Three Dimensional Geometry Test - 40

Given equation of lines are,
$$x - 1 = \dfrac{2y + 3}{3} = \dfrac{z + 5}{2}$$
and $$x = 3r +2; y = - 2r - 1; z = 2$$
$$\Rightarrow \dfrac{x - 1}{1} = \dfrac{y + \dfrac{3}{2}}{\dfrac{3}{2}} = \dfrac{z + 5}{2}$$
and $$\dfrac{x - 2}{3} = r: \dfrac{y + 1}{-2} = r : \dfrac{z - 2}{0} = r$$
$$\Rightarrow \dfrac{x - 1}{1} = \dfrac{y + \dfrac{3}{2}}{\dfrac{3}{2}} = \dfrac{z + 5}{2}$$
and $$\dfrac{x - 2}{3} = \dfrac{y + 1}{-2} = \dfrac{z - 2}{0}$$
DR's of lines Ist and IInd lines are $$\left( 1 , \dfrac{3}{2} , 2 \right )$$ and $$(3, -2, 0)$$.
The angle between two straight lines is
$$\cos \theta = \dfrac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$$
$$= \dfrac{1 \times 3 - \dfrac{3}{2} \times 2 + 2 \times 0}{\sqrt{ 1^2 + \left( \dfrac{3}{2} \right )^2 + 2^2} \sqrt{3^2 + 2^2 + 0^2}}$$
$$= \dfrac{3 - 3 + 0}{\sqrt{1 + \dfrac{9}{4} + 4} \sqrt{9 + 4}}$$
$$= \dfrac{0}{\sqrt{\dfrac{29}{4}} \sqrt{13}}$$
$$\Rightarrow \cos \theta = 0$$
$$\Rightarrow \theta = \dfrac{\pi}{2}$$

Given equation of lines

Since, $$\alpha = \beta = \gamma \Rightarrow \cos^{2}\alpha + \cos^{2}\alpha + \cos^{2}\alpha = 1$$
$$\Rightarrow \alpha + \cos^{-1}\left (\pm \dfrac {1}{\sqrt {3}}\right )$$
So, there are four line whose $$DC's$$ are
$$\left (\dfrac {1}{\sqrt {3}}, \dfrac {1}{\sqrt {3}}, \dfrac {1}{\sqrt {3}}\right ), \left (\dfrac {-1}{\sqrt {3}}, \dfrac {1}{\sqrt {3}}, \dfrac {1}{\sqrt {3}}\right ), \left (\dfrac {1}{\sqrt {3}}, \dfrac {-1}{\sqrt {3}}, \dfrac {1}{\sqrt {3}}\right ) \left (\dfrac {1}{\sqrt {3}}, \dfrac {1}{\sqrt {3}}, \dfrac {-1}{\sqrt {3}}\right )$$.

By solving two equations we get
$$\left( { l }_{ 1 },{ m }_{ 1 },{ n }_{ 1 } \right) =\left( 2\sqrt { 2 } -3,-\sqrt { 2 } ,1 \right) $$
$$\left( { l }_{ 2 },{ m }_{ 2 },{ n }_{ 2 } \right) =\left( -2\sqrt { 2 } -2,\sqrt { 2 } ,1 \right) $$
$$ { l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+{ n }_{ 1 }{ n }_{ 2 }=0$$
Therefore, the angle between them is $$\cfrac { \pi  }{ 2 } $$.

The given line is parallel to the vector $$n=\hat { i } -\hat { j } +2\hat { k } $$. The required plane passes through the point $$(2,3,1)$$ ie., $$2\hat { i } +3\hat { j } +\hat { k } $$
so, its equation is,
$$r-\left( 2\hat { i } -3\hat { j } +\hat { k }  \right) \left( \hat { i } -\hat { j } +2\hat { k }  \right) =0\quad $$
$$\Rightarrow r.\left( \hat { i } -\hat { j } +2\hat { k }  \right) =1$$

Let the cube be of side $$a$$. Then, $$O\left( 0,0,0 \right) , D\left( a,a,a \right) , B\left( 0,a,0 \right) $$ and $$G\left( a,0,a \right) $$
Now, equations of diagonals $$OD$$ and $$BG$$ are
$$\dfrac { x }{ a } =\dfrac { y }{ a } =\dfrac { z }{ a } $$ and $$\dfrac { x }{ a } =\dfrac { y-a }{ -a } =\dfrac { z }{ a } $$, respectively.
Hence, angle between $$OD$$ and $$BG$$ is
$$\cos ^{ -1 }{ \left( \dfrac { { a }^{ 2 }-{ a }^{ 2 }+{ a }^{ 2 } }{ \sqrt { 3{ a }^{ 2 } } \cdot \sqrt { 3{ a }^{ 2 } }  }  \right)  } =\cos ^{ -1 }{ \left( \dfrac { 1 }{ 3 }  \right)  } $$

We need to find value of $$\cos { 2\alpha  } +\cos { 2\beta  } +\cos { 2\gamma  } $$
It is further equal to $$\cos ^{ 2 }{ \alpha  } -1+\cos ^{ 2 }{ \beta  } -1+\cos ^{ 2 }{ \gamma  } -1$$
$$=2\left( \cos ^{ 2 }{ \alpha  } +\cos ^{ 2 }{ \beta  } +\cos ^{ 2 }{ \gamma  }  \right) -3$$
$$=2(1)-3=2=-1\left( \because { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1 \right) $$

Given, $$OP = 12$$ units
and $$\alpha = 45^{\circ}$$ and $$\beta = 60^{\circ}$$
Let direction cosines of line $$OP$$ is $$< l, m, n >$$
$$\therefore l^{2} + m^{2} + n^{2} = 1$$
$$\Rightarrow \cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1$$
$$\Rightarrow \cos^{2} 45^{\circ} + \cos^{2} 60^{\circ} + \cos^{2}\gamma = 1$$
$$\Rightarrow \left (\dfrac {1}{\sqrt {2}}\right )^{2} + \left (\dfrac {1}{2}\right )^{2} = 1 - \cos^{2}\gamma$$
$$\Rightarrow \sin^{2}\gamma = \dfrac {1}{2} + \dfrac {1}{4} = \dfrac {3}{4}$$
$$\Rightarrow \sin \gamma = \dfrac {\sqrt {3}}{2} = \sin 60^{\circ}$$
Thus $$ \gamma = 60^{\circ}$$
So, the coordinates of $$P\equiv (l\cdot OP, m\cdot OP, n\cdot OP)$$
$$\equiv (OP\cos \alpha, OP\cos \beta, OP\cos \gamma)$$
$$\equiv \left (12\cdot \dfrac {1}{\sqrt {2}}, 12\cdot \dfrac {1}{2}, 12\cdot \dfrac {1}{2}\right )$$
$$\equiv (6\sqrt {2}, 6, 6)$$

Since, $$DC's$$ of a line are $$\left (\dfrac {1}{c}, \dfrac {1}{c}, \dfrac {1}{c}\right )$$
$$\because l^{2} + m^{2} + n^{2} = 1$$
$$\therefore \left (\dfrac {1}{c}\right )^{2} + \left (\dfrac {1}{c}\right )^{2} + \left (\dfrac {1}{c}\right )^{2} = 1$$
$$\Rightarrow 1 + 1 + 1 = c^{2} \Rightarrow c^{2} = 3$$
$$\Rightarrow c = \pm \sqrt {3}$$.

A line makes angle $$\theta $$ with $$x$$-axis and $$z$$-axis and $$\beta$$ with $$y$$-axis.
$$ l=\cos { \theta  } , m=\cos { \beta  } , n=\cos { \theta  } $$      ....$$\text{Since}\ { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1$$
$$\Rightarrow \cos ^{ 2 }{ \theta  } +\cos ^{ 2 }{ \beta  } +\cos ^{ 2 }{ \theta  } =1$$
$$\Rightarrow 2\cos ^{ 2 }{ \theta  } =1-\cos ^{ 2 }{ \beta  } =\sin ^{ 2 }{ \beta  } $$
$$\Rightarrow 2\cos ^{ 2 }{ \theta  } =3\sin ^{ 2 }{ \theta  } $$     .....(given, $$\sin ^{ 2 }{ \beta  } =3\sin ^{ 2 }{ \theta  }$$)
$$\Rightarrow 2\cos ^{ 2 }{ \theta  } =3\left( 1-\cos ^{ 2 }{ \theta  }  \right)$$
$$\Rightarrow 5\cos ^{ 2 }{ \theta  } =3$$
$$\Rightarrow \cos ^{ 2 }{ \theta  } =\dfrac { 3 }{ 5 } $$