Maths Solve the Following Question.(3 Marks)

Let M be the foot of the perpendicular drawn from point A (3, 2, 1) to the given line $\frac{x-7}{2}=\frac{y-7}{2}=\frac{z-6}{3}$
therefore x=7-2λ,y=7+2λ,z=6+3λ are the coordinates of any point on the given line M=(7-2λ,y=7+2λ,6+3λ)
The d.r.s of AM are 7 - 2λ - 3, 7 + 2λ - 2, 6 + 3λ-1
i.e. 4 - 2λ, 5 + 2λ, 5 + 3λ
The d.r.s of the given line are -2, 2, 3
Since, AM is perpendicular to the given line
(4 - 2λ) (-2) + (5 + 2λ) (2) + (5 + 3λ) (3) = 0
-8 + 4λ + 10 + 4λ + 15 + 9λ = 0
17λ+ 17 = 0
λ = -1 [1]
The coordinates of the foot of the perpendicular
i.e. M = (9, 5, 3)
The length of the perpendicular distance 
i.e. $AM =\sqrt{(9-3)^2+(5-2)^2+(3-1)^2}$
$=\sqrt{49}$
l(AM) = 7 units

Shortest distance between the lines

shortest distance between the given lines$=\left|-\frac{116}{\sqrt{116}}\right|$
$=\sqrt{116}$
$=2 \sqrt{29}$ units

The Cartesian equation of a line is 
2x -2 =3y + 1 = 6z -2
$2(x-1)=3\left(y+\frac{1}{3}\right)=6\left(z-\frac{1}{3}\right)$....dividing by 6 all side 
$\frac{x-1}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{1}{3}}{1}$
The direction ratios of the line are 3, 2, 1 
Further, the line passes through the point $\left(1,-\frac{1}{3}, \frac{1}{3}\right)$
Let $A =\left(1,-\frac{1}{3}, \frac{1}{3}\right)$
Thus, the line passes through the point having the position vector. 
$\overline{ a }=1 \hat{ i }-\frac{1}{3} \hat{ j }+\frac{1}{3} \hat{ k }$
Let $\widehat{b}=3 \hat{i}+2 \hat{j}+\hat{k}$
Then, the line is parallel to the vector $\bar{b}$
Hence, the vector equation of the line is 
$\overline{ r }=\overline{ a }+\lambda \overline{ b }$, where $\lambda \in R$
i.e. $\bar{r}=\left(1 \hat{i}-\frac{1}{3} \hat{j}+\frac{1}{3} \hat{k}\right)+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})$

Let $\bar{a}$ and $\bar{b}$ be the vectors in the direction of the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and
$\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$ respectively.
$\therefore \bar{a}=4 \hat{i}+\hat{j}+8 \widehat{k}$ and $\bar{b}=2 \hat{i}+2 \hat{j}+\widehat{k}$
$\therefore \bar{a} \cdot \bar{b}=4 \times 2+1 \times 2+8 \times 1=8+2+8=18$
and
$|\bar{a}|=\sqrt{16+1+64}=\sqrt{81}=9$
$\bar{b}=\sqrt{4+1+4}=\sqrt{9}=3$
Let $\Theta$ be the acute angle between the two given lines.
$\therefore \cos \theta=\left\lvert\, \frac{\bar{a} \cdot \bar{b}}{|\bar{a}| \cdot|\bar{b}|}=\frac{18}{9 \times 3}=\frac{2}{3}\right.$
$\theta=\cos ^{-1}\left(\frac{2}{3}\right)$

The lines are
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \ldots(1)$
$\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$...(2)
Here
$x_1=1, y_1=2, z_1=3$ and $a_1=2, b_1=3, c_1=4$
$x_2=2, y_2=4, z_2=5$ and $a_2=3, b_2=4, c_2=5$
Shortest distance between the lines is
$d=\frac{\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|}{\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}$
Now $\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|=\left|\begin{array}{ccc}1 & 2 & 2 \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|$
$=1(15-16)-2(10-12)+2(8-9)=-1+4-2=1$
and
$\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2=(15-16)^2+(12-10)^2+(8-9)^2$
$=1+4+1=6$
Hence, the shortest distance between lines (1) and (2) $=\left|\frac{1}{\sqrt{6}}\right|=\frac{1}{\sqrt{6}}$ units

Equation of lines are.,
$\bar{r}=(4 \overline{ i }-\overline{ j })+\lambda(\overline{ i }+2 \overline{ j }-3 \overline{ k }) \&$
$\bar{r}=(\hat{i}-\hat{j}+2 \widehat{k})+\mu(\hat{i}+4 \hat{j}-5 \widehat{k})$
∴ above lines passes through
$\overline{ a }_1=(4 \overline{ i }-\overline{ j })$ and$\overline{ a }_2=(\overline{ i }-\overline{ j }+2 \overline{ k })$
and parallel to
$\overline{ b }_1=\overline{ i }+2 \overline{ j }-3 \overline{ k } \& \overline{ b }_2=\overline{ i }-4 \overline{ j }-5 \overline{ k }$
Shortest distance $=\left|\frac{\left(\overline{ a }_2-\overline{ a }_1\right) \cdot\left(\overline{ b }_1 \times \overline{ b }_2\right)}{\left|\left(\overline{ b }_1 \times \overline{ b }_2\right)\right|}\right|$
$\Rightarrow \overline{ a }_2-\overline{ a }_1=-3 \overline{ j }+2 \overline{ k }$
$\overline{ b }_1 \times \overline{ b }_2=\left|\begin{array}{ccc}\overline{ i } & \overline{ j } & \overline{ k } \\ 1 & 2 & -3 \\ 1 & 4 & -5\end{array}\right|=2 \overline{ i }+2 \overline{ j }+2 \overline{ k }$
$\therefore\left|\overline{ b }_1 \times \overline{ b }_2\right|=2 \sqrt{3}$
Shortest distance $=\left|\frac{(-3 \overline{ i }+2 \overline{ k }) \cdot(2 \overline{ i }+2 \overline{ j }+2 \overline{ k })}{2 \sqrt{3}}\right|$
$=\left|\frac{-6+4}{2 \sqrt{3}}\right|$
$=\left|-\frac{2}{2 \sqrt{3}}\right|$
$d =\frac{1}{\sqrt{3}}$ units

Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=u$ where is any constant.
So for any point on this line has co-ordinates in the form (2u+1,3u-1,4u+1)
$\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=v$
So for any point on this line has co-ordinates in the form   (v+3,2v+k,v).
Point of intersection of these two lines will have co-ordinates of the form (2u +1, 3u −1,4u +1) and (v +3, 2v + k,v) .
Equating the x, y and z co-ordinates for both the forms we get three equations
2u+1=v+3
2u-v=2.............(1)
3u-1=2v+k
3u-2v=k+1.......(2)
4u+1=v
4u-v=-1...........(3)
Subtracting equation (1)from equation(3) we get,
2u = -3
u=-3/2
Substitute value of u in equation (1) we get,
2(-3/2) - v=2
v=-5
Substitute value of v and in equation (2) we get,
3(-3/2) - 2(-5)=k+1
k=9/2
the value of k is 9/2

Let $p 1$ and $p 2$ be the distances of points $\hat{i}-\hat{j}+3 \hat{k}$ and $3 \hat{i}+4 \hat{j}+3 \widehat{k}$ from $\bar{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+8=0$
The distance of the point A with position vector a from the plane $\bar{r} \cdot \bar{n}= p$ is given by
$d=\frac{|\bar{a} \cdot \bar{n}-p|}{|\bar{n}|}$
$\therefore p_1=\frac{|(\hat{i}-\hat{j}+3 \widehat{k}) \cdot(5 \hat{i}+2 \hat{j}-7 \widehat{k})-(-8)|}{\sqrt{5^2+2^2+(-7)^2}}$
$=\frac{|1(5)-1(2)+3(-7)+8|}{\sqrt{25+4+49}}$
$=\frac{|5-2-21+8|}{\sqrt{78}}=\frac{|-10|}{\sqrt{78}}=\frac{10}{\sqrt{78}}$
and $p_2=\frac{|()()-(-8)|}{\sqrt{5^2+2^2+(-7)^2}}$
$=\frac{|3(5)+4(2)+3(-7)+8|}{\sqrt{25+4+49}}$
$=\frac{|15+8-21+8|}{\sqrt{78}}$
$=\frac{10}{\sqrt{78}}$
$\therefore p _1= p _2$
Hence, points are equidistant from the plane.

The equation $\bar{r}=\bar{a}+s \bar{b}+t \bar{c}$ represents a plane passing through a point having position vector $a$ and parallel to the vectors $b$ and $c$.
Here, $\bar{a}=\hat{i}+\hat{j}, \bar{b}=\hat{i}-\hat{j}+2 \widehat{k}$ and $\bar{c}=\hat{i}+2 \hat{j}+\widehat{k}$
The given plane is perpendicular to the vector$\bar{n}$

Vector equation of the plane in scalar product form is $\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}$

$\therefore x(-5)+y(1)+z(3)=-4$
$\therefore-5 x+y+3 z=-4$
$\therefore 5 x-y-3 z=4$
which is the cartesian form of the equation of the plane

Shortest distance between the lines

shortest distance between the given lines $=\left|-\frac{116}{\sqrt{116}}\right|$
$\begin{aligned} & =\sqrt{116} \\ & =2 \sqrt{29} \text { units }\end{aligned}$