)$$ lies on the
    Solution

  • Question 7
    1 / -0
    What is the length of the segment in the x-y plane with end points at $$(-2, -2)$$ and $$(2, 3)$$?
    Solution

  • Question 8
    1 / -0
    Graph $$x^2+y^2=4$$ in 3D looks like
    Solution
    The given curve is $$x^2+y^2=4$$ 
    So $$x$$ coordinate and y-coordinate are connected by $$x^2+y^2=4$$
    which is locus of a circle with radius $$2$$
    But z-coordinate can be anything, so in three dimension the circle $$x^2+y^2=4$$ will be 
    stretched which will be a cylinder with radius same as the radius of the circle .
  • Question 9
    1 / -0
    If the line joining $$A(1, 3, 4)$$ and $$B$$ is divided by the point $$(-2, 3, 5)$$ in the ratio $$1 : 3$$, then $$B$$ is
    Solution
    Let the co-ordinates of $$B$$ be $$(x, y, z)$$.

    Then applying the section formula gives us
    $$-2=\dfrac{3(1)+x}{3+1}$$
    $$-8=x+3$$
    $$x=-11$$................(i)

    Similarly $$3=\dfrac{3(3)+y}{4}$$
    $$12=9+y$$
    $$y=3$$ ..........(ii)
    $$5=\dfrac{3(4)+z}{4}$$
    $$20=12+z$$
    $$z=8$$ ...........(iii)

    Hence $$B=(-11, 3, 8)$$
  • Question 10
    1 / -0
    Point $$D$$ has coordinates as $$(3,4,5)$$. Referring to the given figure, find the coordinates of point $$E$$.

    Solution
    Point D has coordinates $$(3,4,5)$$
    Point E is in line with D, lying on the y-z plane having x coordinate 0, such that line DE is perpendicular to the y-z plane.
    Thus, the y and  coordinates of point E remain same as that of D.
    The coordinates of point E are therefore $$(0,4,5)$$
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