Maths Solve the Following Question.(2 Marks)

Equation of line is $\frac{x+2}{2}=\frac{2 y-5}{3} ; z=-1$
$\frac{x+2}{2}=\frac{y-\frac{5}{2}}{\frac{3}{2}}=\frac{z+1}{1}$
Direction ratios of the line are 2, 3/2, 1.
∴Direction cosines of the line are
$\frac{2}{\sqrt{4+\left(\frac{9}{4}\right)+1}}, \frac{\frac{3}{2}}{\sqrt{4+\frac{9}{4}+1}}, \frac{1}{\sqrt{4+\frac{9}{4}+1}}$
i.e $\frac{2}{\frac{1}{2} \sqrt{\frac{29}{4}}}, \frac{\frac{3}{2}}{\frac{1}{2} \sqrt{\frac{29}{4}}}, \frac{1}{\frac{1}{2} \sqrt{\frac{29}{4}}}$
$\frac{4}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{2}{\sqrt{29}}$

The plane is perpendicular to $\bar{r}=2 \hat{i}+\hat{j}+2 \widehat{k}$
the normal vector $\bar{n}$ to the plane is
$\bar{n}=2 \hat{i}+\hat{j}+2 \widehat{k}$
∴ unit vector along this normal is 
$\bar{n}=\frac{\bar{n}}{|\bar{n}|}=\frac{2 \hat{i}+\hat{j}+2 \widehat{k}}{\sqrt{2^2+1^2+2^2}}$
$=\frac{2 \hat{i}+\hat{j}+2 \widehat{k}}{3}$
The vector equation of the plane in normal form is $\bar{r}=\bar{n}=p$ where $p$ is the distance of the plane from the origin. Here $p=5$.
$\bar{r}=\frac{2 \hat{i}+\hat{j}+2 \widehat{k}}{3}=5$
$\therefore \bar{r}=(2 \hat{i}+\hat{j}+2 \widehat{k})=15$

The equation of the given plane is 3x+ 4y-2z = 5
Let $\bar{r}=x \hat{i}+y \hat{j}+z \hat{k}$ be any vector in the plane and $3,4,-2$ are direction
ratios of $\bar{n}$, which is normal to the plane
$\therefore \overline{ n }=3 \hat{ i }+4 \hat{ j }-2 \hat{ k }$
$\therefore(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}+4 \hat{j}-2 \hat{k})=5$
$\therefore \hat{ r } \cdot(3 \hat{ i }+4 \hat{ j }-2 \hat{ k })=5$ which is required form.

Let $\bar{a}=4 \hat{i}-\hat{j}+2 \widehat{k}$
$\bar{b}=-\hat{2} i+\hat{j}+\widehat{k}$
Equation of the line passing through point $A(\bar{a})$ and having direction $\bar{b}$ is
$\bar{r}=\bar{a}+\lambda \bar{b}$
$\bar{r}=(4 \hat{i}-\hat{j}+2 \widehat{k})+\lambda(-2 \hat{i}+\hat{j}+\widehat{k})$

$\begin{aligned} & \cos \theta=\left|\frac{3 \times-2+2 \times 1+6 \times 2}{\sqrt{(3)^2+(2)^2+(6)^2} \sqrt{(-2)^2+1^2+2^2}}\right| \\ & =\left|\frac{-6+2+12}{\sqrt{49} \sqrt{9}}\right|=\frac{8}{7 \times 3}=\frac{8}{21} \\ & \Rightarrow \theta=\cos ^{-1}\left(\frac{8}{21}\right)\end{aligned}$

Equation of line passing through the point $A\left(x_1, y_1, z_1\right)$ and $B\left(x_2, y_2, z_2\right)$ is
$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
Equation of line passing through the point A(3, 4,-7) and B(6,-1,1) is
$\begin{aligned} & \frac{x-3}{6-3}=\frac{y-4}{-1-4}=\frac{z-(-7)}{1-(-7)} \\ & \frac{x-3}{3}=\frac{y-4}{-5}=\frac{z+7}{8}\end{aligned}$

QA and QB are the perpendiculars from the point Q( a, b, c) to YZ and ZX planes.
∴ A = ( 0, b, c) and B = (a, 0, c)
The required plane is passing through O(0, 0, 0), A(0, b, c) and B(a, 0, c)
The vector equation of the plane passing through the O,A,B is
$\bar{r} \cdot(\overline{O A} \times \overline{O B})=\overline{0} \cdot(\overline{O A} \times \overline{O B})$
i.e; $\bar{r} \cdot(\vec{a} \times \vec{b})=0$
Now, $\overline{O A}=\bar{a}=0 . \hat{i}+b \hat{j}+c \widehat{k}$
and $\overline{O B}=\bar{b}=a \hat{i}+0 . \hat{j}+c \widehat{k}$
$\therefore \overline{O A} \times \overline{O B}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{k} \\ 0 & b & c \\ a & 0 & c\end{array}\right|$
$=(b c-0) \hat{i}-(0-a c) \hat{j}+(0-a b) \widehat{k}$
$=b c \hat{i}+a c \hat{j}-a b \hat{k}$
∴ from (1), the vector equation of the required plane is
$\bar{r} \cdot(b c \hat{i}+a c \hat{j}-a b \widehat{k})=0$

Given equations of the line are:
Let $\bar{a}$ and $\bar{b}$ be vectors in the direction of lines
$\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}$ respectively
$\therefore \bar{a}=-3 \hat{i}+2 k \hat{j}+2 \widehat{k}$ and $\bar{b}=3 k \hat{i}+\hat{j}-5 \widehat{k}$
$\bar{a} \cdot \bar{b}=-9 k+2 k-10=-7 k-10$
Given lines are at right angle
$\theta=90^{\circ}$
$\cos \theta=\frac{\bar{a} \cdot \bar{b}}{|\bar{a}| \cdot|\bar{b}|}$
$\bar{a} \cdot \bar{b}=0$
$-7 k-10=0$
$k=-\frac{10}{7}$

We know that the vector equation of a plane passing through a point  
$A(\bar{a})$ and normal to $\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}$
Here $\bar{a}=3 \hat{i}-2 \hat{j}+\widehat{k}$ and $\widehat{n}=4 \hat{i}+3 \hat{j}+2 \widehat{k}$
The vector equation of the required plane is
$\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}$
$\bar{r} \cdot(4 \hat{i}+3 \hat{j}+2 \widehat{k})=(3 \hat{i}-2 \hat{j}+\widehat{k}) \cdot(4 \hat{i}+3 \hat{j}+2 \widehat{k})$
$\bar{r} \cdot(4 \hat{i}+3 \hat{j}+2 \widehat{k})=12-6+2$
$\bar{r} \cdot(4 \hat{i}+3 \hat{j}+2 \widehat{k})=8$
The vector equation of the required plane is $\bar{r} \cdot(4 \hat{i}+3 \hat{j}+2 \widehat{k})=8$

Given equations of the line are:
3x -1 = 6y +2 = 1 - z
Rewriting the above equation, we have,
$3\left(x-\frac{1}{3}\right)=6\left(y+\frac{2}{6}\right)=-(z-1)$
$\frac{\left(x-\frac{1}{3}\right)}{\frac{1}{3}}=\frac{\left(y+\frac{1}{3}\right)}{\frac{1}{6}}=\frac{(z-1)}{-1} \ldots$.(1)
Now consider the general equation of the line:
$\frac{x-a}{l}=\frac{y-b}{m}=\frac{z-c}{n} \ldots(2)$
where, l,m and n are the direction ratios of the line and the point (a,b,c) lies on the line.
Compare the equation (1), with the general equation (2),
we have l=1/3, m=1/6 and n=-1
Also, a=1/3, b=-1/3 and c=1
This shows that the given line passes through (1/3, -1/3, 1)
Therefore, the given line passes through the point having position vector $\bar{a}=\frac{1}{3} \hat{i}-\frac{i}{3} \hat{j}+\widehat{k}$ and is parallel to the vector $\bar{b}=\frac{1}{3} \hat{i}+\frac{1}{6} \hat{j}-\widehat{k}$
So its vector equation is
$\bar{r}=\left(\frac{1}{3} \hat{i}-\frac{1}{3} \hat{j}+\widehat{k}\right)+\lambda\left(\frac{1}{3} \hat{i}+\frac{1}{6} \hat{j}-\widehat{k}\right)$

Given planes are:
$\bar{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})=3$ and $\bar{r} \cdot(\hat{i}+2 \hat{j}+\widehat{k})=1$
The angle between two planes with direction ratios,
$\left(a_1, b_1, c_1\right)$ and $\left(a_2, b_2, c_2\right)$ is
$\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$
$\cos \theta=\frac{2 \times 1+1 \times 2-1 \times 1}{\sqrt{2^2+1^2+(-1)^2} \sqrt{1^2+2^2+1^2}}$
$\cos \theta=\frac{3}{\sqrt{6} \cdot \sqrt{6}}$
$\cos \theta=\frac{3}{6}$
$\cos \theta=\frac{1}{2}$
$\cos \theta=\cos \left(\frac{\pi}{3}\right)$
$\theta=\frac{\pi}{3}$

$\cos \theta=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right|$
$=\left|\frac{15-48-65}{\sqrt{25+144+169} \sqrt{9+16+25}}\right|$
$=\left|-\frac{98}{13 \sqrt{2} \times 5 \sqrt{2}}\right|$
$=\left|-\frac{98}{13 \times 5 \times 2}\right|$
$=\frac{49}{65}$
$\theta=\cos ^{-1}\left(\frac{49}{65}\right)$

$\vec{r}(3 \hat{i}-4 \hat{j}+12 \widehat{k})=8 \ldots \ldots \ldots(i)$
$\vec{n}=3 \hat{i}-4 \hat{j}+12 \widehat{k}$
$|\vec{n}|=\sqrt{9+16+144}=13$
$\therefore \vec{n}=\frac{3 \hat{i}-4 \hat{j}+12 \widehat{k}}{13}$
$\therefore \vec{r} \cdot\left(\frac{3 \hat{i}-4 \hat{j}+12 \widehat{k}}{13}\right)=\frac{8}{13} \ldots \ldots \ldots \ldots . . . b y(i)$
$\vec{r} \cdot \vec{n}=p$
$\therefore$ Perpendicular distance from the origin is $\frac{8}{13}$

Let θ be the acute angle between the lines whose direction ratios are 4, –3, 5 and 3, 4, 5.
Then,
$\cos \theta=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \cdot \sqrt{a_2^2+b_2^2+c_2^2}}\right|$
$\cos \theta=\left|\frac{4(3)+(-3)(4)+5(5)}{\sqrt{4^2+(3)^2+5^2} \cdot \sqrt{3^2+4^2+5^2}}\right|$
$=\left|\frac{12-12+25}{\sqrt{16+9+25} \cdot \sqrt{9+16+25}}\right|$
$=\left|\frac{25}{50}\right|=\frac{1}{2}$
$\cos \theta=\frac{1}{2}$
$\theta=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$
The angle between the lines is $\frac{\pi}{3}$

Let $\bar{b}_1$ and $\bar{b}_2$ be the vectors in the direction of the lines

Let θ be the acute angle between the two given lines.
$\therefore \cos \theta=\frac{\bar{b}_1 \bar{b}_2}{\left|\bar{b}_1\right|\left|\bar{b}_2\right|}=\frac{19}{3 \times 7}$
$\therefore \cos \theta=\frac{19}{21}$
$\theta=\cos ^{-1}\left(\frac{19}{21}\right)$