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Three Dimensional Geometry Test - 17

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Three Dimensional Geometry Test - 17
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  • Question 1
    1 / -0
    If points (1,2), (3 , 5) and (0 , b ) are collinear the value of b is  
    Solution
    $$Area=\dfrac{1}{2}| 1(5-b)+3(b-2)+0(2-5)|$$
    As points are collinear , so area =0
    $$\therefore \dfrac{1}{2}| 1(5-b)+3(b-2)+0(2-5)|=0$$
    $$\Rightarrow 5-b+3b-6=0$$
    $$\Rightarrow=1=2b$$
    $$\therefore b=\dfrac{1}{2}$$
  • Question 2
    1 / -0
    The direction angles of the line $$x = 4z + 3, y = 2 - 3z$$ are $$\alpha, \beta$$ and $$\gamma$$, then $$\cos \alpha + \cos \beta + \cos \gamma =$$ ________.
    Solution

    $$\dfrac {x - 3}{4} = \dfrac {y - 2}{-3} = \dfrac {z}{1}$$

    $$\cos\alpha $$is given by $$\cos\alpha=\dfrac{a}{\sqrt{a^2+b^2+c^2}}$$
     $$\cos\beta=\dfrac{b}{\sqrt{a^2+b^2+c^2}}$$
     $$\cos\gamma=\dfrac{c}{\sqrt{a^2+b^2+c^2}}$$
    $$\cos\alpha=\dfrac{4}{\sqrt{9+16+1}}=\dfrac{4}{\sqrt {26}}$$
    $$\cos\beta=\dfrac{4}{\sqrt{9+16+1}}=\dfrac{-3}{\sqrt {26}}$$
    $$\cos\gamma=\dfrac{1}{\sqrt{9+16+1}}=\dfrac{1}{\sqrt {26}}$$
    $$\cos \alpha = \dfrac {4}{\sqrt {26}} ,\cos \beta = \dfrac {-3}{\sqrt {26}} ,\cos \gamma = \dfrac {1}{\sqrt {26}}$$
    $$\therefore \cos \alpha + \cos \beta + \cos \gamma = \dfrac {2}{\sqrt {26}}$$.
  • Question 3
    1 / -0
    The angle between the lines whose direction cosines are $$\left( \dfrac {\sqrt{3}}{4}, \dfrac {1}{4}, \dfrac {\sqrt{3}}{2} \right)$$ and $$\left( \dfrac {\sqrt{3}}{4}, \dfrac {1}{4}, -\dfrac {\sqrt{3}}{2} \right)$$ is :
    Solution
    Let $$(l_{1},m_{1},n_{1})=\left( \dfrac {\sqrt{3}}{4}, \dfrac {1}{4}, \dfrac {\sqrt{3}}{2} \right)$$ and $$(l_{2},m_{2},n_{2})=\left( \dfrac {\sqrt{3}}{4}, \dfrac {1}{4}, -\dfrac {\sqrt{3}}{2} \right)$$ are the direction cosines of two lines

    Angle between two lines with direction cosines $$(l_{1},m_{1},n_{1})$$ and $$(l_{2},m_{2},n_{2})$$ is given by
    $$ \cos \theta = |l_1 l_2 + m_1m_2 + n_1n_2|$$

    $$ = \begin{vmatrix} \dfrac{\sqrt{3}}{4} \times \dfrac{\sqrt{3}}{4} + \dfrac{1}{4}\times \dfrac{1}{4} + \dfrac{\sqrt{3}}{2} \times \left( \dfrac {- \sqrt {3}}{2}\right) \end{vmatrix} $$
    $$ = \begin{vmatrix} \dfrac{3}{16} + \dfrac{1}{16} - \dfrac{3}{4} \end{vmatrix} = \begin{vmatrix} \dfrac {-2}{4} \end{vmatrix} = \dfrac {1}{2} $$
    $$\Rightarrow \theta = \dfrac {\pi}{3} $$
  • Question 4
    1 / -0
    Direction cosines of the line $$\cfrac { x+2 }{ 2 } =\cfrac { 2y-5 }{ 3 } ,z=-1$$ are ____
    Solution
    $$\dfrac { x+2 }{ 2 } =\dfrac { 2y-5 }{ 3 } , z=-1$$
    $$\Rightarrow$$ $$\dfrac { x+2 }{ 4 } =\dfrac { y-2.5 }{ 3 } , z=-1$$
    $$\therefore$$ direction cosine in $$x$$ $$($$ $$\alpha $$ $$) = 4k$$, direction cosine in $$y$$ $$($$ $$\beta$$ $$)$$ $$= 3k$$, direction cosine in $$z ($$ $$\gamma$$ $$)=0$$
    But $${ \cos }^{ 2 }\alpha +{ \cos }^{ 2 }\beta +{ \cos }^{ 2 }\gamma =1$$
    Therefore, $$16{ k }^{ 2 }+9{ k }^{ 2 }=1$$
    $$ \Rightarrow 25{ k }^{ 2 }=1$$
    $$ \Rightarrow k=\dfrac { 1 }{ 5 } $$
    Therefore, direction cosines of the line are $$\left (\dfrac { 4 }{ 5 } ,\dfrac { 3 }{ 5 } ,0\right)$$.
  • Question 5
    1 / -0
    The following lines are $$\hat { r } =\left( \hat { i } +\hat { j }  \right) +\lambda \left( \hat { i } +2\hat { j } -\hat { k }  \right) +\mu \left( -\hat { i } +\hat { j } -\hat { 2k }  \right) $$
    Solution
    Condition for three lines $$\vec { { r }_{ 1 } } $$ , $$\vec { { r }_{ 2 } } $$ , and $$\vec { { r }_{ 3 } } $$ to be collinear is:
    $$\vec { { r }_{ 1 } } +\lambda \vec { { r }_{ 2 } } +\vec { { \mu r }_{ 3 } } =0$$
    where $$\vec { { r }_{ 1 } } =\left( \vec { i } +\vec { j }  \right) $$
    $$\vec { { r }_{ 2 } } =\left( \vec { i } +2\vec { j } -\vec { k }  \right) $$
    $$\vec { { r }_{ 3 } } =\left( -\vec { i } +\vec { j } -2\vec { k }  \right) $$
    and $$\lambda $$ and $$\mu $$ are scalars
    Hence, the answer is collinear.
  • Question 6
    1 / -0
    $$L$$ and $$M$$ are two points with position vectors $$2\overline { a } -\overline { b } $$ and $$a+2\overline { b } $$ respectively. The position vector of the point $$N$$ which divides the line segment $$LM$$ in the ratio $$2:1$$ externally is
    Solution
    Given points are $$L$$ and $$M$$ are $$2\bar { a } -\bar { b } $$ and $$\bar { a } -2\bar { b } $$,
    as $$N$$ divides in the ratio $$2:1$$ externally,
    $$\therefore M$$ is the mid point of $$L$$ and $$N$$,
    $$\Longrightarrow N=2M-L$$,
    $$\Longrightarrow N=2\bar { a } +4\bar { b } -2\bar { a } +\bar { b } $$,
    $$N=5\bar { b } $$
  • Question 7
    1 / -0
    Direction cosines of ray from $$P(1, -2, 4)$$ to $$Q(-1, 1, -2)$$ are
    Solution
    Given the points are $$P(1,-2,4)$$ and $$Q(-1,1,-2)$$.
    Now the direction ratios of the ray $$PQ$$ are $$(-1-1,1+2,-2-4)=(-2,3,-6)$$.
    The direction cosines of the line $$PQ$$ will be $$\left(\dfrac{2}{\sqrt{2^2+3^2+6^2}},\dfrac{3}{\sqrt{2^2+3^2+6^2}},\dfrac{-6}{\sqrt{2^2+3^2+6^2}}\right)=\left(\dfrac{-2}{7},\dfrac{3}{7},\dfrac{-6}{7}\right)$$.
  • Question 8
    1 / -0
    The direction ratios of the line joining the points $$A(4,-3,7)$$ and $$B(1,3,5)$$ are:
    Solution

  • Question 9
    1 / -0
    If the lines $$x=1+a,y=-3-\lambda a,z=1+\lambda a$$ and $$x=\cfrac { b }{ 2 } ,y=1+b,z=2-b$$ are coplanar, then $$\lambda$$ is equal to
    Solution
    The given lines are $$\cfrac { x-1 }{ 1 } =\cfrac { y+3 }{ -\lambda  } =\cfrac { z-1 }{ \lambda  } =\left( a \right) $$ and $$\cfrac { x-0 }{ 1/2 } =\cfrac { y-1 }{ 1 } =\cfrac { z-2 }{ -1 } =(b)$$
    $$\therefore$$ coplanarity, we must have
    $$\begin{vmatrix} -1 & 4 & 1 \\ 1 & -\lambda  & \lambda  \\ 1/2 & 1 & -1 \end{vmatrix}=0$$
    $$\Rightarrow -1\left( \lambda -\lambda  \right) -4\left( -1-\cfrac { \lambda  }{ 2 }  \right) +\left( 1+\cfrac { \lambda  }{ 2 }  \right) =0$$
    $$4+2\lambda +1+\cfrac { \lambda  }{ 2 } =0\quad \Rightarrow 5+\cfrac { 5\lambda  }{ 2 } =0\quad \therefore \lambda =-2$$
  • Question 10
    1 / -0
    The projection of the join of the two points $$(1,4,5), (6,7,2)$$ on the line whose d.s's are $$(4,5,6)$$ is
    Solution
    $$A(1,4,5)$$, $$B(6,7,2)$$
    $$\begin{array}{l}\overline {AB}  = \,(6 - 1)i + (7 - 4)j + (2 - 5)\widehat k\\\overline {AB}  = 5\widehat i + 3\widehat j - 3\widehat k\\\overrightarrow c  = 4i + 5j + 6\widehat k\end{array}$$
    $$\begin{array}{l} \text{proj. AB} = \dfrac{{\overline {AB} .\overline C }}{{\overline {\left| c \right|} }} = \dfrac{{5(4) + 3(5) + ( - 3)(6)}}{{\sqrt {{4^2} + {5^2} + {6^2}} }}\\\dfrac{{20 + 15 - 18}}{{\sqrt {16 + 25 + 36} }} = \dfrac{{17}}{{\sqrt {77} }}\end{array}$$
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