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Introduction to Three Dimensional Geometry Test - 26

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Introduction to Three Dimensional Geometry Test - 26
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  • Question 1
    1 / -0
    Calculate the distance between the points $$(-3,6,7)$$ and $$(2,-1,4)$$ in $$3D$$ space.
    Solution

    We know the distance formula:

    $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

    The coordinates are $$(-3, 6, 7)$$ and$$ (2, -1, 4)$$

    $$d=\sqrt{(2-(-3))^2+((-1)-6)^2+(4-7)^2}$$

    $$d=\sqrt{(5)^2+((-7)^2+(-3)^2}$$

    $$d=\sqrt{25+49+9}$$

    $$d=\sqrt{83}$$

    $$d = 9.11$$

  • Question 2
    1 / -0
    An equation of sphere with centre at origin and radius $$r$$ can be represented as
    Solution
    Sphere is locus of a point in 3D whose distance from a fixed point(center) is constant (radius)
    $$\Rightarrow \sqrt{(x-0)^2+(y-0)^2+(z-0)^2}=|r|$$
    $$\Rightarrow x^2+y^2+z^2=r^2$$, square both sides 
  • Question 3
    1 / -0
    The ratio in which the line joining $$(2, -4, 3)$$ and $$(-4, 5, -6)$$ is divided by the plane $$3x+2y+z-4=0$$ is
    Solution
    A line passes through the points $$(2,-4,3)$$ and $$(-4,5,-6)$$
    The equation of the line can be written as $$t = \cfrac{x - 2}{-4 - 2} = \cfrac{y + 4}{5 + 4} = \cfrac{z - 3}{-6 - 3}$$
    $$\therefore x = -6t + 2, y = 9t - 4, z = -9t + 3$$
    Substituting this in equation of the plane, we get $$-18t + 6 + 18t - 8 - 9t + 3 - 4 = 0$$
    $$\Rightarrow t = \cfrac{-1}{3}$$ and the point becomes $$(2 + 2, -3 - 4, 3 + 3)$$ i.e. $$(4,-7,6)$$
    Let this point divide the line joining $$(2,-4,3)$$ and $$(-4,5,-6)$$ in the ratio $$m:n$$, 
    $$\therefore \cfrac{-4m + 2n}{m + n} = 4$$
    $$\therefore -4m + 2n = 4m + 4n $$
    $$\therefore -8m = 2n$$
    $$\therefore m:n = -1:4$$
  • Question 4
    1 / -0
    If the distance between the points $$(7,1,-3)$$ and $$(4,5,\lambda)$$ is $$13$$ units, then what is one of the values of $$\lambda$$?
    Solution
    Given two points A(7,1,-3) and B(4,5,$$\lambda$$)
    total distance=13units ,$$\lambda$$ =?
    distance of AB $$=\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-{ y }_{ 1 })^{ 2 }+({ z }_{ 2 }-{ z }_{ 1 })^{ 2 } } \\ 13=\sqrt { (4-7)^{ 2 }+(5-1)^{ 2 }+(\lambda +3)^{ 2 } } \\ 169=9+16+\left[ \lambda ^{ 2 }+6\lambda +9 \right] \\ { \lambda  }^{ 2 }+6\lambda +9=144\\ \Rightarrow { \lambda  }^{ 2 }+6\lambda -135=0\\ \therefore \lambda =9,-15\\ \therefore \lambda =9$$
  • Question 5
    1 / -0
    The planes $$2x - y + 4z = 5$$ and $$5x - 2.5y + 10z = 6$$ are
    Solution
    Planes are $$2x-y+4z=5$$ 
    and $$5x-2.5y+10z=6$$
    Multiply both sides by 2 to the second equation
    $$\Rightarrow 10x-5y+20=12$$
    Now divide both sides by $$2$$
    $$\Rightarrow 2x-y+4z=\dfrac{12}{5}$$

    Clearly both planes are parallel 
  • Question 6
    1 / -0
    The equation of plane passing through $$(-1,0,-1)$$ parallel to $$xz$$ plane is
    Solution
    Given that the plane is parallel to $$xz$$ plane and the plane passes through $$(-1,0,-1)$$
    Since the plane is parallel to $$xz$$ plane , the $$y$$ coordinate should be constant
    Given that it passes through point $$(-1,0,-1)$$ , therefore the plane lies on $$xz$$ plane
    Therefore the equation of plane is $$y=0$$
    The correct options are $$B$$
  • Question 7
    1 / -0
    The distance of the point $$P(a, b, c)$$ from the $$x$$-axis is.
    Solution
    $$\textbf{Step 1: Find Co-ordinate of Point on X axis and Find its Distance from Point P}$$
                   $$\text{Given:Co-ordinate of Point $P=(x_1,y_1,z_1)$=(a,b,c)}$$ 

                  $$\text{ On X-axis, y and z coordinate will be zero.}$$

                  $$\therefore \text{Coordinate on the X-axis where perpendicular from point will meet is $(x_2,y_2,z_3)$= (a,0,0)}$$

                  $$\text{By Distance Formula,}$$
                  $$\text{Distance Between 2 Points=}$$ $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

                  $$\Rightarrow \text{Distance Between P and X axis}$$ $$=\sqrt{(a-a)^2+(b-0)^2+(c-0)^2}$$

                                                                                  $$ =\sqrt{b^2+c^2}$$

    $$\textbf{Therefore,Distance Between Point P and X axis is }$$$$ \sqrt{b^2+c^2}$$
  • Question 8
    1 / -0
    If a line $$OP$$ of length $$r$$ (Where '$$O$$' is the origin) makes an angle $$\alpha$$ with x-axis and lies on the xz-plane, then what are the coordinates of $$P$$?
    Solution
    As OP lies on XZ plane so y co-ordinate is 0.
    QP=$$r\cos { \alpha  } $$
    PR=$$r\sin { \alpha  } $$
    Co-ordinate of P=$$(r\cos { \alpha  },0,r\sin{ \alpha  })$$
    Option A is correct

  • Question 9
    1 / -0
    Points $$A(3,2,4),B\left( \cfrac { 33 }{ 5 } ,\cfrac { 28 }{ 5 } ,\cfrac { 38 }{ 5 }  \right) $$, and $$C(9,8,10)$$ are given. The ratio in which $$B$$ divides $$\overline { AC } $$ is
    Solution
    $$B$$ divides $$AC$$ in the ratio is $${x}_{1}-{x}_{2}:{x}_{2}-{x}_{3}$$
    $$3-\cfrac { 33 }{ 5 } :\cfrac { 33 }{ 5 } -9$$
    $$3:2$$
  • Question 10
    1 / -0
    The locus of a point, which is equidistant from the points $$(1, 1)$$ and $$(3, 3)$$, is
    Solution
    Let $$P(h, k)$$ be a point, which is equidistant from the points $$A(1, 1)$$ and $$B(3, 3)$$.
    i.e., $$PA = PB \Rightarrow (PA)^{2} = (PB)^{2}$$ (by distance formula)
    $$\Rightarrow 1 - 2h + 1 - 2k = 9 - 6h + 9 - 6k$$
    $$\Rightarrow 4h + 4k = 16$$
    $$\Rightarrow h + k = 4$$
    So, required locus is $$x + y = 4$$.
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