Tag: three dimensional geometry - ii

The D.C. of the line are $\begin{pmatrix}\displaystyle\frac{1}{\sqrt{2}},\,\displaystyle\frac{1}{\sqrt{2}},\,0\end{pmatrix}$
any point on the line at a distance $\lambda$ from $A(0,\,1,\,2)$
is $\begin{pmatrix}0+\displaystyle\frac{\lambda}{\sqrt{2}},\,1+\displaystyle\frac{\lambda}{\sqrt{2}},\,2+\lambda.0\end{pmatrix}$
which lies on $x+y+z=0$
$\therefore\;\displaystyle\frac{\lambda}{\sqrt{2}}+\begin{pmatrix}1+\displaystyle\frac{\lambda}{\sqrt{2}}\end{pmatrix}+2=0$
$\Rightarrow\;\displaystyle\frac{2\lambda}{\sqrt{2}}=-3\;\;\;\;\Rightarrow\;\;\;\;\lambda=\displaystyle\frac{-3}{\sqrt{2}}$
$\therefore\;B=\begin{pmatrix}-\displaystyle\frac{3}{2},\,-\displaystyle\frac{1}{2},\,2\end{pmatrix}$
$A=(0,\,1,\,2)\;&\;B\equiv\;\begin{pmatrix}-\displaystyle\frac{3}{2},\,-\displaystyle\frac{1}{2},\,2\end{pmatrix}$
$AB=\sqrt{\displaystyle\frac{9}{4}+\displaystyle\frac{9}{4}}=\displaystyle\frac{3\sqrt{2}}{2}=\displaystyle\frac{3}{\sqrt{2}}$

Equation of plane in cartesian form is, $x-2y+3z-17=0.....(1)$
Assume this plane (1) divide the line segment joining the points $(-2,4,7)$ and $(3,-5,8)$ in $m:n$ ratio.
Therefore,  $\dfrac{m}{n} = \dfrac{-2-2(4)+3(7)-17}{3-2(-5)+3(8)-17}= \dfrac{-3}{10} < 0$
Hence plane (1) divides the given line segment externally  $3:10$.  

A straight line $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } $ meets the plane $\overrightarrow { r } .\overrightarrow { n } =0$ at $P$ for which $\lambda$ is given by,

$\displaystyle \left( \overrightarrow { a } +\lambda \overrightarrow { b }  \right) .\overrightarrow { n } =0\Rightarrow \lambda =-\dfrac { \overrightarrow { a } .\overrightarrow { n }  }{ \overrightarrow { b } .\overrightarrow { n }  } $

Thus, the position vector of $P$ is

$\displaystyle \overrightarrow { r } =\overrightarrow { a } -\dfrac { \overrightarrow { a } .\overrightarrow { n }  }{ \overrightarrow { b } .\overrightarrow { n }  } \overrightarrow { b } $  $[$ putting the value of $\lambda$ in $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } ]$

Given equation of straight line $\displaystyle \dfrac { x-4 }{ 1 } =\dfrac { y-2 }{ 1 } =\dfrac { z-k }{ 2 } $

Given line is $\displaystyle\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ any point on the line is $(t, 2t, 3t)$. It lies in the given plane, if $t-2(2t)+3t-6=0$
i.e., $0\cdot t=6$, which is not true for any real $t$. So, the line and plane do not meet. i.e., the line is parallel to the plane.

We have,

$\dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{4}=1$

Compare this equation of plane,

$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

Then,

$a=2,\,b=3,\,c=4$

Then,

The coordinate of X-axis is $=A\left( a,0,0 \right)$$=A\left( 2,0,0 \right)$

The coordinate of Y-axis is $=B\left( 0,b,0 \right)$$=A\left( 0,3,0 \right)$

The coordinate of X-axis is $=C\left( 0,0,c \right)$$=C\left( 0,0,2 \right)$

We know that, the area of  \[\Delta ABC\] is

$ Area\,of\,\Delta ABC=\dfrac{1}{2}\sqrt{{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}} $

$ =\dfrac{1}{2}\sqrt{{{2}^{2}}\times {{3}^{2}}+{{3}^{2}}\times {{4}^{2}}+{{4}^{2}}\times {{2}^{2}}} $

$ =\dfrac{1}{2}\sqrt{36+144+64} $

$ =\dfrac{1}{2}\sqrt{180+64} $

$ =\dfrac{1}{2}\sqrt{244} $

$ =\dfrac{1}{2}\sqrt{2\times 2\times 61} $

$ =\sqrt{61} $

Hence, this is the answer.

We have point $(0,3,4)$ and we have direction's ratio also of line so we can write equation of line
$\dfrac{x-0}{2}=\dfrac{y-3}{-2}=\dfrac{z-4}{1}=k$
So consider a point which is lied on line $(2k,-2k+3,k+4)$
This point lies on plane also so it will satisfy the equation of plane:
$\Rightarrow 2(2k)-2(-2k+3)+k+4-10=0$
$\Rightarrow k=\dfrac{4}{3}$
Substitute the value of $k$ in point $\left (\dfrac{8}{3},\dfrac{1}{3},\dfrac{16}{3}\right)$

The line is $\displaystyle \frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}2{}$

Given  lines
 $x -2y + 4z + 4 = 0$    ....(1)
 $x + y + z - 8 = 0$       .....(2)
Subtracting (2) from (1), we get
$\Rightarrow y-z=4$        .....(3)
Given equation of plane $x-y+2z+1=0$
Since, the line intersects the plane, so using (3), we get 
$\Rightarrow x+z=3$      ......(4)
Hence, $y=5$, $z=1 $ and $x=2$
Hence, option C is correct.