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 $(-4,1,3)$