Three Dimensional Geometry Test - 44

$$\overrightarrow { a } =0\hat { i } +\hat { j } +\hat { k } ,\overrightarrow { b } =\hat { i } +\hat { j } +2\hat { k } ,\overrightarrow { c } =-\hat { i } +2\hat { j } -2\hat { k } $$

$$P(6,-7,-1)$$ and  $$Q(2,-3,1)$$

 We know that ,

The direction cosines of a line making angle $$\alpha $$ with x-axis, $$\beta $$ with y- axis and $$\gamma $$ with z-axis are $$l,m,n.$$

Then,

$$l=\cos \alpha ,\,\,\,m=\cos \beta ,\,\,\,n=\cos \gamma $$

Given that $$\alpha ={{90}^{0}},\,\,\,\,\beta ={{60}^{0}}\,and\,\,\,\,\gamma ={{60}^{0}}$$

So, direction cosines of a line

$$l=\cos {{90}^{0}},\,\,\,m=\cos {{60}^{0}},\,\,\,n=\cos {{60}^{0}}$$

$$l=0,\,\,\,m=\dfrac{1}{2},\,\,\,n=\dfrac{1}{2}$$

Therefore direction cosines are  $$\left( 0,\dfrac{1}{2},\dfrac{1}{2} \right)$$.

Option (D) is correct answer.

$$\begin{array}{l} According\, to\, \, the\, \, question: \\ direction\, ratio\, equ:a,b,c\, \, and\, \, two\, line\, is{ {  } }perpendicular: \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (1,-2,-2,)\, \, \& \left( { 0,2,1 } \right)  \\ relation\, of\, direction\, ratio\, \, with\, perpendicular\, line\, is: \\ \, \Rightarrow { a_{ 1 } }{ a_{ 2 } }+{ b_{ 1 } }{ b_{ 2 } }+{ c_{ 1 } }{ c_{ 2 } }=0 \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, a-2b-2c=0 \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \underline { \, \, 0a+2b+c=0 }  \\ the\, ratio\, of\, 3direction\, po{ { int } }\, is: \\ \frac { a }{ 2 } =\frac { { -b } }{ 0 } =\frac { c }{ 2 }  \\ so\, the\, ratio\, is\, \left( { \frac { 2 }{ 3 } ,\frac { { -1 } }{ 3 } ,\frac { 2 }{ 3 }  } \right)  \\ and\, the\, \, { { correct } }\, option\, is\, A. \end{array}$$

Let $$A\equiv(3,4,2);\quad B\equiv (5,1,8)$$

angles made with $$x,y,z$$ axis are $$\cfrac { \alpha  }{ 2 } ,\cfrac { \beta  }{ 2 } ,\cfrac { \gamma  }{ 2 } $$

Let the angle subtended with each of the coordinate axis be Q

$$L:\cfrac { x-1 }{ 1 } =\cfrac { y-2 }{ 2 } =\cfrac { z+1 }{ 3 } =k\\ \therefore (k+1,2k+3,3k-1)\\ \sqrt { { (k+1) }^{ 2 }+{ (2k+2) }^{ 2 }+{ (3k-1) }^{ 2 } } =\sqrt { 6 } \\ { (k+1) }^{ 2 }+{ (2k+2) }^{ 2 }+{ (3k-1) }^{ 2 }=6\\ { k }^{ 2 }+1+2k+4{ k }^{ 2 }+4+8k+9{ k }^{ 2 }+1-6k=6\\ 14{ k }^{ 2 }+4k=0\\ \therefore \Rightarrow k=0,\cfrac { -2 }{ 7 } $$

According to question.................

$$\begin{array}{l} \Rightarrow l+m+n=0----(i) \\ \, \, \, \, \, \, l=-\, (m+n) \\ \, \Rightarrow { l^{ 2 } }={ m^{ 2 } }+{ n^{ 2 } }---(ii) \\ \, \, \, \, \, \, \, \, \Rightarrow \, { \left( { -\, (m+n) } \right) ^{ 2 } }={ m^{ 2 } }+{ n^{ 2 } } \\ \, \, \, \, \, \, \, \, \, \Rightarrow { (m+n)^{ 2 } }={ m^{ 2 } }+{ n^{ 2 } } \\ \, \, \, \, \, \, \Rightarrow { m^{ 2 } }+{ n^{ 2 } }+2mn={ m^{ 2 } }+{ n^{ 2 } } \\ \, \, \, \, \, \, \, \, \, \, \, \Rightarrow 2mn=0 \\ \, \, \, \, \, \, \, \, \, \, \, \therefore \, \, \, \, mn=0 \\ there\, \, are\, \, two\, \, \, case:\, \, \, m=0,n=0 \\ case\, \, 1:\, \, (m=0) \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, l=-n\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| { u{ { sing } }\, equ\, :l+m+n=0----(i) } \right.  \\ \, suppose: \\ \, l=k,\, \, m=o,\, \, n=-k \\ \Rightarrow { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }=1 \\ \Rightarrow { k^{ 2 } }+0+{ k^{ 2 } }=1 \\ \Rightarrow k=\dfrac { 1 }{ { \sqrt { 2 }  } } \, \,  \\ Apply, \\ \Rightarrow l=\dfrac { 1 }{ { \sqrt { 2 }  } } \, ,\, \, \, \, m=o,\, \, \, \, n=\dfrac { { -1 } }{ { \sqrt { 2 }  } } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (we\, Asume:\, { l_{ 1 } }=\dfrac { 1 }{ { \sqrt { 2 }  } } \, ,\, \, \, \, { m_{ 1 } }=o,\, \, \, \, { n_{ 1 } }=\dfrac { { -1 } }{ { \sqrt { 2 }  } } \,  \\ case\, \, 2:\, \, \, (n=0) \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, l=-m\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| { u{ { sing } } } \right. \, \, equ:l+m+n=0----(i) \\suppose:\, \,  \\ \, \, \Rightarrow \, \, l=k,\, \, \, \, \, m=-k,\, \, \, n=0\, \, \, \, \, \, \, ( \\ \, then, \\ { l^{ 2 } }+{ m^{ 2 } }+{ n^{ 2 } }=1 \\ { k^{ 2 } }+{ k^{ 2 } }+0=1 \\ k=\dfrac { 1 }{ { \sqrt { 2 }  } }  \\ apply, \\ \Rightarrow l=\dfrac { 1 }{ { \sqrt { 2 }  } } \, ,\, \, m=\dfrac { { -1 } }{ { \sqrt { 2 }  } } ,\, \, n=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (we\, Assume:{ l_{ 2 } }=\dfrac { 1 }{ { \sqrt { 2 }  } } \, ,\, \, { m_{ 2 } }=\dfrac { { -1 } }{ { \sqrt { 2 }  } } ,\, \, { n_{ 2 } }=0\, \,  \\ Angle\, \, between\, \, \, 2\, lines=\theta  \\ \cos  \theta =\dfrac { { \, \, \, \, \, \, \, ({ l_{ 1 } }+{ m_{ 1 } }+{ n_{ 1 } })\, .({ l_{ 2 } }+{ m_{ 2 } }+{ n_{ 2 } })\, \, \, \,  } }{ { \sqrt { ({ l_{ 1 } }^{ 2 }+{ m_{ 1 } }^{ 2 }+{ n_{ 1 } }^{ 2 })\, .\, \, ({ l_{ 2 } }^{ 2 }+{ m_{ 2 } }^{ 2 }+{ n_{ 2 } }^{ 2 })\,  }  } }  \\ \, \, \, =\dfrac { { { l_{ 1 } }\, { l_{ 2 } }+{ m_{ 1 } }{ m_{ 2 } }+{ n_{ 1 } }{ n_{ 2 } } } }{ { \sqrt { ({ l_{ 1 } }^{ 2 }+{ m_{ 1 } }^{ 2 }+{ n_{ 1 } }^{ 2 })\, \, (\, { l_{ 2 } }^{ 2 }+{ m_{ 2 } }^{ 2 }+{ n_{ 2 } }^{ 2 }) }  } }  \\ \, =\dfrac { { \dfrac { 1 }{ { \sqrt { 2 }  } } . \dfrac { 1 }{ { \sqrt { 2 }  } } + 0 .\dfrac { { -1 } }{ { \sqrt { 2 }  } } +\dfrac { { -1 } }{ { \sqrt { 2 }  } } . 0 } }{ { \sqrt { \left( { \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 }  +0} \right) \, \, \left( { \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } \ +0 } \right)  }  } } =\dfrac { { \dfrac { 1 }{ 2 }  } }{ { \sqrt { 1 }  } }  \\ \cos  \theta =\dfrac { 1 }{ 2 } ,\, \, \, \, \, \, \, \, \therefore \, \, \, \theta =\dfrac { \pi  }{ 3 }  \end{array}$$

So,that the correct option is  B.