Three Dimensional Geometry Test - 23

Let $$ \cos\alpha, \cos\beta $$ and $$\cos\gamma$$  be the direction cosines.
We know that $$\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1$$
Now rewrite it as $$\cos^{2}\alpha = 1 - \sin^{2}\alpha$$
$$\Rightarrow \cos^{2}\beta = 1 - \sin^{2}\beta$$
$$\Rightarrow \cos^{2}\gamma = 1 - \sin^{2}\gamma$$
Then we get, $$\sin^{2}\alpha + \sin^{2}\beta + \sin^{2}\gamma = 2$$

The $$3$$ given points are $$A(0,0,1), B(0,1,2)$$ and $$C(1,2,3)$$

Let the equation of the plane be $$Ax+By+Cz=D.$$

Dot product of dc's of two lines gives,
$$(l_{ 1 }\hat i+m_{ 1 }\hat j+n_{ 1 }\hat k).(l_{ 2 }\hat i+m_{ 2 }\hat j+n_{ 2 }\hat k)=\sqrt { { l_{ 1 } }^{ 2 }+{ m_{ 1 } }^{ 2 }+{ n_{ 1 } }^{ 2 } } \times \sqrt { { l_{ 2 } }^{ 2 }+{ m_{ 2 } }^{ 2 }+{ n_{ 2 } }^{ 2 } } \times \cos\theta $$
We know $${ l_{ 1 } }^{ 2 }+{ m_{ 1 } }^{ 2 }+{ n_{ 1 } }^{ 2 }=1,\quad { l_{ 2 } }^{ 2 }+{ m_{ 2 } }^{ 2 }+{ n_{ 2 } }^{ 2 }=1$$
 $$\Rightarrow l_{ 1 }l_{ 2 }+m_{ 1 }m_{ 2 }+n_{ 1 }n_{ 2 }=1.1.\cos\theta \Rightarrow { (l_{ 1 }l_{ 2 }+m_{ 1 }m_{ 2 }+n_{ 1 }n_{ 2 }) }^{ 2 }={ \cos }^{ 2 }\theta $$
Cross product of dc's of two lines
$$\left| (l_{ 1 }\hat i+m_{ 1 }\hat j+n_{ 1 }\hat k)\times (l_{ 2 }\hat i+m_{ 2 }\hat j+n_{ 2 }\hat k) \right| =\left| \sqrt { { l_{ 1 } }^{ 2 }+{ m_{ 1 } }^{ 2 }+{ n_{ 1 } }^{ 2 } } \times \sqrt { { l_{ 2 } }^{ 2 }+{ m_{ 2 } }^{ 2 }+{ n_{ 2 } }^{ 2 } } \times \sin\theta  \right| $$
 $$\Rightarrow \sqrt { \Sigma (m_{ 1 }n_{ 2 }-m_{ 2 }n_{ 1 })^{ 2 } } =\sin\theta \Rightarrow \Sigma (m_{ 1 }n_{ 2 }-m_{ 2 }n_{ 1 })^{ 2 }={ \sin }^{ 2 }\theta $$
 $${ (l_{ 1 }l_{ 2 }+m_{ 1 }m_{ 2 }+n_{ 1 }n_{ 2 }) }^{ 2 }+\Sigma (m_{ 1 }n_{ 2 }-m_{ 2 }n_{ 1 })^{ 2 }={ \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta =1$$

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.

$$\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma)=1$$
$$\Rightarrow 3\cos^2(\alpha)=1$$
$$ \displaystyle \Rightarrow \cos \alpha = \pm \dfrac{1}{\sqrt{3}}$$
Product will be $$\cos^3(\alpha)= \pm 1/3(\sqrt3)$$
Hence, option C is correct.

$$\left (\cos\dfrac {\pi}{12}\right)^{2}+\left (\cos \dfrac {5\pi}{12}\right)^{2}+(\cos(\gamma))^{2}=1$$
$$\left (\cos  \dfrac {\pi}{12}\right)^{2}+\left (\sin \dfrac {\pi }{12}\right)^{2}+(\cos(\gamma))^{2}=1$$     ..... $$\left(\cos\theta=\sin\left (\dfrac {\pi}{2}-\theta\right)\right)$$

Given points are $$O(0,0,0)=(x_1,y_1,z_1)$$ and $$P(3,12,4)=(x_2,y_2,z_2)$$ 

$$(\cos\alpha )^2+(\cos\beta )^2+(\cos\gamma )^2=1$$
$$2 \times [(\cos\alpha )^2+(\cos\beta )^2+(\cos\gamma )^2]=2$$
$$2 \times [(\cos\alpha )^2+(\cos\beta )^2+(\cos\gamma )^2]-3=-1$$
$$2(\cos\alpha )^2-1+2(\cos\beta )^2-1+2(\cos\gamma )^2-1]=-1$$
$$(\cos2\alpha )+(\cos2\beta )+(\cos2\gamma )=-1$$