CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 7 - MCQExams.com

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

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 $$\left( \pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{3}},\pm \dfrac{1}{\sqrt{3}} \right)$$

 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  $$\left( 0,\dfrac{1}{2},\dfrac{1}{2} \right)$$.

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.

We have,

Given points are $$\left( 3,4,7 \right)\,\,and\,\,\left( 5,1,6 \right)$$.

We know that,

Vector equation of a line passing through the two points with position vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ is

$$\overrightarrow{r}=\overrightarrow{a}+\lambda \left( \overrightarrow{b}-\overrightarrow{a} \right)$$

Let the given points are $$A\left( 3,4,7 \right)\,\,and\,\,\left( 5,1,6 \right)$$

Now,

Equation of vector,

$$ \overrightarrow{r}=\left( 3\overrightarrow{i}+4\overrightarrow{j}+7\overrightarrow{k} \right)+\lambda \left( 5\overrightarrow{i}+\overrightarrow{j}+6\overrightarrow{k}-3\overrightarrow{i}-4\overrightarrow{j}-7\overrightarrow{k} \right) $$

$$ \overrightarrow{r}=\left( 3\overrightarrow{i}+4\overrightarrow{j}+7\overrightarrow{k} \right)+\lambda \left( 2\overrightarrow{i}-3\overrightarrow{j}-\overrightarrow{k} \right) $$

Hence, this is the answer.