Three Dimensional Geometry - Class 12 Engineering Maths - Extra Questions

$$l+m+n=0\quad -(i)\\ { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1\quad -(ii)\\ { l }^{ 2 }+{ m }^{ 2 }-{ n }^{ 2 }=0\quad -(iii)\\ (ii)+(iii)\\ 2({ l }^{ 2 }+{ m }^{ 2 })=1\\ 2{ n }^{ 2 }=1\Rightarrow n=\pm \cfrac { 1 }{ \sqrt { 2 }  } \\ { l }^{ 2 }+{ m }^{ 2 }=\cfrac { 1 }{ 2 } \\ l+m=\cfrac { 1 }{ \sqrt { 2 }  } \\ { l }^{ 2 }+{ (\cfrac { 1 }{ \sqrt { 2 }  } -l) }^{ 2 }=\cfrac { 1 }{ 2 } \\ { l }^{ 2 }+{ l }^{ 2 }+\cfrac { 1 }{ 2 } -\sqrt { 2 } l=\cfrac { 1 }{ 2 } \\ 2{ l }^{ 2 }-\sqrt { 2 } l=0\\ l=0,\cfrac { 1 }{ \sqrt { 2 }  } \\ \Rightarrow m=\cfrac { 1 }{ \sqrt { 2 }  } ,0\\ { l }^{ 2 }+{ m }^{ 2 }=\cfrac { 1 }{ 2 } \\ l+m=\cfrac { -1 }{ \sqrt { 2 }  } \\ { l }^{ 2 }+{ (\cfrac { -1 }{ \sqrt { 2 }  } -l) }^{ 2 }=\cfrac { 1 }{ 2 } \\ { l }^{ 2 }+\cfrac { 1 }{ 2 } +{ l }^{ 2 }+\sqrt { 2 } l=\cfrac { 1 }{ 2 } \\ 2{ l }^{ 2 }+\sqrt { 2 } l=0\\ l=0,\cfrac { -1 }{ \sqrt { 2 }  } \\ \Rightarrow \cfrac { -1 }{ \sqrt { 2 }  } ,0$$

Possible values of $$(l,m,n)$$ are:

$$(0,\cfrac { 1 }{ \sqrt { 2 }  } ,\cfrac { -1 }{ \sqrt { 2 }  } ),(\cfrac { 1 }{ \sqrt { 2 }  } ,0,\cfrac { -1 }{ \sqrt { 2 }  } ),(0,\cfrac { -1 }{ \sqrt { 2 }  } ,\cfrac { 1 }{ \sqrt { 2 }  } ),(\cfrac { -1 }{ \sqrt { 2 }  } ,0,\cfrac { 1 }{ \sqrt { 2 }  } )$$

$$\overline a  = 4\widehat i - 3\widehat j - \widehat k$$

Direction ratios of first line are $$2,2,1$$

Direction ratios of second line joining the points $$(3,1,4)(7,2,12)$$ and $$(7-3),(2-1),(12-4)$$

i.e    $$4,1,8$$

Let $$\theta$$ be the angle between the two lines.

$$\begin{array}{l} \therefore \cos  \theta =\frac { { 2\left( 4 \right) +2\left( 1 \right) +1\left( 8 \right)  } }{ { \sqrt { { { \left( 2 \right)  }^{ 2 } }+{ { \left( 2 \right)  }^{ 2 } }+{ { \left( 1 \right)  }^{ 2 } } } \sqrt { { { \left( 4 \right)  }^{ 2 } }+{ { \left( 1 \right)  }^{ 2 } }+{ 8^{ 2 } } }  } }  \\ \left[ { \cos  \theta =\frac { { { a_{ 1 } }{ a_{ 2 } }+{ b_{ 1 } }{ b_{ 2 } }+{ c_{ 1 } }{ c_{ 2 } } } }{ { \sqrt { a_{ 1 }^{ 2 }+b_{ 1 }^{ 2 }+c_{ 1 }^{ 2 } } \sqrt { a_{ 2 }^{ 2 }+b_{ 2 }^{ 2 }+c_{ 2 }^{ 2 } }  } }  } \right]  \\ =\frac { { 8+2+8 } }{ { 3\times 9 } } =\frac { { 18 } }{ { 3\times 9 } } =\frac { 2 }{ 3 }  \\ \therefore \theta ={ \cos ^{ 1 }  }\left( { \frac { 2 }{ 3 }  } \right)  \end{array}$$

$$\begin{array}{l} y=\sqrt { 3x } -5=0\, \, \, \, \, \, \, \, \, \, { m_{ 1 } }=\sqrt { 3 }  \\ \sqrt { 3y } =x-6\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, { m_{ 2 } }=\frac { 1 }{ { \sqrt { 3 }  } }  \\ \tan  \theta =\left| { \dfrac { { { m_{ 1 } }-{ m_{ 2 } } } }{ { 1+{ m_{ 1 } }{ m_{ 2 } } } }  } \right| =\dfrac { 1 }{ { \sqrt { 3 }  } }  \\ \therefore \, \, \theta ={ 30^{ 0 } } \end{array}$$

$$\therefore$$ Angle b/w llines is $$30^0$$ .

Parallel vector of second line = $$3\hat{i}-2\hat{j}-\hat{k}$$

Let the angle between two lines be $$\theta$$

Then, Cos $$\theta = \dfrac{(-2\hat{i}+\hat{j}+3\hat{k}).(3\hat{i}-2\hat{j}-\hat{k})}{|-2\hat{i}+\hat{j}+3\hat{k}||3\hat{i}-2\hat{j}-\hat{k}|} = \dfrac{-11}{\sqrt{14}\sqrt{14}} = \dfrac{-11}{14}$$

Thus, $$\theta = cos^{-1}\dfrac{-11}{14}$$

Parallel vector of second line = $$-\hat{i}+8\hat{j}+4\hat{k}$$

Let the angle between two lines be $$\theta$$

Then, Cos $$\theta = \dfrac{(2\hat{i}+3\hat{j}-3\hat{k}).(-\hat{i}+8\hat{j}+4\hat{k})}{|2\hat{i}+3\hat{j}-3\hat{k}||-\hat{i}+8\hat{j}+4\hat{k}|} = \dfrac{10}{\sqrt{22}\sqrt{81}} = \dfrac{10}{9\sqrt{22}}$$

Thus, $$\theta = cos^{-1}\dfrac{10}{9\sqrt{22}}$$

Therefore, $$\left| \vec { { b }_{ 1 } }  \right| =\sqrt { { \left( 1 \right)  }^{ 2 }+{ \left( -1 \right)  }^{ 2 }+{ \left( -2 \right)  }^{ 2 } } =\sqrt { 6 }$$
and $$\left| \vec { { b }_{ 2 } }  \right| =\sqrt { { \left( 3 \right)  }^{ 2 }+{ \left( -5 \right)  }^{ 2 }+{ \left( -4 \right)  }^{ 2 } } =5\sqrt { 2 }$$
Thus $$\vec { { b }_{ 1 } } .\vec { { b }_{ 2 } } =\left( \hat { i } -\hat { j } -2\hat { k }  \right) .\left( 3\hat { i } -5\hat { j } -4\hat { k }  \right)$$
$$= 1.3-1(-5)-2(-4)$$

$$=3+5+8=16$$
If $$\theta$$ is the angle between the given lines then,
$$\cos { \theta } =\left| \cfrac { \vec { { b }_{ 1 } } .\vec { { b }_{ 2 } }  }{ \left| \vec { { b }_{ 1 } }  \right| \left| \vec { { b }_{ 2 } }  \right|  }  \right|$$
$$\Rightarrow$$ $$\cos { \theta } =\cfrac { 16 }{ \sqrt { 6 } .5\sqrt { 2 }  } =\cfrac { 16 }{ \sqrt { 2 } .\sqrt { 3 } .5\sqrt { 2 }  } =\cfrac { 16 }{ 10\sqrt { 3 }  }$$
$$\Rightarrow$$ $$\cos { \theta } =\cfrac { 8 }{ 5\sqrt { 3 }  }$$
$$\Rightarrow$$ $$\theta=\cos ^{ -1 }{ \left( \cfrac { 8 }{ 5\sqrt { 3 }  }  \right)  }$$

This the required angle between the given pair of lines.

If $$\theta$$ be the angle between the given lines.
Then  $$\cos { \theta } =\left| \cfrac { \vec { { b }_{ 1 } } .\vec { { b }_{ 2 } }  }{ \left| \vec { { b }_{ 1 } }  \right| \left| \vec { { b }_{ 2 } }  \right|  }  \right|$$
where $$\vec { { b }_{ 1 } } =\left( 3\hat { i } +2\hat { j } +6\hat { k }  \right)$$ and $$\vec { { b }_{ 2 } } =\left( \hat { i } +2\hat { j } +2\hat { k }  \right)$$
$$\therefore$$ $$\left| \vec { { b }_{ 1 } }  \right| =\sqrt { { 3 }^{ 2 }+{ 2 }^{ 2 }+{ 6 }^{ 2 } } =7$$
$$\left| \vec { { b }_{ 2 } }  \right| =\sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 2 }^{ 2 } } =3$$
$$\vec { { b }_{ 1 } } .\vec { { b }_{ 2 } } =\left( 3\hat { i } +2\hat { j } +6\hat { k }  \right) .\left( \hat { i } +2\hat { j } +2\hat { k }  \right)$$
$$=3\times 1+2\times 2+6\times 2=3+4+12=19$$
$$\Rightarrow$$ $$\cos { \theta } =\cfrac { 19 }{ 7\times 3 }$$
$$\Rightarrow$$ $$\theta=\cos ^{ -1 }{ \left( \cfrac { 19 }{ 21 }  \right)  }$$

Let $$\vec { { b }_{ 1 } }$$ and $$\vec { { b }_{ 2 } }$$ be the  vectors parallel to the pair of lines,
$$\cfrac{x-2}{2}=\cfrac{y-1}{5}=\cfrac{z+3}{-3}$$ and
$$\cfrac{x+2}{-1}=\cfrac{y-4}{8}=\cfrac{z-5}{4}$$ respectively,
then $$\vec { { b }_{ 1 } } =\left( \hat { 2i } +5\hat { j } -3\hat { k }  \right)$$ and $$\vec { { b }_{ 2 } } =\left( -\hat { i } +8\hat { j } +4\hat { k }  \right)$$
$$\left| \vec { { b }_{ 1 } }  \right|= \sqrt { { \left( 2 \right)  }^{ 2 }+{ \left( 5 \right)  }^{ 2 }+{ \left( -3 \right)  }^{ 2 } } =\sqrt { 38 }$$
$$\left| \vec { { b }_{ 2 } }  \right| =\sqrt { { \left( -1 \right)  }^{ 2 }+{ \left( 8 \right)  }^{ 2 }+{ \left( 4 \right)  }^{ 2 } } =\sqrt { 81 } =9$$
$$\vec { { b }_{ 1 } } .\vec { { b }_{ 2 } } =\left( \hat { 2i } +5\hat { j } -3\hat { k }  \right) .\left( -\hat { i } +8\hat { j } +4\hat { k }  \right)$$
$$=2(-1)+5\times 8+(-3).4=-2+40-12=26$$
The angle $$\theta$$ between the given pair of lines is given by the relation,
$$\Rightarrow \cos { \theta } =\left| \cfrac { \vec { { b }_{ 1 } } .\vec { { b }_{ 2 } }  }{ \left| \vec { { b }_{ 1 } }  \right| \left| \vec { { b }_{ 2 } }  \right|  }  \right|$$
$$\Rightarrow$$ $$\cos { \theta } =\cfrac { 26 }{ 9\sqrt { 38 }  }$$
$$\Rightarrow$$ $$\theta=\cos ^{ -1 }{ \left( \cfrac { 26 }{ 9\sqrt { 38 }  }  \right)  }$$

(ii)

We have,
$$\vec { { b }_{ 1 } } =\left( \hat { 2i } +2\hat { j } +\hat { k }  \right)$$
$$\vec { { b }_{ 2 } } =\left( 4\hat { i } +\hat { j } +8\hat { k }  \right)$$
$$\therefore$$   $$\left| \vec { { b }_{ 1 } }  \right| =\sqrt { { \left( 2 \right)  }^{ 2 }+{ \left( 2 \right)  }^{ 2 }+{ \left( 1 \right)  }^{ 2 } } =\sqrt { 9 } =3$$
$$\left| \vec { { b }_{ 2 } }  \right| =\sqrt { { \left( 4  \right)  }^{ 2 }+{ \left( 1 \right)  }^{ 2 }+{ \left( 8 \right)  }^{ 2 } } =\sqrt { 81 } =9$$
$$\vec { { b }_{ 1 } } .\vec { { b }_{ 2 } }  =\left( \hat { 2i } +2\hat { j } +\hat { k }  \right) .\left( 4\hat { i } +\hat { j } +8\hat { k }  \right)$$
$$=2\times 4+2\times 1+1\times 8=8+2+8=18$$
If  $$\theta$$ is the angle between the given pair of lines, then $$\cos { \theta }  =\left| \cfrac { \vec { { b }_{ 1 } } .\vec { { b }_{ 2 } }  }{ \left|  \vec { { b }_{ 1 } }  \right| \left| \vec { { b }_{ 2 } }  \right|  }   \right|$$
$$\Rightarrow$$ $$\cos { \theta } =\cfrac { 18 }{ 3\times 9 } =\cfrac { 2 }{ 3 }$$
$$\Rightarrow$$ $$\theta =\cos ^{ -1 }{ \left( \cfrac { 2 }{ 3 }  \right)  }$$

Let the angle between the lines be $$\alpha$$

 

Also given that    $$2 l+2 m-n=0\implies n=2 l+2 m\cdots(1)$$

$$m n+n l+l m=0\cdots(2)$$

$$\implies m(2 l+2 m)+l(2 l+2 m)+l m=0$$                       $$(\because \text{from}(1)\ )$$

$$\implies 2 m^2+5 l m+2 l^2=0$$

$$\implies 2 m^2+4 l m+l m+2 l^2=0$$

$$\implies 2 m(m+2 l)+l(m+2 l)=0$$

$$\implies (2 m+l)(m+2 l)=0$$

$$\implies m=-\dfrac{l}{2},-2 l$$

$$n=2 l+2 m=2 l+2\bigg(-\dfrac{l}{2}\bigg),2 l+2(-2 l)=2 l-l,2 l-4 l=l,-2 l$$

So the direction cosines of two lines are $$<l,-\dfrac{l}{2},l>$$ and $$<l,-2 l,-2 l>$$

So the direction ratios will be $$<2,-1,2>$$ and $$<1,-2,-2>$$

As we know that

If $$\theta$$ is the angle between the two lines having direction ratios as $$<a,b,c>$$ and $$<d,e,f>$$  

                                                        then $$\cos \theta=\dfrac{a d+be +c f}{\sqrt{a^2+b^2+c^2}\sqrt{d^2+e^2+f^2}}$$

Hence $$\cos \alpha=\dfrac{(2)(1)+(-1)(-2)+2(-2)}{\sqrt{2^2+(-1)^2+2^2}\sqrt{1^2+(-2)^2+(-2)^2}}=\dfrac{2+2-4}{3\times 3}=0$$

$$\implies \alpha=\dfrac{\pi}{2}$$

So the angle between the two lines is $$\dfrac{\pi}{2}$$

$$\dfrac{x-2}{3} = \dfrac{y+1}{-2} = \dfrac{z-2}{0}$$

$$\dfrac{x-1}{1}=\dfrac{y+\dfrac{3}{2}}{\dfrac{3}{2}}=\dfrac{z+5}{2}$$

Now, angle b/w the lines

$$cos\theta = |\dfrac{3+(-3)+0}{\sqrt{9+4} \sqrt{1+\dfrac{9}{4}+4}}|$$

$$cos\theta=0$$

$$\theta=90^{o}$$

Parallel vector of second line = $$\hat{i}+\hat{j}+2\hat{k}$$

Let the angle between two lines be $$\theta$$

Then, Cos $$\theta = \dfrac{(3\hat{i}+5\hat{j}+4\hat{k}).(\hat{i}+\hat{j}+2\hat{k})}{|3\hat{i}+5\hat{j}+4\hat{k}||\hat{i}+\hat{j}+2\hat{k}|} = \dfrac{16}{\sqrt{50}\sqrt{6}} = \dfrac{16}{5\sqrt{12}}$$

Thus, $$\theta = cos^{-1}\dfrac{16}{5\sqrt{12}}$$

$$=\dfrac{3\times 1+2\times 2+6\times 2}{\sqrt{3^2+2^2+6^2}.\sqrt{1^2+2^2+2^2}}$$

$$=\dfrac{3+4+12}{7.3}$$

$$\cos\theta=\dfrac{19}{21}\Rightarrow\theta=\cos^{-1}\left(\dfrac{19}{21}\right)$$