Three Dimensional Geometry Test - 24

Equation of line bisecting $$XOY$$ is $$x=y$$
Therefore, d.r. s are $$(1,1,0)$$
And thus d.c. s are $$\left( \dfrac { 1 }{ \sqrt { 2 }  } ,\dfrac { 1 }{ \sqrt { 2 }  } ,0 \right) $$
Ans: B

Given points are $$P(2,3,-1)=(x_1,y_1,z_1)$$ and $$(0,0,0)=(x_2,y_2,z_2)$$

Line passing through $$A(2,4,5)$$ and $$B(-7,-2,8)$$ is
$$L: \dfrac{x-2}{9} = \dfrac{y-4}{6} = \dfrac{z-5}{-3}$$
Lets suppose $$O$$ is collinear with $$A$$ and $$B$$, then $$L = k$$
$$\therefore$$ Coordinates of $$O$$ are $$(2+9k,4+6k,5-3k)$$
Looking at options, only option C satisfies the coordinates of $$O$$ for $$k = -\dfrac{1}{3}$$

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 $$l=\dfrac{1}{2}, m=\dfrac{1}{2} , n$$ as the d.c's of a line.
Let $$\gamma $$ be the angle which the line makes with $$OZ$$ axis i.e. $$\cos\gamma=n$$
We know that 
$$l^2+m^2+n^2=1$$
$$\Rightarrow n^2=\dfrac{1}{2}$$
$$\Rightarrow n=\pm \dfrac{1}{\sqrt{2}}$$
Since, $$n<0$$
$$\Rightarrow n=-\dfrac{1}{\sqrt{2}}$$
$$\Rightarrow \cos \gamma =-\dfrac{1}{\sqrt{2}}$$
$$\Rightarrow \gamma =-\cos {45^0}$$
$$\Rightarrow \gamma =135^{0}$$

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

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

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

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

Projections of the line segment $$AB$$ on coordinate axes are $$2,3,$$ and $$6$$
$$\therefore$$ the line can be $$L: 2\hat i + 3\hat j + 6\hat k$$
Angle of line $$L$$ with $$x = 0\>i.e.\>\hat j$$ is:

$$(2, 1, 3)$$ and $$(-1, 2, 4)$$ are extremities of a diagonal of a rhombus. 
Hence, the direction ratios of that diagonal are equal to $$2-(-1), 1-2, 3-4$$.
So, the direction ratios are $$3, -1, -1$$. 
In rhombus the diagonals are perpendicular to each other. Suppose the direction ratios of the other diagonal are $$l. m , n$$. So, $$3l -m - n$$ should be equal to zero.
$$3l-m-n = 0$$
Only option B satisfies this condition. Hence the direction ratios are $$-2,\; 3\;, -9\;$$.