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Three Dimensional Geometry Test - 25

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Three Dimensional Geometry Test - 25
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  • Question 1
    1 / -0
    If the d.rs of $$OA$$ and $$OB$$ are $$1, -1, -1$$ and $$2, -1, 1$$, then the d.cs of the line perpendicular to both $$OA$$ and $$OB$$ are
    Solution
    The Directionsratios of $$OA$$ will be,
    $$(\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}})$$
    The Direction cosines of $$OB$$ will be,
    $$(\dfrac{2}{\sqrt{6}},\dfrac{-1}{\sqrt{6}}, \dfrac{1}{\sqrt{6}})$$
    Now the lines should be such that the product of Dcs will be $$0$$.
    This is only satisfied by Option C
  • Question 2
    1 / -0
    $$A = (1, 2, 3), B = (4, 5, 7), C = (-4, 3, -6), D =(2, k, 2)$$ are four points. If the lines $$AB$$ and $$CD$$ are parallel, then $$k =$$
    Solution
    Given, $$ A=(1,2,3), B=(4,5,7), C=(-4,3,-6), D=(2,k,2)$$
    For $$AB$$ to be parallel to $$CD$$, their direction cosines should be equal.
    Therefore, Dcs of $$AB$$ will be
    $$(\dfrac{3}{\sqrt{34}},\dfrac{ 3}{\sqrt{34}},\dfrac{4}{\sqrt{34}})$$

    D.C.s of $$CD$$ will be
    $$\left (\dfrac{6}{\sqrt{100+(k-3)^{2}}},\dfrac{k-3}{\sqrt{100+(k-3)^{2}}},\dfrac{8}{\sqrt{100+(k-3)^{2}}}\right)$$
    Now $$\dfrac{6}{\sqrt{100+(k-3)^{2}}}=\dfrac{3}{\sqrt{34}}$$

    $$\dfrac{k-3}{\sqrt{100+(k-3)^{2}}}=\dfrac{ 3}{\sqrt{34}}$$

    $$\dfrac{8}{\sqrt{100+(k-3)^{2}}}=\dfrac{4}{\sqrt{34}}$$

    Solving any of the above can give us the value of $$k$$:
    $$\dfrac{6}{\sqrt{100+(k-3)^{2}}}=\dfrac{3}{\sqrt{34}}$$

    $$4(34)=100+(k-3)^{2}$$

    $$136=100+(k-3)^{2}$$
    $$36=(k-3)^{2}$$
    $$k-3=\pm6$$
    $$k=9$$ and $$k=-3$$
  • Question 3
    1 / -0
    lf $${P}(x,y,z)$$ is a point on the line segment  joining $${A}(2,2,4)$$ and $${B}(3,5,6)$$ such that projection of $$\overline{OP}$$ on axes are $$\displaystyle \dfrac{13}{5},\dfrac{19}{5},\dfrac{26}{5}$$ respectively, then $${P}$$ divide $${A}{B}$$ in the ratio
    Solution
    $$P\left( x,y,z \right) $$ lies on line joining $$A\left( 2,2,4 \right) ,B\left( 3,5,6 \right) $$
    Equation of $$AB:\dfrac { x-2 }{ 3-2 } =\dfrac { y-2 }{ 5-2 } =\dfrac { z-4 }{ 6-4 } $$
    $$AB:\dfrac { x-2 }{ 1 } =\dfrac { y-2 }{ 3 } =\dfrac { z-4 }{ 2 } =t$$
    $$\Rightarrow x=t+2,\quad y=3t+2,\quad z=2t+4$$
    $$\overrightarrow { OP } =\left( t+2 \right) \hat { i } +\left( 3t+2 \right) \hat { j } +\left( 2t+4 \right) \hat { k } $$
    Projection of $$OP$$ on axes are $$\dfrac { 13 }{ 5 } ,\dfrac { 19 }{ 5 } ,\dfrac { 26 }{ 5 } $$
    $$\overrightarrow { OP } .\hat { i } =t+2=\dfrac { 13 }{ 5 } $$
    $$\Rightarrow t=\dfrac { 3 }{ 5 } $$
    $$\overrightarrow { OP } =\left( \dfrac { 13 }{ 5 } ,\dfrac { 19 }{ 5 } ,\dfrac { 26 }{ 5 }  \right) $$
    $$P$$ divides $$AB$$ in $$m:n$$ ratio
    $$\Rightarrow \dfrac { 13 }{ 5 } =\dfrac { m(3)+n(2) }{ m+n }$$
    $$13m+13n=15m+10n$$
    $$3n=2m$$
    $$m:n=3:2$$
  • Question 4
    1 / -0
    The projections of a line segment on $$x,y\ and\ z$$ axes are respectively $$\sqrt{2},3,5$$. The length of the line segment is
    Solution
    Let the line segment be represented in vector form as $$\vec a = a_1\vec i+a_2\vec j+a_3\vec k$$
    Vector along coordinate axes are $$\vec i,\vec j,\vec k$$
    Given that projection of $$\vec a$$ on x axis is $$ \sqrt3$$, that with y axis is $$3$$ and that with z axis is $$5$$.
    $$\Longrightarrow \vec a . \vec i$$ =  $$\sqrt2$$, $$ \therefore \ a_1=\sqrt2$$
    similarily, $$a_2=3,a_3=5$$
    $$\therefore \vec a=\sqrt2\vec i +3\vec j+5\vec k$$
     the length of the line segment=$$ |\vec a| =\sqrt{2+9+25}=\sqrt36=6$$
  • Question 5
    1 / -0
    If the d.rs of two lines are $$1, -2, 3$$ and $$2, 0, 1$$, then the d.rs of the line perpendicular to both the given lines is
    Solution
    $$OA$$ and $$OB$$ are given by $$(1,-2,3), (2,0,1)$$.
    A line that will be perpendicular to both $$OA$$ and $$OB$$ can be obtained by doing the cross product of $$OA$$ with $$OB$$.
    Then, $$n= OA\times OB$$
    Hence, $$n=-2i+5j+4k$$
    Hence, the dr's are $$(-2,5,4)$$.
  • Question 6
    1 / -0
    lf $${A}=(3,1, -2), {B}=(-1,0,1)$$ and $$l,m$$ are the projections of $${A}{B}$$ on the $${y}$$-axis, $$zx$$-plane respectively, then $$3l^{2}-m+1=$$
    Solution
    Given $$A=\left( 3,1,-2 \right) ,B=\left( -1,0,1 \right) $$
    DR's of $$AB$$ are $$\left( 3-(-1),1-0,-2-1 \right) $$
    $$\overrightarrow { AB } =\left( 4,1,-3 \right)$$ or $$ \left( 4\hat { i } +\hat { j } -3\hat { k }  \right) $$
    Projection of $$AB$$ on $$x-$$ axis is $$\overrightarrow { AB } .\hat { i } $$
    $${ a }_{ x }=4$$
    Projection of $$AB$$ on $$y-$$ axis is $$\overrightarrow { AB } .\hat { j } $$
    $$l={ a }_{ y }=1$$
    Projection of $$AB$$ on $$z-$$ axis is $$\overrightarrow { AB } .\hat { k } $$
    $${ a }_{ z }=-3$$
    Projections of $$\overrightarrow { AB } $$ on $$zx$$ plane $$=\sqrt { { { a }_{ x } }^{ 2 }+{ { a }_{ z } }^{ 2 } } $$
    $$=\sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } =5$$
    $$3{ l }^{ 2 }-m+1=3\times 1-5+1=-1$$
  • Question 7
    1 / -0
    The d.rs of the lines $$x=  ay + b$$, $$z= cy + d$$ are:
    Solution
    Given $$x=ay+b$$ and $$z=cy+d$$
    $$\Rightarrow \dfrac{x-b}{a}=y$$ and $$\dfrac{z-d}{c}=y$$
    $$\Rightarrow \dfrac{x-b}{a}=\dfrac{y}{1}=\dfrac{z-d}{c}$$
    Therefore Drs of given line is $$a,1,c$$
  • Question 8
    1 / -0
    Find the angles between the lines, whose direction cosines are give by the equation $${ l }^{ 2 }-{ m }^{ 2 }+{ n }^{ 2 }=0,l+m+n=0$$
    Solution

    Lines are $$l+m+n=0\Rightarrow -l=\left( m+n \right) $$

    and $${ l }^{ 2 }-{ m }^{ 2 }+{ n }^{ 2 }=0\Rightarrow { l }^{ 2 }={ m }^{ 2 }-{ n }^{ 2 }$$

    Solving them gives,

    $${ \left( -\left( m+n \right)  \right)  }^{ 2 }={ m }^{ 2 }+{ n }^{ 2 }+2mn={ m }^{ 2 }-{ n }^{ 2 }\Rightarrow 2n\left( n+m \right) =0$$

    Now, For $$n=0$$ 

    $$m=-l$$, so d.c's $$\displaystyle \left( \dfrac { 1 }{ \sqrt { 2 }  } ,-\dfrac { 1 }{ \sqrt { 2 }  } ,0 \right) $$

    And for $$n=-m$$

    $$l=0$$ so d.c's $$\displaystyle \left( 0,\dfrac { 1 }{ \sqrt { 2 }  } ,-\dfrac { 1 }{ \sqrt { 2 }  }  \right) $$

    Angle between the lines $$\displaystyle \left| \cos { \theta  }  \right| =\left| \dfrac { 1 }{ 2 }  \right| \Rightarrow \theta =\dfrac { \pi  }{ 3 } $$

  • Question 9
    1 / -0
    The projection of the line segment joining $$(0, 0, 0)$$ and $$(5, 2, 4)$$ on the line whose direction ratios are $$2, -3, 6$$ is
    Solution
    The line segment passing through the points $$(0,0,0)$$ and $$(5,2,4)$$ is parallel to $$\vec{a}=5i+2j+4k$$. The dr's of the line (as given ) is $$\vec{l}=2i-3j+6k$$
    Hence the required projection is
    $$=\dfrac{\vec{a}.\vec{l}}{|\vec{l}|}$$

    $$=\dfrac{10-6+24}{\sqrt{4+36+9}}$$

    $$=\dfrac{28}{7}$$

    $$=4$$.
  • Question 10
    1 / -0
    lf $$AB \perp BC$$, then the value of $$\lambda$$ equal, where $$ A(2k,2,3), B(k,1,5),C(3+k,2,1)$$
    Solution
    The dr's of $$AB$$ are $$(k,1,-2)$$
    The dr's of $$BC$$ are $$(3,1,-4)$$
    Since, they are perpendicular, $$AB.BC = 0$$
    $$\Rightarrow 3k + 1 +8 = 0$$
    $$\Rightarrow k = -3$$
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