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Three Dimensional Geometry Test - 18

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Three Dimensional Geometry Test - 18
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  • Question 1
    1 / -0
    Vector equation of the line $$6x - 2 = 3y + 1 = 2z - 2$$ is 
    Solution

    $$6x - 2 = 3y + 1 = 2z - 2$$


    $$\dfrac{{x - \dfrac{1}{3}}}{{\dfrac{1}{6}}} = \dfrac{{y + \dfrac {1}{3}}}{{{\dfrac {1}{3}}}} = \dfrac{{z - 1}}{{\dfrac{1}{2}}}$$


    $$\overrightarrow r  = \overrightarrow a  + \lambda \,\overrightarrow b $$


    $$a = \left( {{1 \over 3}, - {1 \over 3},1} \right),b = \left( {{1 \over 6},{1 \over 3},{1 \over 2}} \right)$$


    $$velue\,e{q^n}\,is\, = \left( {1,2,3} \right)$$


    $$\mathop {{\text{ }}r}\limits^ \to   = \dfrac{1}{3}\widehat i - \dfrac{1}{3}\widehat j + \widehat k + \lambda \left( {\dfrac{1}{6}\widehat i + \dfrac{1}{3}\widehat j + \dfrac{1}{2}\widehat k} \right)$$


    $$\mathop {{\text{ }}r}\limits^ \to   = \dfrac{1}{3}\widehat i - \dfrac{1}{3}\widehat j + \widehat k + \lambda \left( {\widehat i + 2\widehat j + 3\widehat k} \right)$$


  • Question 2
    1 / -0
    The equation of the plane passing through $$(2, -3, 1)$$ and is normal to the line joining the points $$(3, 4, -1)$$ and $$(2, -1, 5)$$ is given by
    Solution
    $$A(3,4,-1)$$ and  $$B(2,-1,5)$$
    $$ \overrightarrow { AB } =-\hat { i } -5\hat { j } +6\hat { k } $$
    $$\overrightarrow { AB } $$ is normal to the required plane $$\Rightarrow $$ Directions of normal $$(-1,-5,6)$$
    $$\therefore$$ Equation , $$-x-5y+6z=k$$
    Point $$(2,-3,1)$$ passes through the plane,
    $$\therefore -2+15+6=k\quad \Rightarrow k=19\\ \therefore -x-5y+6z=19\\ x+5y-6z+19=0$$
  • Question 3
    1 / -0
    The points with position vectors $$60\hat{i}+3\hat{j}$$, $$40\hat{i}-8\hat{j}$$, $$a\hat{i}-52\hat{j}$$  are collinear if
    Solution
    suppose $${60i + 3j}$$ , $${40i - 8j}$$ and $${ai - 52j}$$ is the three position of vector $$A,B,C$$

    $$\begin{array}{l} \overrightarrow { AB } =\left( { 40i-8j } \right) -\left( { 60i+3j } \right)  \\ \overrightarrow { AB } =-20i-11j \\ \overrightarrow { BC } =\left( { ai-52j } \right) -\left( { 40i-8j } \right)  \\ \overrightarrow { BC } =\left( { a-40 } \right) i-44j \\ \left( { a-40 } \right) i-44j=m\left( { -20i-11j } \right)  \\ \left( { a-40 } \right) i-44j=-20im-11jm \\ -44=-11m \\ m=\frac { { -44 } }{ { -11 } }  \\ m=4 \\ a-40=-20m \\ a-40=-20\left( 4 \right)  \\ a=-80+40 \\ a=-40 \end{array}$$
  • Question 4
    1 / -0
    Direction cosines $$l, m, n$$ of two lines are connected by the equation $$l-5m+3n=0$$ and $$7l^{2}+5m^{2}-3n^{2}=0$$. The direction cosines of one of the lines are
    Solution
    Consider the problem 
    Given 

    $$l-5m+3n=0$$
    $$l=5m-3m$$   ----  $$(i)$$

    And 
    $$7l^2+5m^2-3n^2=0$$  ----  $$(ii)$$

    From $$(i)$$ put the value of $$l$$ in equation $$(ii)$$
    Then,
    $$7(5m-3n)^2+5m^2-3n^2=0$$
    Hence 
    $$30(2m-n)(3m-2n)=0$$
    thus,
    $$2m=n$$ and $$3m=2n$$
    therefore, 
    $$\dfrac m 1=\dfrac n 2=\dfrac{(5m-3n)}{5-2.3}=\dfrac{1}{\sqrt 6}$$
    And
    $$\dfrac{m}{2}=\dfrac n 3=\dfrac{(5m-3n)}{5.2-3.3}=\dfrac{1}{\sqrt {14}}$$

    Hence the required Directions cosines of line are 

    $$ - \dfrac{1}{\sqrt 6 },\dfrac{1}{{\sqrt 6 }},\dfrac{2}{{\sqrt 6 }}$$
  • Question 5
    1 / -0
    Can $$\dfrac{1}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}, \dfrac{-2}{\sqrt{3}}$$ be the direction cosines of any directed line?
    Solution
    No, they can not be the direction cosines of any directed line.
    As the sum of square of them is not $$1$$.
    As $$\left(\dfrac{1}{\sqrt{3}}\right)^2+$$$$\left(\dfrac{2}{\sqrt{3}}\right)^2$$$$+\left(\dfrac{-2}{\sqrt{3}}\right)^2$$
    $$=\dfrac{1+4+4}{3}$$
    $$=3$$.
  • Question 6
    1 / -0
    If the $$d.c's$$ of two lines are connected by the equations $$l + m + n = 0, l^2 + m^2 - n^2 = 0$$, then angle between the lines is
    Solution
    Given equation are,
    $$\begin{array}{l} l+m+n=0----------(1) \\ { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }=0---------(2) \\ from\, \, \, \, equ\, \, (1)\, ,\, \, \, l=-\, \, (m+n) \\ value\, \, of\, \, \, l....... \\ substituting\, \, in\, \, (2) \\ { (m+n)^{ 2 } }+{ m^{ 2 } }-{ n^{ 2 } }=0 \\ \Rightarrow { m^{ 2 } }+{ n^{ 2 } }+2mn+{ m^{ 2 } }-{ n^{ 2 } }=0 \\ \Rightarrow 2{ m^{ 2 } }+2mn=0 \\ \Rightarrow 2m(m+n)=0 \\ \Rightarrow m=0\, \, \, \, and\, \, \, (m+n)=0 \\ Now, \\ m=0\, ,\, substituting\, in\, \, (1) \\ l+m+n=0 \\ l+n=0,\Rightarrow l=-n \\ Dr.s\, of\, the\, \, first\, line\, are\, \, (1,0,-1) \\ and, \\ m+n=0\, \, \, \Rightarrow \, m=-n \\ substituting\, \, \, in\, \, (1)\,  \\ l+m+n=0 \\ l=0 \\ D'rs\, of\, the\, \, { { second } }\, \, line\, are\, \, (0,1,-1) \end{array}$$
    $$\begin{array}{l} Lets\, \, \theta \, be\, the\, angle\, \, between\, the\, \, line, \\ \cos  \theta =\dfrac { { { a_{ 1 } }{ a_{ 2 } }+{ b_{ 1 } }{ b_{ 2 } }+{ c_{ 1 } }{ c_{ 2 } } } }{ \begin{array}{l} \sqrt { { a_{ 1 } }^{ 2 }+{ b_{ 1 } }^{ 2 }+{ c_{ 1 } }^{ 2 } } \, \, \sqrt { { a_{ 2 } }^{ 2 }+{ b_{ 2 } }^{ 2 }+{ c_{ 2 } }^{ 2 } }  \\ =\dfrac { { 0+0+1 } }{ { \sqrt { 2 } .\sqrt { 2 }  } } =\frac { 1 }{ 2 }  \\ \, \cos  \theta =\dfrac { 1 }{ 2 }  \\ \therefore \, \, \, \theta =\dfrac { \pi  }{ 3 } \,  \end{array} }  \end{array}$$
    so that the  correct option is A.
  • Question 7
    1 / -0
    The direction cosines of a line which is equally inclined to axes, is given by
    Solution
    Consider the problem 

    Let, $$l=m=n=\cos \alpha$$
    then, 
    $$l^2+m^2+n^2=\cos ^2\alpha +\cos ^2\alpha+\cos ^2\alpha  $$

    $$1=3\cos ^2 \alpha $$

    $$\cos ^2\alpha =\dfrac{1}{3}$$

    $$\cos \alpha =\pm \dfrac{1}{\sqrt 3}$$
    thus, 
    $$l=m=n=\pm\dfrac{1}{\sqrt 3}$$
  • Question 8
    1 / -0
    The equation of the plane containing the line 
    $$\vec r = \hat i + \hat j + t\left( {2\hat i + \hat j + 4\hat k} \right)$$, is
    Solution
    The line 
    $$\vec r = \hat i + \hat j + t(2\hat i + \hat j + 4\hat k)$$
    passes through the point $$(1,1,0)$$ and parallel to vector
    $$\vec b = 2\hat i + \hat j + 4\hat k$$
    taking plane 
    $$\vec r \cdot (\hat i + 2\hat j - \hat k) = 3$$          (i)
    putting 
    $$\vec r =( \hat i + \hat j + 0\hat k)$$         in equation   (i)
    $$(\hat i + \hat j + 0\hat k) \cdot (\hat i + 2j\hat k - k) = 3$$
    $$1+2+0=3$$
    $$3=3$$
    therefore the plane (1) contains point $$(1,1,0)$$ 
    also 
    $$2 \times 1 + 1 \times 2 + 4 \times ( - 1)$$
    $$=2+2-4$$
    $$=4-4$$
    $$=0$$
    i.e perpendicular to plans(1) is perpendicular to the given line
    therefore the plane is  $$\vec r \cdot (\hat i + 2\hat j - \hat k) = 3$$
  • Question 9
    1 / -0
    A line passes through the points $$(6, -7, -1)$$ and $$(2, -3, 1)$$. The direction cosines of the line so directed that the angle made by it with the positive direction of x-axis is acute, is?
    Solution
    Consider the problem 
    Let $$l,m,n$$ are direction cosines of the given line. 
    then as it made an acute angle with $$x-axis$$,
    Therefore, $$l>0$$
    The line passes through $$(6,-7,-1)$$ and $$(2,-3,1)$$
    Therefore, its direction ratios are 
    $$6-2,-7+3,-1-1$$ or $$2,-2,-1$$
    Hence direction cosines of the line are given by $$\dfrac{2}{3},-\dfrac{2}{3},-\dfrac{1}{3}$$.
  • Question 10
    1 / -0
    If $$P$$ be the point $$(2,6,3)$$ then the equation of the plane trough $$P$$, at right angles to $$OP$$, where  $$'O'$$ is the origin is
    Solution
    $$P(2,6,3)$$, $$\overrightarrow { OP } =2\hat { i } +6\hat { j } +3\hat { k } $$(direction of normal 
    $$\therefore$$ Equation of plane $$\Rightarrow \overrightarrow { r } .\overrightarrow { n } =\overrightarrow { p } .\overrightarrow { n } \\ \overrightarrow { r } .(2\hat { i } +6\hat { j } +3\hat { k } )=49$$
    $$ \therefore 2x+6y+3z=49$$ -Equation of Plane.
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