Maths Solve the Following Question.(4 Marks)

$\bar{r} \cdot \bar{n}_1=d_1$ and $\bar{r}, \bar{n}_2=d_2$ is given by

$\cos \theta=\left|\frac{\bar{n}_1 \cdot \bar{n}_2}{\left|\bar{n}_1\right|\left|\bar{n}_2\right|}\right|$

Here,

$\begin{aligned} & \bar{n}_1=-2 \hat{i}+\hat{j}+2 \hat{k} \\ & \bar{n}_2=2 \hat{i}+2 \hat{j}+\hat{k} \\ & \therefore \bar{n}_1 \cdot \bar{n}_2 \\ & =(2 \hat{i}+\hat{j}+2 \hat{k}) \cdot(2 \hat{i}+\hat{j}+\hat{k})\end{aligned}$

= (1)(2) + (1)(1) + (2)(1) = 2 + 1 + 2 = 5 Also,

$\begin{aligned} & \left|\bar{n}_1\right|=\sqrt{1^2+1^2+2^2}=\sqrt{6} \\ & \left|\bar{n}_2\right|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}\end{aligned}$

$\therefore$ from $(1)$, we have

$\begin{aligned} & \cos \theta=\left|\frac{3}{\sqrt{6} \sqrt{6}}\right| \\ & =\frac{3}{6} \\ & =\frac{1}{2} \cos 90^{\circ} \\ & \therefore \theta=90^{\circ} .\end{aligned}$

$\begin{aligned} & \therefore \frac{\hat{i}}{\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|}=\frac{\hat{j}}{\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|}=\frac{\hat{k}}{\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|} \\ & \therefore \frac{\hat{i}}{1}=\frac{\hat{j}}{1}=\frac{\hat{k}}{1} \\ & \text { i.e. } \frac{\hat{i}}{1}=\frac{\hat{j}}{1}=\frac{\hat{k}}{1}\end{aligned}$

$\therefore \hat{\mathrm{i}}, \mathrm{j}, \hat{\mathrm{k}}$ are proprtional to $1,1,1$

∴ from (1), the required cartesian equation is x – y + z = 0 Case 2 : Let he plane make non zero intercept p on each axis.

then its equation is $\frac{\hat{\mathrm{i}}}{p}+\frac{\hat{\mathrm{j}}}{p}+\frac{\hat{\mathrm{k}}}{p}=1=1$

i.e. $\hat{i}+\hat{j}+\hat{k}=p=p$...(2)

Since this plane pass through (1, 1, 1) ∴ 1 + 1 + 1 = p ∴ p = 3

$\therefore$ from (2), the required cartesian equation is $\hat{i}+\hat{j}+\hat{k}=3$

Hence, the cartesian equations of required planes are $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=3$

$\therefore \frac{a}{\left|\begin{array}{ll}2 & 3 \\ 2 & 1\end{array}\right|}=\frac{b}{\left|\begin{array}{ll}3 & 1 \\ 1 & 3\end{array}\right|}=\frac{c}{\left|\begin{array}{ll}1 & 2 \\ 3 & 2\end{array}\right|}$

$\therefore \frac{a}{-4}=\frac{b}{8}=\frac{c}{-4}$

i.e. $\frac{a}{1}=\frac{b}{-2}=\frac{c}{1}$

∴ a, b, c are proportional to 1, -2, 1 ∴ from (1), the required cartesian equation is x – 2y + z = 0. Case 2 : Let the plane make non zero intercept p on each axis.

then its equation is $\frac{x}{p}+\frac{y}{p}+\frac{z}{p}=1$

i.e. x + y + z = p …(2) Since this plane pass through (1, 2, 3) and (3, 2, 1) ∴ 1 + 2 + 3 = p and 3 + 2 + 1 = p ∴ p = 6 ∴ from (2), the required cartesian equation is x + y + z = 6 Hence, the cartesian equations of required planes are x + y + z = 6 and x – 2y + z = 0.

position vector $\overline{\mathrm{a}}$ and parallel to vectors $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$.

Here,

$\begin{aligned} & \bar{b}=\hat{i}+\hat{j}-\hat{k} \\ & \bar{c}=\hat{i}+2 \hat{j}+3 \hat{k}\end{aligned}$

$\begin{aligned} & \therefore \overline{\mathrm{b}} \times \overline{\mathrm{c}}=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3\end{array}\right| \\ & =(3+2) \hat{\mathrm{i}}-(3-1) \hat{j}+(-2-1) \hat{\mathrm{k}} \\ & =5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\ & =\bar{a}\end{aligned}$

Also,

$\begin{aligned} & \bar{a} \cdot(\bar{b} \times \bar{c}) \\ & =\bar{a} \cdot \bar{a}=|\bar{a}|^2 \\ & =(5)^2+(4)^2+(0)^2 \\ & =0\end{aligned}$

The vector equation of the plane passing through $\mathrm{A}(\bar{a})$ and parallel to $\bar{b}$ and $\bar{c}$ is

$\bar{r} \cdot(\bar{b} \times \bar{c})=\bar{a} \cdot(\bar{b} \times \bar{c})$

$\therefore$ the vector equation of the given plane is

$\bar{r} \cdot(5 \hat{i}-4 \hat{j}-\hat{k})=0$

If $\overline{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}$, then this equation becomes

$(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(5 \hat{i}-4 \hat{j}-\hat{k})=0$

$\therefore 5 x-4 y+z=0$.

This is the cartesian equation of the required plane.

$\begin{aligned} & 4 x=3 z-5=3\left(z-\frac{5}{3}\right) \\ & \therefore \frac{4 x}{12}=\frac{3\left(z-\frac{5}{3}\right)}{12} \\ & \therefore \frac{x}{3}=\frac{z-\frac{5}{3}}{4}\end{aligned}$

$\therefore$ the cartesian equations of the line are

$\frac{x}{3}=\frac{z-\frac{5}{3}}{4}, y=2$

This line passes through the point $\mathrm{A}\left(0,2, \frac{5}{3}\right)$ position vector is $\bar{a}=2 \hat{j}+\frac{5}{3} \hat{k}$

Also the line has direction ratio 3, 0, 4.

If $\bar{b}$ is a vector parallel to the line, then $\bar{b}=3 \hat{i}+4 \hat{k}$

The vector equation of the line passing through $\mathrm{A}(\bar{a})$ and parallel to $b$ is $\bar{r}=\bar{a}+\lambda b$

where $\lambda$ is $\bar{a}$ scalar,

∴ the vector equation of the required line is

$\bar{r}=\left(2 \hat{j}+\frac{5}{3} \hat{k}\right)+\lambda(3 \hat{i}+4 \hat{k})$

Then $\bar{a}=-\hat{i}-\hat{j}+2 \hat{k}$

The equation of given line is x – 2 = 3y + 1 = 6z – 2.

$\begin{aligned} & \therefore 2(x-1)=3\left(y+\frac{1}{3}\right)=6\left(z-\frac{1}{3}\right) \\ & \therefore \frac{x-1}{\left(\frac{1}{2}\right)}=\frac{y+\frac{1}{3}}{\left(\frac{1}{3}\right)}=\frac{z-\frac{1}{3}}{\left(\frac{1}{6}\right)}\end{aligned}$

The direction ratios of this line are

$\frac{1}{2}, \frac{1}{3}, \frac{1}{6}$ i.e. $3,2,1$

Let $\bar{b}$ be the vector parallel to this line.

Then $\bar{b}=3 \hat{i}+2 \hat{j}+\hat{k}$

The vector equation of the line passing through $\mathrm{A}(\bar{a})$ and

parallel to $\bar{b}$ is

$\bar{r}=\bar{a}+\lambda \bar{b}$, where $\lambda$ is a scalar

$\therefore$ the vector equation of the required line is

$\bar{r}=(-\vec{i}-\vec{j}+2 \hat{k})+2(3 \hat{i}+2 \hat{j}+\hat{k})$.

The line passes through $(-1,-1,2)$ and has direction

ratios $3,2,1$

$\therefore$ the cartesian equations of the line are

$\begin{aligned} & \frac{x-(-1)}{3}=\frac{y-(-1)}{2}=\frac{z-2}{1} \\ & \text { i.e. } \frac{x+1}{3}=\frac{y+1}{2}=\frac{z-2}{1}\end{aligned}$

$\bar{r}=\bar{a}_1+\lambda \bar{b}_1$ and $\bar{r}=\bar{a}_2+\mu \bar{b}_2$ is given by

$d=\left|\frac{\left(\bar{a}_2-\bar{a}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)}{\left|\bar{b}_1 \times \bar{b}_2\right|}\right|$.

Here, $\bar{a}_1=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \bar{a}_2=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$,

$\begin{aligned} & \overline{\mathrm{b}}_1=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}_2=\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\ & \therefore \overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -1 & 1 \\ 1 & 1 & -2\end{array}\right| \\ & =(2-1) \hat{\mathrm{i}}-(-4-1) \hat{j}+(4+1) \hat{k} \\ & =\hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\end{aligned}$

$\begin{aligned} & \overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1=(2 \hat{\mathrm{i}}+2 \hat{j}-3 \hat{k})-(\hat{i}+\hat{j}-\hat{k}) \\ & \therefore\left(\bar{a}_2-\overline{\mathrm{a}}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=\hat{i} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k}) \\ & =1(-1)+0(3)+0(2) \\ & =-1\end{aligned}$

and

$\begin{aligned} & \left|\bar{b}_1 \times \bar{b}_2\right|=\sqrt{(-1)^2+3^2+2^2} \\ & =\sqrt{1+9+4} \\ & =\sqrt{14}\end{aligned}$

Shortest distance between the lines is 0. ∴ the lines intersect each other.

$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda$

……(Say) The coordinates of any point on the line are given by x = 5λ – 3, y = 2λ + 1, z = 3λ – 4 Let M = (5λ – 3, 2λ + 1, 3λ – 4) …(1) The direction ratios of PM are 5λ – 3 – 0, 2λ + 1 – 2, 3λ – 4 – 3 i.e. 5λ – 3, 2λ – 1, 3λ – 7 Since, PM is perpendicular to the line whose direcction ratios are 5, 2, 3, 5(5λ – 3) + 2(2λ – 1) + 3(3λ – 7) = 0 25λ – 15 + 4λ – 2 + 9λ – 21 =0 38λ – 38 = 0 ∴ λ = 1 Substituting λ = 1 in (1), we get. M = (5 – 3, 2 + 1, 3 – 4) = (2, 3, -1). Hence, the coordinates of the foot of perpendicular are (2, 3, – 1).

∴ the direction ratios of the line are 1, 1, 0. The direction ratios of the line x = 0, z = 0, i.e., Y-axis are 0, 1, 0. ∴ its direction ratios are 0, 1, 0.

Let $\bar{a}$ and $\bar{b}$ be the vectors in the direction of the lines $x=y, z=0$ and $x=0, z=0$

$\begin{aligned} & \text { Then } \bar{a}=\hat{i}+\hat{j}, \bar{b}=\hat{j} \\ & \therefore \bar{a} \cdot \bar{b}=(\hat{i}+\hat{j}) \cdot \hat{j} \\ & =(1)(0)+(1)(1)+(0)(0) \\ & =1\end{aligned}$

$\begin{aligned} & |\bar{a}|=\sqrt{1^2+1^2}=\sqrt{2} \\ & |\bar{b}|=|\hat{j}|=1\end{aligned}$

If $\theta$ is the acute angle between the lines, then

$\cos \theta=\left|\frac{\bar{a} \cdot \bar{b}}{|\bar{a}| \bar{b} \mid}\right|=\left|\frac{1}{\sqrt{2} \times 1}\right|=\frac{1}{\sqrt{2}}=\cos 45^{\circ}$.

$\therefore \theta=45^{\circ}$.

It is perpendicular to the vector $\bar{b}=\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=3 \hat{i}+2 \hat{j}-\hat{k}$.

∴ it is perpendicular to lines whose direction ratios are 1, 2, 1 and 3, 2, -1. ∴ p + 2q + r = 0, 3 + 2q – r = 0

$\begin{aligned} & \therefore \frac{p}{\left|\begin{array}{cc}2 & 1 \\ 2 & -1\end{array}\right|}=\frac{q}{\left|\begin{array}{cc}1 & 1 \\ -1 & 3\end{array}\right|}=\frac{r}{\left|\begin{array}{ll}1 & 2 \\ 3 & 2\end{array}\right|} \\ & \therefore \frac{p}{-4}=\frac{q}{4}=\frac{r}{-1} \\ & \therefore \frac{p}{2}=\frac{q}{-7}=\frac{r}{4}\end{aligned}$

∴ the required line has direction ratios 2, -7, 4. The cartesian equations of the line passing through (x1, y1, z1) and having direction ratios a, b, c are

$\frac{\boldsymbol{x}=\boldsymbol{x}_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

∴ the cartesian equation of the line passing through the point (2, -7, 4) and having directions ratios 2, -7, 4 are

$\frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-2}{4}$

It is perpendicular to the vectors $\bar{b}=\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=3 \hat{i}+2 \hat{j}-\hat{k}$.

∴ it is perpendicular to lines whose direction ratios are 1, 2, 1 and 3, 2, -1. ∴ p + 2q + r = 0, 3p + 2q – r = 0

$\therefore \frac{p}{\left|\begin{array}{rr}2 & 1 \\ 2 & -1\end{array}\right|}=\frac{q}{\left|\begin{array}{rr}1 & 1 \\ -1 & 3\end{array}\right|}=\frac{r}{\left|\begin{array}{ll}1 & 2 \\ 3 & 2\end{array}\right|}$

$\therefore \frac{p}{-4}=\frac{q}{4}=\frac{r}{-4}$

$\therefore \frac{p}{-1}=\frac{q}{1}=\frac{r}{-1}$

∴ the required line has direction ratios -1, 1, -1. The cartesian equations of the line passing through (x1, y1, z1) and having direction ratios a, b, c are

$\frac{x-\boldsymbol{x}_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

∴ the cartesian equations of the line passing through the point (1, 1, 2) and having direction ratios -1, 1, -1 are

$\frac{x-1}{-1}=\frac{y-1}{1}=\frac{z-2}{-1}$

vectors $\bar{b}$ and $\bar{c}$ is

$\bar{r} \cdot(\bar{b} \times \bar{c})=\bar{a} \cdot(\bar{b} \times \bar{c}) \ldots(1)$

Here, $\bar{a}=-2 \hat{i}+7 \hat{j}+5 \hat{k}$

$\bar{b}=4 \hat{i}-\hat{j}+3 \hat{k}, \bar{c}=\hat{i}+\hat{j}+\hat{k}$

$\therefore \bar{b} \times \bar{c}=\left|\begin{array}{ccc}\vec{i} & \hat{j} & \vec{k} \\ 4 & -1 & 3 \\ 1 & 1 & 1\end{array}\right|$

$=(-1-3) \hat{i}-(4-3) \hat{j}+(4+1) \hat{k}$

$=-4 \hat{i}-\hat{j}+5 \hat{k}$

$\begin{aligned} \therefore \bar{a} \cdot(\bar{b} \times \bar{c}) & =(-2 \hat{i}+7 \hat{j}+5 \hat{k}) \cdot(-4 \hat{i}-\hat{j}+5 \hat{k}) \\ & =(-2)(-4)+(7)(-1)+(5)(5) \\ & =8-7+25=26\end{aligned}$

$\therefore$ from (1), the vector equation of the required plane is

$\bar{r} \cdot(-4 \hat{i}-\hat{j}+5 \hat{k})=26$.

normal and p is the length of perpendicular drawn from origin to the plane.

Given plane is $\vec{r} \cdot(3 \hat{i}+4 \hat{j}+12 \hat{k})=78$

$\ldots(1)$

$\bar{n}=3 \hat{i}+4 \hat{j}+12 \hat{k}$ is normal to the plane

$\therefore|\bar{n}|=\sqrt{3^2+4^2+12^2}=\sqrt{169}=13$

Dividing both sides of (1) by 13 , we get

$\bar{r} \cdot\left(\frac{3 \hat{i}+4 \hat{j}+12 \hat{k}}{13}\right)=\frac{78}{13}$

i.e. $\bar{t} \cdot\left(\frac{3}{13} \hat{i}+\frac{4}{13} \hat{j}+\frac{12}{13} \vec{k}\right)=6$

This is the normal form of the equation of plane. Comparing with $\bar{r} \cdot \hat{n}=\mathrm{p}$.

(i) the length of the perpendicular from the origin to plane is 6.

(ii) direction cosines of the normal are $\frac{3}{13}, \frac{4}{13}, \frac{12}{13}$.

$\sqrt{2^2+6^2+(-3)^2}=\sqrt{49}=7$, we get

$\frac{2}{7} x+\frac{6}{7} y-\frac{3}{7} z=\frac{63}{7}=9$

This is the normal form of the equation of plane. ∴ the direction cosines of the perpendicular drawn from the origin to the plane are

$1=\frac{2}{7}, m=\frac{6}{7}, n=-\frac{3}{7}$

and length of perpendicular from origin to the plane is p = 9. ∴ the coordinates of the foot of the perpendicular from the origin to the plane are (lp, mp,

np) i.e. $\left(\frac{18}{7}, \frac{54}{7},-\frac{27}{7}\right)$.

$\bar{r} \cdot(6 \hat{i}-2 \hat{j}+3 \hat{k})=7 \ldots(1)$

where $\bar{r}=x \hat{i}+u \hat{i}+z \hat{k}$

$\therefore \bar{n}=6 \hat{i}-2 \hat{j}+3 \hat{k}$ is normal to the plane

$|\bar{n}|=\sqrt{6^2+(-2)^2+3^2}=\sqrt{49}=7$

Unit vector along $\bar{n}$ is

$\hat{n}=\frac{\bar{n}}{|\bar{n}|}=\frac{6 \hat{i}-2 \hat{j}+3 \hat{k}}{7}$

Dividing both sides of (1) by 7 , we get

$\bar{r} \cdot\left(\frac{6 \hat{i}-2 \hat{j}+3 \hat{k}}{7}\right)=\frac{7}{7}$

$\therefore \bar{r} \cdot \hat{n}=1$

Comparing with normal form of equation of the plane $\bar{r} \cdot \hat{n}=p_1$ it follows that length of

perpendicular from origin is 1 unit. Alternative Method: The equation of the plane is 6x – 2y + 3z – 7 = 0 i.e. 6x – 2y + 3z =

i.e. $\left(\frac{6}{6^2+(-2)^2+3^2}\right) x-\left(\frac{2}{\sqrt{6^2+(-2)^2+3^2}}\right) y+$

$\left(\frac{3}{\sqrt{6^2+(-2)^2+3^2}}\right) z=\frac{7}{\sqrt{6^2+(-2)^2+3^2}}$

i.e. $\frac{6}{7} x-\frac{2}{7} y+\frac{3}{7} z=\frac{7}{7}=1$

This is the normal form of the equation of plane. ∴ perpendicular distance of the origin from the plane is p = 1 unit.