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Three Dimensional Geometry Test - 43

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Three Dimensional Geometry Test - 43
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  • Question 1
    1 / -0
    If the direction cosine of a directed line be $$a, 3a, 7a$$ then $$a =$$
    Solution
    Give $$a,3a,7a$$ be the direction cosines of a directed line.
    Then from the property of direction cosines we get
    $$a^2+(3a)^2+(7a)^2=1$$
    or, $$59a^2=1$$
    or, $$a=\pm\dfrac{1}{\sqrt{59}}$$.
  • Question 2
    1 / -0
    The direction ratios of the line $$6x - 2 = 3y + 1 = 2z - 2$$ are 
    Solution
    $$6x-2=3y+1=2x-2$$

    $$6(x-\dfrac26)=3(y+\dfrac13)=2(x-\dfrac22)$$

    $$\dfrac{(x-\dfrac13)}1=\dfrac{(y+\dfrac13)}2=\dfrac{(x-1)}3$$


    Line will be passing through the poits, $$(\dfrac13,-\dfrac13,1)$$

    and parallel to the line having direction ratios is  $$1,2,3$$

    option $$C$$ will be the correct answer
  • Question 3
    1 / -0
    If $$l,m,n$$ are d.c's of vector $$\overline {OP}$$ then maximum value of $$lmn$$ is
    Solution
    Let the $$3$$ number be $$l^2, m^2, n^2$$
    we know that
    $$AM \geq GM$$
    i.e. Arithemetic Mean $$\geq$$ Geometric Mean
    $$\therefore \dfrac{l^2 + m^2 + n^2}{3} \geq \sqrt[3]{(lmn)^2}$$
    $$(lmn)^\dfrac{2}{3} \leq \dfrac{1}{3}$$          $$(l^2 + m^2 + n^2 =1)$$
    $$lmn \leq \dfrac{1}{3\sqrt{3}}$$
    $$\therefore$$ option C is the correct answer
  • Question 4
    1 / -0
    If O is the origin and the coordinates of P is $$(1, 2, -3)$$, then find the equation of the plane passing through P and perpendicular to OP.
    Solution
    Consider the problem 

    Origin $$O(0,0,0)$$ and $$P(1,2,-3)$$
    then direction ratios of $$OP$$

    $$\begin{array}{l} ({ x_{ 2 } }-{ x_{ 1 } }),({ y_{ 2 } }-{ y_{ 1 } }),({ z_{ 2 } }-{ z_{ 1 } }) \\ (1-0),(2-0),(-3-0) \\ \Rightarrow (1,2,-3) \end{array}$$

    Equation of plane passing through $$({x_1},{y_1},{z_1})$$
    $$a(x - {x_1}) + b(y - {y_1}) + c(z - {z_1}) = 0$$

    where $$a,b,c$$ are direction ratios of normal are $$1,2,-3$$
    And the point $$P(1,2,-3)$$
    therefore equation of required plane 

    $$1(x - 1) + 2(y - 2) - 3(z - ( - 3)) = 0$$

    $$x - 1 + 2y - 4 - 3z - 9 = 0$$

    $$x + 2y - 3z - 14 = 0$$

    $$x + 2y - 3z  = 14$$


  • Question 5
    1 / -0
    The direction cosines of two lines are related by $$l+m+n=0$$ and $$al^2+bm^2+cn^2=0$$. The lines are parallel if
    Solution
    Let $${ l }_{ 1 }{ m }_{ 1 }{ n }_{ 1 }$$ and  $${ l }_{ 2 }{ m }_{ 2 }{ n }_{ 2 }$$ be the direction cosines.
    $$l+m+m=0\Rightarrow l=-m-n\\ a{ l }^{ 2 }+b{ m }^{ 2 }+c{ n }^{ 2 }=0\\ a({ m }^{ 2 }+{ n }^{ 2 }+2mn)+b{ m }^{ 2 }+c{ n }^{ 2 }=0\\ (a+b){ m }^{ 2 }+(a+c){ n }^{ 2 }+2amn=0$$
    $$ (a+b){ (\cfrac { m }{ n } ) }^{ 2 }+2a(\cfrac { m }{ n } )+(a+c)=0$$ -Quadratic in $$\cfrac { m }{ n } $$
    $$ \therefore$$ product of roots$$=\cfrac { { m }_{ 1 } }{ { n }_{ 1 } } \times \cfrac { { m }_{ 2 } }{ { n }_{ 2 } } =\cfrac { a+c }{ a+b } $$
    Similarly,
    $$m=-l-n\\ a{ l }^{ 2 }+b({ l }^{ 2 }+{ n }^{ 2 }+2ln)+c{ n }^{ 2 }=0\\ (a+b){ l }^{ 2 }+(b+c){ n }^{ 2 }+2bln=0$$
    $$ (a+b){ (\cfrac { l }{ n } ) }^{ 2 }+2b(\cfrac { l }{ n } )+(b+c)=0-$$ Quadratic in $$\cfrac { l }{ n } $$
    Product of roots $$ =\cfrac { { l }_{ 1 } }{ { n }_{ 1 } } \times \cfrac { { l }_{ 2 } }{ { n }_{ 2 } } =\cfrac { b+c }{ a+b } $$
    Since the two lines are perpendicular,
    $$\therefore { l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+{ n }_{ 1 }{ n }_{ 2 }=cos{ 90 }\\ { l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+{ n }_{ 1 }{ n }_{ 2 }=0\\ \cfrac { { l }_{ 1 }{ l }_{ 2 } }{ { n }_{ 1 }{ n }_{ 2 } } +\cfrac { { m }_{ 1 }{ m }_{ 2 } }{ { n }_{ 1 }{ n }_{ 2 } } +1=\cfrac { 0 }{ { n }_{ 1 }{ n }_{ 2 } } \\ \cfrac { a+c }{ a+b } +\cfrac { b+c }{ a+b } +1=0\\ 2(a+b+c)=0\\ a+b+c=0$$
  • Question 6
    1 / -0
    The direction cosines of a line segment AB are $$ - \dfrac{2}{{\sqrt {17} }},\dfrac{3}{{\sqrt {17} }}, - \dfrac{2}{{\sqrt {17} }}$$. If $$AB=\sqrt {17} $$ and the coordinates of A are $$(3,-6,10)$$, then the coordinates of B are 
    Solution
    DC of the line $$AB <\dfrac{-2}{\sqrt{17}}, \dfrac{3}{\sqrt{17}}, \dfrac{-2}{\sqrt{17}}>$$
    Since $$AB = \sqrt{17}$$
    $$\therefore$$ DR of the line AB  $$=<-12, 3, -2>$$
    DR of the line $$AB$$ $$=$$ Co-ordinates of $$B$$ - co-ordinates $$A$$
    Let co-ordinates of $$B$$ be $$(x, y, z)$$
    $$\Rightarrow \therefore <-2, 3, -2> = <x - 3, y + 6, z - 10>$$
    Co-ordinates of $$A$$ :
    $$x - 3 = -2$$
    $$\therefore x = 1$$
    $$y + 6 = 3$$
    $$\therefore y = -3$$
    $$z - 10 = -2$$
    $$\therefore z = 8$$
    $$\therefore <x, y, z> = <1, -3, 8>$$
    option D is the correct option
  • Question 7
    1 / -0
    The foot of the perpendicular drawn from the origin to a plane is $$(1, 2, -3)$$. Find the equation of the plane.
    Solution

    Given:

    Let the equation of plane passing through $$\left ( 1, 2, 3 \right ) $$ and perpendicular to $$ OA $$.

    $$ A = \hat{i}+2\hat j-3\hat k $$

    $$\vec{n} = \overrightarrow{OA}=\hat i+2\hat j-3\hat k $$

    $$a=|\hat i+2\hat j-3\hat k|=\sqrt{1+4+9}=\sqrt{14}$$

    $$\hat{n}=\dfrac{\hat i+2\hat j-3\hat k}{\sqrt{14}}$$

    The equation of the required plane is  $$ \vec{r}\cdot \vec{n}=a$$

    $$\Rightarrow \vec{r}\cdot \left (\dfrac{\hat i+2\hat j-3\hat k}{\sqrt{14}} \right ) =\sqrt{14}$$

    $$\Rightarrow \vec{r} \cdot \left (\hat i+2\hat j-3\hat k \right ) =14 $$

    $$\Rightarrow  x + 2y - 3z = 14 $$

    Thus, the equation of the plane is $$ x + 2y - 3z = 14 $$.

     

  • Question 8
    1 / -0
    If a  line has the direction ratios $$4, -12,18$$ then find its direction cosines.
    Solution
    DR of a line $$<4, -12, 18>$$
    DC  $$=\dfrac{4}{\sqrt{4^2 + (-12)^2 + (18)^2}}, \dfrac{-12}{\sqrt{4^2 + (-12)^2 + (18)^2}}, \dfrac{18}{\sqrt{4^2 + (-12)^2 + (18)^2}}$$
    $$= \dfrac{4}{22}, \dfrac{-12}{22}, \dfrac{18}{22}$$
    $$= <\dfrac{2}{11}, \dfrac{-6}{11}, \dfrac{9}{11}>$$
  • Question 9
    1 / -0
    A mirror and a source of light are situated at the origin $${\rm O}$$ and at a point on $${\rm O}X$$, respectively. A ray of light from the source strikes the mirror and is reflected. If the direction ratios of the normal to the plane are $$1,\, - 1,\,1$$, then find the DCs of the reflected ray.
    Solution
    Suppose the source of light be situated at $$A(a,0,0)$$
    where $$a\neq 0$$, $$AO=$$ incident ray, $$OB=$$ reflected ray.
    $$ON=$$ normal to the mirror at $$0$$.
    $$\therefore \angle AON=\angle NOB=\dfrac{\theta}{2}$$   (say)
    Direction ratios of normal place are $$(1,-1,1)$$
    $$\therefore $$ Direction cosine of normal : $$\left(\dfrac{1}{\sqrt{1^2+(-1)^2+1^2}},\dfrac{1}{\sqrt{1^2+(-1)^2+1^2}},\dfrac{1}{\sqrt{1^1(-1)^2+1^2}}\right)$$
    $$=\left(\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)$$
    Hence, $$\cos \left(\dfrac{\theta}{2}\right)=\dfrac{1}{\sqrt{3}}$$
    Direction ratio of $$AO$$ are $$(\alpha,0,0)$$ and direction cosines are $$(1,0,0)$$
    $$\therefore$$ Now let $$l, m, n$$ be direction cosines of reflected ray $$OB$$.
    $$\therefore \dfrac{1+1}{2\cos\dfrac{\theta}{2}}=\dfrac{1}{\sqrt{3}}$$
    $$\Rightarrow 1+1=\dfrac{2}{\sqrt{3}}\cos \dfrac{\theta}{2}=\dfrac{2}{3}$$
    $$\Rightarrow 1=\dfrac{2}{3}-1=\dfrac{-1}{3}$$
    $$\dfrac{m+o}{2\cos\left(\dfrac{\theta}{2}\right)}=\dfrac{-1}{\sqrt{3}}$$
    $$\Rightarrow m=\dfrac{-2}{\sqrt{3}}\cos\dfrac{\theta}{2}=\dfrac{-2}{3}$$
    $$\dfrac{n+o}{2\cos\left(\dfrac{\theta}{2}\right)}=\dfrac{1}{\sqrt{3}}$$
    $$\Rightarrow n=\dfrac{2}{\sqrt{3}}\cos\dfrac{\theta}{2}=\dfrac{2}{3}$$
    $$\therefore$$ Direction cosines of reflected ray are $$\left(\dfrac{-1}{3},\dfrac{-2}{3},\dfrac{2}{3}\right)$$

  • Question 10
    1 / -0
    The direction`cosines of a line equally inclined to three mutually perpendicular lines having D.C.'s as $${\ell _1}{m_1}{n_1}:{\ell _2}{m_2}{n_2}:{\ell _3}{m_3}{n_3}\,\,$$ are 
    Solution

    We know that, $${{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1$$

    But given that,

      $$ \alpha$$ =$$\beta$$ = $$\gamma$$

     $$ {{\cos }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\cos }^{2}}\alpha =1 $$

     $$ 3{{\cos }^{2}}\alpha =1 $$

     $$ \cos \alpha =\pm \dfrac{1}{\sqrt{3}} $$

    Hence, direction cosine are $$\left( \pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{3}} \right)$$


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