Introduction to Three Dimensional Geometry Test - 22

Given, $$A(1,2,-1), B(2,5,-2), C(4,4,-3),$$ and $$D(3,1,-2)$$

Given points are $$A(2,-3,-1)$$ and $$B(8,-1,2)$$
Therefore direction  ratio of $$AB$$ are $$l = \dfrac{6}{7}, m=\dfrac{2}{7},n= \dfrac{3}{7}$$ or $$l=\dfrac{-6}{7},m= \dfrac{-2}{7},n= \dfrac{-3}{7}$$
Hence, coordinates of a point $$14$$ unit from point $$A$$ is given as,$$(2+14l,-3+14m,-1+14n)$$
$$\Rightarrow (14,1,5)$$ or $$(-10,-7,-7)$$
Hence, option 'A' is correct.

Let $$P\equiv \left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$Q\equiv \left( { x }_{ 2 },{ y }_{ 2 } \right) $$
Let the equation of given circle be $${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$
The equation of chord of contact of tangent drawn from the point $$P\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ to the given circle is $${ xx }_{ 1 }+{ yy }_{ 1 }={ a }^{ 2 }$$.
Since it passes through $$Q\left( { x }_{ 2 },{ y }_{ 2 } \right) $$
$$\therefore { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 }={ a }^{ 2 }$$ ...(1)
Now $${ l }_{ 1 }=\sqrt { { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }-{ a }^{ 2 } } ,{ l }_{ 2 }=\sqrt { { x }_{ 2 }^{ 2 }+{ y }_{ 2 }^{ 2 }-{ a }^{ 2 } } $$
and $$\\ PQ=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \\ =\sqrt { \left( { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 } \right) +\left( { x }_{ 2 }^{ 2 }+{ y }_{ 2 }^{ 2 } \right) -2\left( { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 } \right) } \\ =\sqrt { \left( { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 } \right) +\left( { x }_{ 2 }^{ 2 }+{ y }_{ 2 }^{ 2 }-{ a }^{ 2 } \right) } \\ =\sqrt { \left( { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }-{ a }^{ 2 } \right) \left( { x }_{ 2 }^{ 2 }+{ y }_{ 2 }^{ 2 }-{ a }^{ 2 } \right) } \\ =\sqrt { { l }_{ 1 }^{ 2 }+{ l }_{ 2 }^{ 2 } } $$

Let $$R(x,0,0)$$ be the point on $$x$$-axis which is equidistant from $$P(4,3,1)$$ and $$Q(-2,-6,-2)$$

Let the fourth vertex be $$(h,k,l)$$

Since $$\overrightarrow { OP } $$ has projection $$\displaystyle \dfrac { 13 }{ 5 } ,\dfrac { 19 }{ 5 } $$ and $$\displaystyle \dfrac { 26 }{ 5 } $$ on the coordinate axes, therefore $$\displaystyle \overrightarrow { OP } =\dfrac { 13 }{ 5 } i+\dfrac { 19 }{ 5 } j+\dfrac { 26 }{ 5 } k$$

Let $$P$$ divides the join of $$Q\left( 2,2,4 \right) $$ and $$R\left( 3,5,6 \right) $$ in the ratio $$\lambda :1$$.

The position vector of $$P$$ is $$\displaystyle \left( \dfrac { 3\lambda +2 }{ \lambda +1 }  \right) i+\left( \dfrac { 5\lambda +2 }{ \lambda +1 }  \right) j+\left( \dfrac { 6\lambda +4 }{ \lambda +1 }  \right) k$$

$$\displaystyle \therefore \dfrac { 13 }{ 5 } i+\dfrac { 19 }{ 5 } j+\dfrac { 26 }{ 5 } k=\left( \dfrac { 3\lambda +2 }{ \lambda +1 }  \right) i+\left( \dfrac { 5\lambda +2 }{ \lambda +1 }  \right) j+\left( \dfrac { 6\lambda +4 }{ \lambda +1 }  \right) k$$

$$\displaystyle \Rightarrow \dfrac { 3\lambda +2 }{ \lambda +1 } =\dfrac { 13 }{ 5 } ,\dfrac { 5\lambda +2 }{ \lambda +1 } =\dfrac { 19 }{ 5 } ,\dfrac { 6\lambda +4 }{ \lambda +1 } =\dfrac { 26 }{ 5 }$$

$$ \Rightarrow \lambda =\dfrac { 3 }{ 2 } $$

Equation of the line through $$\left( 1,-2,3 \right) $$ parallel to the line $$\displaystyle \dfrac { x }{ 2 } =\dfrac { y }{ 3 } =\dfrac { z-1 }{ -6 } $$ is

$$\displaystyle \dfrac { x-1 }{ 2 } =\dfrac { y+2 }{ 3 } =\dfrac { z-1 }{ -6 } =r$$ (say)   ...$$(1)$$

Then any point on $$(1)$$ is $$\left( 2r+1,3r-2,-6r+3 \right) $$.

If this point lies on the plane $$x-y+z=5$$, then 

$$\displaystyle \left( 2r+1 \right) -\left( 3r-2 \right) +\left( -6r+3 \right) =5\Rightarrow -7r+6=5\Rightarrow r=\dfrac { 1 }{ 7 } $$

Hence, the point is $$\displaystyle \left( \dfrac { 9 }{ 7 } ,-\dfrac { 11 }{ 7 } ,\dfrac { 15 }{ 7 }  \right) $$

Distance between $$\left( 1,-2,3 \right) $$ and $$\displaystyle \left( \dfrac { 9 }{ 7 } ,-\dfrac { 11 }{ 7 } ,\dfrac { 15 }{ 7 }  \right) $$

$$\displaystyle =\sqrt { \left( \dfrac { 4 }{ 49 } +\dfrac { 9 }{ 49 } +\dfrac { 36 }{ 49 }  \right)  } =\sqrt { \dfrac { 49 }{ 49 }  } =1$$

Let $$\vec {r}$$ be a point on a plane $$\vec {r} .(\vec i-2\vec j+3\vec k)=17$$ which divides line joining the points $$-2\vec i+4\vec j+7\vec k$$ & $$3\vec i-5\vec j+8\vec k$$ in the ratio $$n:m$$
Therefore, $$\vec {r}=\vec { r } =\dfrac { m\left( -2\vec i+4\vec j+7\vec k \right) +n\left( 3\vec i-5\vec j+8\vec k \right)  }{ m+n } =\dfrac { \left( -2m+3n \right) i+\left( 4m-5n \right) j+\left( 7m+8n \right) k }{ m+n } $$

Substituting $$\vec {r}$$ in equation of plane, we get
$$\left( \dfrac { \left( -2m+3n \right) \vec i+\left( 4m-5n \right) \vec j+\left( 7m+8n \right) \vec k }{ m+n }  \right) .\left( \vec i-2\vec j+3\vec k \right) =17$$
$$\Rightarrow -2m+3n-8m+10n+21m+24n=17m+17n$$
$$\Rightarrow -6m+20n=0$$
$$\Rightarrow n:m=3:10$$

Ans: B

Let $$C$$ be $$(a,b,c)$$
By using the formula of centroid of triangle, we get
$$G(2,0,2) = \left(\dfrac{-1+2+a}{3},\dfrac{2-3+b}{3},\dfrac{4+0+c}{3}\right)$$

The three vertices of tetrahedron is $$(0,0,0)$$ , $$(6,-5,-1)$$ and $$(-4,1,3)$$