Three Dimensional Geometry Test - 34

Equation of the plane parallel to the given plane is
$$\displaystyle   \overline{r} \cdot \left ( 2\hat{i}+4\hat{j}-7\hat{k} \right )+\lambda=0$$...........................................................$$(i)$$
This plane passes through the point with P.V. $$\displaystyle 2\hat{i}-\hat{j}-4\hat{k}$$
$$\displaystyle \therefore \left ( 2\hat{i}-\hat{j}-4\hat{k} \right ) \cdot \left ( 2\hat{i}-\hat{j}-4\hat{k} \right )+\lambda=0$$
$$\displaystyle  \Rightarrow 4-4+28+\lambda=0$$
$$\displaystyle \therefore \lambda= -28 $$
Putting $$\displaystyle  \lambda= -28 $$ in $$(i)$$ we get the required plane given
by $$\displaystyle   \overline{r} \cdot \left ( 2\hat{i}+4\hat{j}-7\hat{k} \right )=28$$
Hence (d) is the correct choice.

Let $$\displaystyle P\left ( a,b,c \right )$$ be the foot of the  perpendicular from the origin to the plane, then direction ratios of
$$OP$$ are $$\displaystyle a-0,b-0,c-0$$ i.e. $$\displaystyle a,b,c$$.
So, the equation of the plane passing through $$\displaystyle P\left (
a,b,c \right )$$ the direction ratios of the normal to which
are $$\displaystyle a,b,c$$ is 
 $$\displaystyle a\left ( x-a \right )+b\left ( y-b \right )+c\left ( z-c \right )=0$$
$$\displaystyle \Rightarrow$$   $$\displaystyle ax+by+cz=a^2+b^2+c^2$$

given direction cosines of the line normal to the plane are $$ l, m, n$$
$$\Rightarrow \hat{n} = l\hat{i}+m\hat{j}+n\hat{n}$$
we know equation of plane whose distance is p from the origin is given by,
$$ \vec{r} \cdot \hat{n} = p$$
$$\Rightarrow lx + my + nz = p$$

Given $$\displaystyle \overrightarrow{r}=\left ( s-2t \right )\hat{i}+\left ( 3-t \right )\hat{j}+\left ( 2s-t \right )\hat{k}$$
$$\displaystyle  \Rightarrow \overrightarrow{r}=3\hat{j}+s\left ( \hat{i}+2\hat{k} \right )+t\left ( -2\hat{i}+\hat{j}-\hat{k} \right )$$
This equation represent the equation ofthe plane through a point having position vector $$\displaystyle \overrightarrow{a}=3\hat{j}$$ parallel to the vectors
 $$\displaystyle \overrightarrow{b}=\hat{i}+2\hat{k}$$ and $$\overrightarrow{c}=-2\hat{i}-\hat{j}+2\hat{k}$$.
Normal Vector to the plane is given by
$$\displaystyle \overrightarrow{n}=\left | \overrightarrow{b} \times \overrightarrow{c} \right |=2\hat{i}-5\hat{j}-\hat{k}$$
Now using scalar product form of equation of plane
$$\displaystyle  \overrightarrow{r} \cdot \overrightarrow{n}= \overrightarrow{a} \cdot \overrightarrow{n}$$
$$\displaystyle  \Rightarrow \overrightarrow{r} \cdot \left ( 2\hat{i}-5\hat{j}-\hat{k} \right )= (3\hat{j})\left ( 2\hat{i}-5\hat{j}-\hat{k} \right )$$
$$\displaystyle  \Rightarrow \overrightarrow{r} \cdot \left ( 2\hat{i}-5\hat{j}-\hat{k} \right )= -15 $$
Hence choice (a) is correct.

As the points are collinear, the slope of the line joining any two points, should be same as the slope of the line joining two other points. 
Slope of the line passing through points $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ $$ $$=$$ $$ $$\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } } $$
So, slope of the line joining $$ (3,5) , (m,6) = $$ Slope of the line joining $$ (3,5) $$ and $$\left  (\dfrac {1}{2}, \dfrac {15}{2}\right ) $$ 

Therefore, $$ \dfrac { 6 - 5 }{ m - 3 } = \dfrac { \frac {15}{2} - 5 }{ \frac {1}{2} - 3 } $$
$$\Rightarrow  \dfrac { 1 }{ m - 3 } = -1 $$

$$\Rightarrow  m - 3 = -1 $$

$$\Rightarrow  m = 2 $$

Let $$OA$$ and $$OB$$ be two lines with $$DC's{ l }_{ 1 },{ m }_{ 1 },{ n }_{ 1 }$$ and $$\displaystyle { l }_{ 2 },{ m }_{ 2 },{ n }_{ 2 }.$$

Let $$l,m,n$$ be the  d.c's of  the given line segment $$PQ.$$

The given points are $$A(-5,1)$$, $$B(1,p)$$ and $$C(4,-2)$$
We have $$ (x_1= -5, y_1= 1), (x_2= 1, y_2= p)$$ and $$ (x_3= 4, y_3= -2)$$
The given points $$A, B$$ and $$ C$$ are collinear
Therefore, $$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$$
$$\Rightarrow (-5)\cdot (p+2)+1\cdot (-2-1)+4\cdot (1-p)=0$$
$$\Rightarrow (-5p-10-3+4-4p)=0$$
$$\Rightarrow -9p= -9$$
$$\Rightarrow p= -1$$
Hence, $$p= -1$$

The position vectors of two given points are $$a=\hat { i } -\hat { j } +3\hat { k } $$ and $$b=3\hat { i } +3\hat { j } +3\hat { k } $$ and the equation of the given plane is $$r.\left( 5\hat { i } +2\hat { j } +7\hat { k }  \right) +9=0$$ or $$r.n+d=0.$$
We have, $$a.n.+d=\left( \hat { i } -\hat { j } -\hat { k }  \right) .\left( 5\hat { i } +2\hat { j } -7\hat { k }  \right) +9=5-2-21+9<0$$
and $$b.n+d=\left( 3\hat { i } +3\hat { j } +3\hat { k }  \right) .\left( 5\hat { i } +2\hat { j } +7\hat { k }  \right) +9=15+6-21+9>0$$
So, the points  $$a$$ and $$b$$ are on the opposite sides of the plane. 

Let $$ax+by+cz+d=0$$ be the plane, then $$O(0,0,0);A(1,-2,-1);B(3,-2,3)$$
$$\Rightarrow d=0$$ and $$a-2b-c=0$$ also $$3a-2b+3c=0$$
Putting $$c=(a-2b),$$ we get $$6a=8b$$
$$\displaystyle a=\frac { 4 }{ 3 } b\therefore a=\frac { 4 }{ 3 } b,b=b$$ and $$\displaystyle c=-\frac { 2b }{ 3 } $$
And the plane is $$\displaystyle b\left( \frac { 4 }{ 3 } x+y-\frac { 2 }{ 3 } z \right) =0$$
$$4x+3y-2z=0$$
The normal $$\pm \left( \overrightarrow { n } =4\hat { i } +3\hat { j } -2\hat { k }  \right) $$
$$\displaystyle \hat { n } =\pm \left( \frac { 4 }{ \sqrt { 29 }  } \hat { i } +\frac { 3 }{ \sqrt { 29 }  } \hat { j } +\frac { 2 }{ \sqrt { 29 }  } \hat { k }  \right) $$
D.C' s of normal vector $$\displaystyle \left< \pm \frac { 4 }{ \sqrt { 29 }  } \pm \frac { 3 }{ \sqrt { 29 }  } \pm \frac { -2 }{ \sqrt { 29 }  }  \right> $$