Three Dimensional Geometry Test - 55

Given vector
$$\vec a=3\hat i+0\hat j+0\hat k$$
Whose direction ratio's are $$3,0,0$$.
Direction cosines of $$\vec a$$ are
$$\dfrac{3}{\sqrt{(3)^2+0+0}},\dfrac{0}{\sqrt{(3)^2+0+0}},\dfrac{0}{\sqrt{(3)^2+0+0}}$$
or $$\dfrac{3}{3},\dfrac{0}{3},\dfrac{0}{3}$$
or $$1,0,0$$

Let H(a,b,c) be the orthocenter of $$\Delta ABC$$
AH is perpendicular to BC $$\Rightarrow (a-1)0+b(2-0)+c(0-3)=0$$
$$2b=3c\Rightarrow b=\dfrac{3}{2}c$$
Also BH is perpendicular to CA 
$$\Rightarrow (a-0)(0-1)+(b-2)0+(c-0)3=0$$
$$\Rightarrow -a+3c=0$$
$$\Rightarrow a=3c$$ 
Now (a,b,c) lies on the plane $$\dfrac{x}{1}+\dfrac{y}{2}+\dfrac{z}{3}=1$$
$$3c+\dfrac{1}{2}\dfrac{3}{2}c+\dfrac{1}{3}c=1$$
$$\Rightarrow c=\dfrac{12}{49}$$
$$\therefore a=\dfrac{36}{49},b=\dfrac{18}{49}$$
$$\therefore $$ orthocenter =$$\left ( \dfrac{36}{49} ,\dfrac{18}{49},\dfrac{12}{49}\right )$$
Centroid of $$\Delta ABC$$ is $$\left ( \dfrac{1}{3} ,\dfrac{2}{3},1\right )$$
Dr s of the line joining orthocentre and crirumcentre 
$$=$$ Dr s of the line joining orthocenter and centroid .
$$=\dfrac{36}{49}-\dfrac{1}{3},\dfrac{18}{49}-\dfrac{2}{3},\dfrac{12}{49}-1$$

$$=59,-44-111$$ 

If a ray makes angles $$\alpha, \beta, \gamma$$ and $$\delta$$ with the four diagonals of a cube.

Then we know that $$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta=\displaystyle\dfrac{4}{3}$$ 

 

$$\mathrm{A}:$$

$$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta=\displaystyle\dfrac{4}{3}$$

 

$$\mathrm{B}:$$

$$\sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma+\sin^{2}\delta$$

$$=1-\cos^{2}\alpha+1-\cos^{2}\beta+1-\cos^{2}\gamma+1-\cos^{2}\delta=4-\displaystyle\dfrac{4}{3}=\displaystyle\dfrac{8}{3}$$

 

$$\mathrm{C}:$$

$$\cos 2\alpha+\cos 2\beta+\cos 2\gamma+\cos 2\delta$$

$$=1-2\sin^{2}\alpha+1-2\sin^{2}\beta+1-2\sin^{2}\gamma+1-2\sin^{2}\delta=4-\displaystyle\dfrac{16}{3}=\displaystyle-\dfrac{4}{3}$$

 

Descending order is $$B,A,C$$

Hence, option A is correct answer.

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

and $$\overrightarrow { r } =5i-2k+\mu \left( 3i+2j+6k \right) $$

We know, angle between $$\overrightarrow { r } =\overrightarrow { { a }_{ 1 } } +\lambda \overrightarrow { { b }_{ 1 } } $$ and $$\overrightarrow { r } =\overrightarrow { { a }_{ 2 } } +\lambda \overrightarrow { { b }_{ 2 } } $$ is given by,

$$\displaystyle \cos { \theta  } =\dfrac { \overrightarrow { { b }_{ 1 } } .\overrightarrow { { b }_{ 2 } }  }{ \left| \overrightarrow { { b }_{ 1 } }  \right| \left| \overrightarrow { { b }_{ 2 } }  \right|  } $$

 

$$\Rightarrow \cos { \theta  } =\dfrac { \left( i+2j+2k \right) .\left( 3i+2j+6k \right)  }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 2 }^{ 2 } } \sqrt { { 3 }^{ 2 }+{ 2 }^{ 2 }+{ 6 }^{ 2 } }  } $$ 

 

$$\displaystyle =\dfrac { 3+4+12 }{ \sqrt { 9 } \sqrt { 49 }  } =\dfrac { 19 }{ 21 } $$

$$\Rightarrow \theta =\cos ^{ -1 }{ \left( \dfrac { 19 }{ 21 }  \right)  } $$

Let us consider a unit cube with one of the vertices as the origin and the coordinate axes along the edges of the cube. We can find the angles between the given lines using dot product of vectors along the lines.  
The angle between the two diagonals of a cube is $$\theta_{1}=\cos^{-1} {\dfrac{1}{3}}$$
The angle between two diagonals of a face of cube is $$\theta_{2} = \dfrac{\pi}{2} $$
The angle between a diagonal of a cube and diagonal of face of a cube is $$\theta_{3}=\cos^{-1} {\sqrt{\dfrac{2}{3}}}$$
Arranging in ascending order, 
$$\theta_{3},\theta_{1},\theta_{2}$$
Hence, option D is the correct answer.

If $$ \alpha,\; \beta$$ and $$\gamma$$ are angles made by a line with the coordinate axes in the positive direction, then

Since, the line joining the $$2$$ points is a normal to the plane, the equation of the plane is of the form,
$$4x-2y-2z = d$$
Since, the midpoint of the $$2$$ points lies on the plane,
$$-4-6-8 = d$$
$$4x-2y-2z = -18$$
The respective intercepts are,
$$\left ( \dfrac{-9}{2} , 9 , 9 \right )$$

Let the length of the cube be $$a$$ units.
Let $$AA', BB', CC' , OP$$ be the diagonals of the cube.
Coordinates of vertices are $$O(0,0,0),A(a,0,0), A'(0,a,a),B(0,b,0),B'(b,0,b),C(0,0,c),C'(c,c,0), P(a,a,a)$$

D.Rs of $$AA'$$ are  $$(-a,a,a)$$.
D.Rs of $$BB'$$ are $$ (a,-a,a)$$
D.Rs of $$CC'$$ are  $$(a,a,-a)$$
D.Rs of $$OP$$ are  $$(a,a,a)$$

D.c's of $$AA'$$ are $$\displaystyle \frac{-a}{\sqrt{a^2+a^2+a^2}},\frac{a}{\sqrt{a^2+a^2+a^2}},\frac{a}{\sqrt{a^2+a^2+a^2}}=-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$$

D.c's of $$BB'$$ are $$\displaystyle \frac{1}{\sqrt{3}} ,-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$$
D.c's of $$CC'$$ are $$\displaystyle \frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}$$
D.c's of $$OP$$ are $$\displaystyle \frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$$

Let the d.c's of given line be $$l,m,n$$ such that $$l^2+m^2+n^2=1$$
Given , the line makes angles $$60^{o}, 45^{o}, 45^{o}$$ and $$\theta$$ with the four diagonals of a cube.
$$\displaystyle \cos {60}^{0}=\frac{-l+m+n}{\sqrt{3}}$$
$$\Rightarrow \displaystyle \frac{1}{2}=\frac{-l+m+n}{\sqrt{3}}$$   .....(1)
$$\displaystyle \cos {45}^{0}=\frac{l-m+n}{\sqrt{3}}$$
$$\Rightarrow \displaystyle \frac{1}{\sqrt{2}}=\frac{l-m+n}{\sqrt{3}}$$   .....(2)
$$\displaystyle \cos {45}^{0}=\frac{l+m-n}{\sqrt{3}}$$
$$\Rightarrow \displaystyle \frac{1}{\sqrt{2}}=\frac{l+m-n}{\sqrt{3}}$$   .....(3)
$$\displaystyle \cos {\theta}=\frac{l+m+n}{\sqrt{3}}$$    ....(4)

Squaring and adding (1),(2),(3),(4), we get
$$\Rightarrow \displaystyle \frac{5}{4}+\cos^{2}\theta=\frac{4}{3}$$   ($$\because l^2+m^2+n^2=1$$)
$$\Rightarrow \cos^{2}\theta=\displaystyle \frac{1}{12}$$
$$\Rightarrow \displaystyle \sin^{2}\theta=\frac{11}{12}$$

Let the length of the sides of cube is a, then coordinates of corner (P) opposite to origin are (a,a,a)