CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 3 - MCQExams.com

Let $$l,m,n$$ be the direction cosines of $$\overrightarrow { r } $$.

Since $$\overrightarrow { r } $$ is equally inclined with $$x,y$$ and $$z$$ axis, $$l=m=n$$

$$\displaystyle \therefore { l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }\Rightarrow 3{ l }^{ 3 }=1\Rightarrow l=\pm \dfrac { 1 }{ \sqrt { 3 }  } $$

$$\therefore$$ direction cosines of $$\overrightarrow { r } $$ are $$\displaystyle \pm \dfrac { 1 }{ \sqrt { 3 }  } ,\pm \dfrac { 1 }{ \sqrt { 3 }  } ,\pm \dfrac { 1 }{ \sqrt { 3 }  } $$

Now, $$\displaystyle \overrightarrow { r } =\left| \overrightarrow { r }  \right| \left( li+mj+nk \right) =\overrightarrow { r } =\left| \overrightarrow { r }  \right| \left( \pm \dfrac { 1 }{ \sqrt { 3 }  } i\pm \dfrac { 1 }{ \sqrt { 3 }  } j\pm \dfrac { 1 }{ \sqrt { 3 }  } k \right) $$

Since $$+$$ and $$-$$ signs can be arranged at three places,

$$\Rightarrow$$ there are eight vectors, i.e $$2\times 2\times 2$$ which are equally inclined to axes.

Let the direction cosines of the line be $$l_{1},m_{1},n_{1}$$.

Hence, $$l_{1}=\cos\alpha$$, $$m_{1}=\cos\beta$$, $$n_{1}=\cos\gamma$$

Similarly, for another line

$$l_{2}=\cos\alpha'$$, $$m_{2}=\cos\beta'$$, $$n_{2}=\cos\gamma'$$

Now the direction cosines of the angle bisector will be, 

$$\displaystyle \cos\left(\dfrac{\alpha-\alpha'}{2}+\alpha'\right),\cos \left(\dfrac{\beta-\beta'}{2}+\beta' \right),\cos\left(\dfrac{\gamma-\gamma'}{2}+\gamma'\right)$$

$$=\cos\left(\dfrac{\alpha+\alpha'}{2}\right),\cos\left(\dfrac{\beta+\beta'}{2}\right),\cos\left(\dfrac{\gamma+\gamma'}{2}\right)$$

$$=l,m,n$$ 

Now consider $$\cos\left(\dfrac{\alpha+\alpha'}{2}\right)$$

Multiplying and dividing by $$\displaystyle 2\cos\left(\dfrac{\alpha-\alpha'}{2}\right)$$, we get

$$\displaystyle \dfrac{2\cos\left(\dfrac{\alpha-\alpha'}{2}\right)\cos\left(\dfrac{\alpha+\alpha'}{2}\right)}{2\cos\left(\dfrac{\alpha-\alpha'}{2}\right)}$$

$$=\displaystyle \dfrac{\cos\alpha+\cos\alpha'}{2\cos(\dfrac{\alpha-\alpha'}{2})}$$

$$=\displaystyle \dfrac{l_{1}+l_{2}}{2\cos(\dfrac{\alpha-\alpha'}{2})}$$

$$=l$$

Hence, $$ l = \dfrac{l_1+l_2}{2 \cos \left(\dfrac{\theta}{2} \right) }$$

Similarly, 

$$m=\displaystyle \dfrac{m_{1}+m_{2}}{2\cos(\dfrac{\beta-\beta'}{2})}$$

$$n=\displaystyle \dfrac{n_{1}+n_{2}}{2\cos(\dfrac{\gamma-\gamma'}{2})}$$

Hence, option B is correct.

We have, $$\left (\cos\left (\dfrac {\pi}{3}\right)\right)^{2}+\left (\cos\left (\dfrac {\pi}{4}\right)\right)^{2}+(\cos(\gamma))^{2}=1$$
$$\Rightarrow \cos^{2}\gamma = \dfrac {1}{4}$$

$$ \displaystyle \Rightarrow \cos \gamma = \pm \frac{1}{2}$$
$$\Rightarrow \gamma=\dfrac {\pi}{3}$$ or $$ \dfrac {2 \pi}{3}$$

Given $$3lm-4ln+mn=0$$    ...$$(1)$$

 and $$l+2m+3n=0$$    ....$$(2)$$

From equation $$(2)$$, $$l=-(2m+3n)$$, putting in equation $$(1)$$, we get

$$-3\left( 2m+3n \right) m+4\left( 2m+3n \right) n+mn=0\\ \Rightarrow -6{ m }^{ 2 }+12{ n }^{ 2 }=0\Rightarrow m=\pm \sqrt { 2 } n$$

Now, $$m=\pm \sqrt { 2 } n\Rightarrow l=-\left( 2\sqrt { 2 } n+3n \right) =-\left( 2\sqrt { 2 } +3 \right) n$$

$$\therefore l:m:n=-\left( 3+2\sqrt { 2 }  \right) n:\sqrt { 2 } n:n=-\left( 3+2\sqrt { 2 }  \right) :-\sqrt { 2 } :1$$

Also, $$m=-\sqrt { 2 } n\Rightarrow l=-\left( -2\sqrt { 2 } +3 \right) n$$

$$\therefore l:m:n=-\left( 3-2\sqrt { 2 }  \right) n:-\sqrt { 2 } n:n=-\left( 3-2\sqrt { 2 }  \right) :-\sqrt { 2 } :1$$

$$\displaystyle \Rightarrow \cos { \theta  } =\dfrac { \left( 3+2\sqrt { 2 }  \right) \left( 3-2\sqrt { 2 }  \right) +\left( \sqrt { 2 }  \right) \left( -\sqrt { 2 }  \right) +1.1 }{ \sqrt { { \left( 3+2\sqrt { 2 }  \right)  }^{ 2 }+{ \left( \sqrt { 2 }  \right)  }^{ 2 }+{ 1 }^{ 2 } } \sqrt { { \left( 3-2\sqrt { 2 }  \right)  }^{ 2 }+{ \left( -\sqrt { 2 }  \right)  }^{ 2 }+{ 1 }^{ 2 } }  } $$

$$\displaystyle \Rightarrow \theta =\dfrac { \pi  }{ 2 } $$

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

$$x=12,y=4,z=3$$

Direction cosines $$\displaystyle =\dfrac{x}{\sqrt{x^{2}+y^{2}+z^{2}}} ,\displaystyle \dfrac{y}{\sqrt{x^{2}+y^{2}+z^{2}}},\dfrac{z}{\sqrt{x^{2}+y^{2}+z^{2}}} $$ 

$$\displaystyle = \dfrac{12}{13},\dfrac{4}{13},\dfrac{3}{13}$$

Hence, option 'B' is correct.

$$  {\textbf{Step - 1: Find direction ratio}} $$

                  $$  {\text{Let the foot of the perpendicular be P(a,b,c)}}{\text{.}} $$

                 $$  {\text{Then, direction ratios of OP are }} {\text{a - 0,b - 0,c - 0 }}\;{\text{i}}{\text{.e}}{\text{ a,b,c}} $$

$$  {\textbf{Step - 2: Make equation}} $$

                 $$  {\text{So, the equation of the plane passing through P(a,b,c), }} $$

                 $$  {\text{the direction ratios of the normal to which are a,b,c is}} $$

                 $$  {\text{a(x - a) + b(y - b) + c(z - c) = 0}} $$

                 $$   \Rightarrow {\text{ax + by + cz = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + }}{{\text{c}}^2}{\text{.}} $$

$$  {\textbf{Hence,correct answer is option C}}{\text{.}} $$