Maths Solve the Following Question.(5 Marks)

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c} \ldots(1)$

This line is perpendicular to the line x – 1 = y – 2 = z – 1 whose direction ratios are 1, 1, 1.

$\therefore a+b+c=0$

$\ldots(2)$

The lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and

$\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_1}$ intersect, if

$\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|=0$

Applying this condition for the lines $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$ and

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$, we get

$\left|\begin{array}{rrr}1-0 & -1-0 & 1-0 \\ a & b & c \\ 2 & 3 & 4\end{array}\right|=0$

∴ 1(4b – 3c) + 1(4a – 2c) + 1(3a – 2b) = 0 ∴ 4b – 3c + 4a – 2c + 3a – 2b = 0 ∴ 7a + 2b – 5c = 0 From (2) and (3), we get

$\frac{a}{\left|\begin{array}{rr}1 & 1 \\ 2 & -5\end{array}\right|}=\frac{b}{\left|\begin{array}{rr}1 & 1 \\ -5 & 7\end{array}\right|}=\frac{a}{\left|\begin{array}{ll}1 & 1 \\ 7 & 2\end{array}\right|}$

$\therefore \frac{a}{-7}=\frac{b}{12}=\frac{c}{-5}$

$\therefore$ the required line has direction ratios $-7,12,-5$.

From (1), cartesian equation of required line are

$\begin{aligned} & \frac{x}{-7}=\frac{y}{12}=\frac{z}{-5} \\ & \text { i.e. } \frac{x}{7}=\frac{y}{-12}=\frac{z}{5} .\end{aligned}$

$\frac{x-5}{4}=\frac{y-7}{5}=\frac{z+3}{5}$ and $x-6=y-8=z+2$ is given by

$d=\frac{\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{array}\right|}{\sqrt{\left(m_1 n_2-m_2 n_1\right)^2+\left(l_2 n_1-1_1 n_2\right)^2+\left(l_1 m_2-l_2 m_1\right)^2}}$

The equation of the given lines are

$\frac{x-5}{4}=\frac{y-7}{5}=\frac{z+3}{5}$ and $x-6=y-8=z+2$

$\begin{aligned} & \therefore x_1=5, y_1=7, z_1=3, x_2=6, y_2=8, z_2=2 \text {, } \\ & I_1=4, m_1=5, n_1=1, r_2=1, m_2=-2, n_2=1 \\ & \left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{array}\right|=\left|\begin{array}{ccc}4 & 6 & 8 \\ 4 & 5 & 5 \\ -6 & -8 & 2\end{array}\right| \\ & \end{aligned}$

= 4(-6 + 2) – 6(7 – 1) + 8(-14 + 6) = -16 – 36 – 64 = -116 and

$\begin{aligned} & \left(m_1 n_2-m_2 n_1\right)^2+\left(l_2 n_1-l_1 n_2\right)^2+\left(l_1 m_2-l_2 m_1\right)^2 \\ & =(-6+2)^2+(1-7)^2+(1-7)^2+(-14+6)^2 \\ & =16+36+64 \\ & =116\end{aligned}$

Hence, the required shortest distance between the given lines

$\begin{aligned} & =\left|\frac{-116}{\sqrt{116}}\right| \\ & =\sqrt{116} \\ & =2 \sqrt{29} \text { units }\end{aligned}$

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

$\frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{2-3}{2}$

... (1)

and $\frac{7-7 x}{3 \lambda}=\frac{y-5}{1}=\frac{6-z}{5}$

$\ldots(2)$

Equation (1) can be written as :

$\frac{-(x-1)}{3}=\frac{7(y-2)}{2 \lambda}=\frac{z-3}{2}$

i.e. $\frac{x-1}{-3}=\frac{y-2}{(2 \lambda / 7)}=\frac{z-3}{2}$

The direction ratios of this line are

$a_1=-3, b_1=\frac{2 \lambda}{7}, c_1=2$

Equation (2) can be written as :

$\frac{-7(x-1)}{3 \lambda}=\frac{y-5}{1}=\frac{-(z-6)}{5}$

i.e. $\frac{x-1}{-(3 \lambda / 7)}=\frac{y-5}{1}=\frac{z-6}{-5}$

The direction ratios of this line are

$a_2=\frac{-3 \lambda}{7}, b_2=1, c_2=-5$

Since the lines (1) and (2) are at right angles,

$\begin{aligned} & a_1 a_2+b_1 b_2+c_1 c_2=0 \\ \therefore & (-3)\left(\frac{-3 \lambda}{7}\right)+\left(\frac{2 \lambda}{7}\right)(1)+2(-5)=0 \\ \therefore & \frac{9 \lambda}{7}+\frac{2 \lambda}{7}-10=0 \\ \therefore & \frac{11 \lambda}{7}=10 \quad \therefore \lambda=\frac{70}{11} .\end{aligned}$

∴ coordinates of any point on the line are x = λ + 1, y = λ + 2, z = λ + 3 ∴ position vector of any point on the line is

$(\lambda+1) \hat{i}+(\lambda+2) \hat{j}+(\lambda+3) \hat{k} \ldots(1)$

If $\bar{b}$ is parallel to the given line whose direction ratios are $1,1,1$, then $\bar{b}=\hat{i}+\hat{j}+\hat{k}$.

Let the required line passing through O meet the given line at M. ∴ position vector of M

$=\bar{m}=(\lambda+1) \hat{i}+(\lambda+2) \hat{j}+(\lambda+3) \hat{k} \ldots[$ By (1)]

The required line is perpendicular to given line

$\begin{aligned} & \therefore \overline{\mathrm{OM}} \cdot \bar{b}=0 \\ & \therefore \bar{m} \cdot \bar{b}=0\end{aligned}$

$\begin{aligned} & \therefore[(\lambda+1) \hat{i}+(\lambda+2) \hat{j}+(\lambda+3) \hat{k}] \cdot(\hat{i}+\hat{j}+\hat{k})=0 \\ & \therefore(\lambda+1) \times 1+(\lambda+2) \times 1+(\lambda+3) \times 1=0 \\ & \therefore 3 \lambda+6=0 \quad \therefore \lambda=-2 \\ & \therefore \bar{m}=(-2+1) \hat{i}+(-2+2) \hat{j}+(-2+3) \hat{k} \\ & \quad=-\hat{i}+\hat{k}\end{aligned}$

The vector equation of the line passing through $\mathrm{A}(\bar{a})$ and $\mathrm{B}(\bar{b})$ is $\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a}), \lambda$ is a scalar.

$\therefore$ the vector equation of the line passing through $o(\bar{o})$ and $M(\bar{m})$ is

$\bar{r}=\overline{0}+\lambda(\bar{m}-\overline{0})=\lambda \bar{m}=\lambda(-\hat{i}+\hat{k})$ where $\lambda$ is a scalar.

Hence, vector equation of the required line is .

coplanar if $\bar{a}_1 \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=\bar{a}_2 \cdot\left(\bar{b}_1 \times \bar{b}_2\right)$

Here $\bar{a}_1=2 \hat{j}-3 \hat{k}, \bar{a}_2=2 \hat{i}+6 \hat{j}+3 \hat{k}$

$\bar{b}_1=\hat{i}+2 \hat{j}+3 \hat{k}, \bar{b}_2=2 \hat{i}+3 \hat{j}+4 \hat{k}$

$\begin{aligned} \therefore \bar{a}_2-\bar{a}_1 & =(2 \hat{i}+6 \hat{j}+3 \hat{k})-(2 \hat{j}-3 \hat{k}) \\ & =2 \hat{i}+4 \hat{i}+6 \hat{k}\end{aligned}$

$\bar{b}_1 \times \bar{b}_2=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 3 & 4\end{array}\right|$

$=(8-9) \hat{i}-(4-6) \hat{j}+(3-4) \hat{k}$

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

$\begin{aligned} \therefore \bar{a}_1 \cdot\left(\bar{b}_1 \times \bar{b}_2\right) & =(2 \hat{j}-3 \hat{k}) \cdot(-\hat{i}+2 \hat{j}-\hat{k}) \\ & =0(-1)+2(2)+(-3)(-1) \\ & =0+4+3=7\end{aligned}$

and $\bar{a}_2 \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=(2 \hat{i}+6 \hat{j}+3 \hat{k}) \cdot(-\hat{i}+2 \hat{j}-\hat{k})$

$\begin{aligned} & =2(-1)+6(2)+3(-1) \\ & =-2+12-3=7 \\ & \therefore \bar{a}_1 \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=\bar{a}_2 \cdot\left(\bar{b}_1 \times \bar{b}_2\right)\end{aligned}$

Hence, the given lines are coplanar. The plane determined by these lines is given by

$\therefore \bar{r} \cdot\left(\overline{b_1} \times \overline{b_2}\right)=\overline{a_1} \cdot\left(\overline{b_1} \times \overline{b_2}\right)$

i.e. $\bar{r} \cdot(-\hat{i}+2 \hat{j}-\hat{k})$

Hence, the given lines are coplanar and the equation of the plane determined by these lines is

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

collinear is $\bar{r} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})=\bar{a} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}) \ldots$ (1)

The required plane makes intercepts 1, 1, 1 on the coordinate axes. ∴ it passes through the three non-collinear points A (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)

$\begin{aligned} & \therefore \bar{a}=\hat{i}, \bar{b}=\hat{j}, \bar{c}=\hat{k} \\ & \overline{\mathrm{AB}}=\bar{b}-\bar{a}=\hat{j}-\hat{i}=-\hat{j}+\hat{j} \\ & \therefore \overline{\mathrm{AC}}=\bar{c}-\bar{a}=\hat{k}-\hat{i}=-\hat{i}+\hat{k}\end{aligned}$

$\therefore \overline{\mathrm{AB}} \times \overline{\mathrm{AC}}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right|$

$=(1-0) \hat{i}-(-1-0) \hat{j}+(0+1) \hat{k}$

$=\hat{i}+\vec{j}+\hat{k}$

Also, $\bar{a} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})=\hat{i} \cdot(\hat{i}+\hat{j}+\hat{k})$

$\begin{aligned} & =1 \times 1+0 \times 1+0 \times 1 \\ & =1\end{aligned}$

$\therefore$ from (1), the vector equation of the required plane is $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1$.

parallel to vectors $\bar{b}$ and $\bar{c}$.

Here, $\bar{a}=5 \hat{i}-2 \hat{j}-3 \hat{k}, \bar{b}=\hat{i}+\hat{j}+\hat{k}, \bar{c}=\hat{i}-2 \hat{j}+3 \hat{k}$

$\therefore \vec{b} \times \vec{c}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3\end{array}\right|$

$=(3+2) \hat{i}-(3-1) \hat{j}+(-2-1) \hat{k}$

$=5 \hat{i}-2 \hat{j}-3 \hat{k}=\bar{a}$

Also, $\bar{a} \cdot(\bar{b} \times \bar{c})=\bar{a} \cdot \bar{a}=|\vec{a}|^2$

$=(5)^2+(-2)^2+(3)^2=38$

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

$\vec{r} \cdot(5 \vec{i}-2 \hat{j}-3 \hat{k})=38$

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

$(x \hat{i}+y \hat{i}+z \hat{k}) \cdot(5 \hat{i}-2 \hat{j}-3 \hat{k})=38$

∴ 5x – 2y – 3z = 38. This is the cartesian equation of the required plane.

$\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ and $\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$

intersect, if $\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{array}\right|=0$

The equations of the given lines are

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$

$\begin{aligned} & \therefore x_1=1, y_1=-1, z_1=1, x_2=3, y_2=k_i z_2=0, \\ & l_1=2, m_1=3, n_1=4, l_2=1, m_2=2, n_2=1 .\end{aligned}$

Since these lines intersect, we get

$\left|\begin{array}{ccc}2 & k+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1\end{array}\right|=0$

∴ 2 (3 – 8) – (k + 1)(2 – 4) – 1 (4 – 3) = 0 ∴ -10 + 2(k + 1) – 1 = 0 ∴ 2(k + 1) = 11

$\begin{aligned} & \therefore k+1=\frac{11}{2} \\ & \therefore k=\frac{9}{2}\end{aligned}$

$\frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}$ and $\frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3}$ is give $\mathrm{n}$ by

$\mathrm{d}=\frac{\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{array}\right|}{\sqrt{\left(m_1 n_2-m_2 n_1\right)^2+\left(l_2 n_1-1_1 n_2\right)^2+\left(l_1 m_2-l_2 m_1\right)^2}}$

The equation of the given lines are

$\frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}$ and $\frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3}$

$\therefore x_1=-1, y_1=-1, z_1=-1, x_2=3, y_2=5, z_2=7$,

$l_1=7, m_1=-6, n_1=1, l_2=1, m_2=-2, n_2=1$

$\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{array}\right|=\left|\begin{array}{ccc}4 & 6 & 8 \\ 4 & -5 & -5 \\ 7 & 1 & 3\end{array}\right|$

= 4(- 6 + 2) – 6(7 – 1) + 8(-14 + 6) = -16 – 36 – 64 = -116 and

$\begin{aligned} & \left(m_1 n_2-m_2 n_1\right)^2+\left(l_2 n_1-l_1 n_2\right)^2+\left(l_1 m_2-l_2 m_1\right)^2 \\ & =(-6+2)^2+(1-7)^2+(-14+6)^2\end{aligned}$

= 16 + 36 + 64 = 116 Hence, the required shortest distance between the given lines

$\begin{aligned} & =\left|\frac{-116}{\sqrt{116}}\right| \\ & =\sqrt{116} \\ & =2 \sqrt{29} \text { units }\end{aligned}$

or The shortest distance between the lines

$=\frac{282}{\sqrt{3830}}$ units

Hence, the gives lines do not intersect.

The shortest distance between the lines

$\bar{r}=\overline{a_1}+\lambda \overline{b_1}$ and $\bar{r}=\overline{a_2}+\mu \overline{b_2}$ is given by

$d=\left|\frac{\left(\overline{a_2}-\overline{a_1}\right) \cdot\left(\overline{b_1} \times \overline{b_2}\right)}{\left|\overline{b_1} \times \overline{b_2}\right|}\right|$

Here, $\overline{a_1}=\hat{i}-\hat{j}, \overline{a_2}=2 \hat{i}-\hat{j}, \overline{b_1}=2 \hat{i}+\hat{k}, \overline{b_2}=\hat{i}+\hat{j}-\hat{k}$.

$\therefore \overline{b_1} \times \overline{b_2}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & -1\end{array}\right|$

$=(0-1) \hat{i}-(-2-1) \hat{j}+(2-0) \hat{k}$

$=-\vec{i}+3 \hat{j}+2 \hat{k}$

and $\overline{a_2}-\overline{a_1}=(2 \hat{i}-\hat{j})-(\hat{i}-\hat{j})=\hat{i}$

$\therefore\left(\overline{a_2}-\overline{a_1}\right) \cdot\left(\overline{b_1} \times \overline{b_2}\right)=\hat{i} \cdot(-\hat{i}+3 \hat{j}+2 \hat{k})$

$=1(-1)+0(3)+0(2)=-1$

and $\left|\overline{b_1} \times \overline{b_2}\right|=\sqrt{(-1)^2+3^2+2^2}$

$=\sqrt{1+9+4}=\sqrt{14}$

$\therefore$ the shortest distance between the given lines

$=\left|\frac{-1}{\sqrt{14}}\right|=\frac{1}{\sqrt{14}}$ unit

Hence, the given lines do not intersect.

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

$\therefore$ the equation of the line $B C$ passing through the points

$B(0,-11,13)$ and $C(2,-3,1)$ is

$\frac{x-0}{2-0}=\frac{y+11}{-3+11}=\frac{z-13}{1-13}$

i.e. $\frac{x}{2}=\frac{y+11}{8}=\frac{z-13}{-12}=\lambda$

... (Say)

AD is the perpendicular from the point A (1, 0, 4) to the line BC. The coordinates of any point on the line BC are given by x = 2λ, y = -11 + 8λ, z = 13 – 12λ Let the coordinates of D be (2λ, -11 + 8λ, 13 – 12λ) … (1) ∴ the direction ratios of AD are 2λ – 1, -1λ + 8λ – 0, 13 – 12λ – 4 i.e. 2λ – 1, -11 + 8λ, 9 – 12λ The direction ratios of the line BC are 2, 8, -12. Since AD is perpendicular to BC, we get 2(2λ – 1) + 8(-11 + 8λ) – 12(9 – 12λ) = 0 ∴ 42λ – 2 – 88 + 64λ – 108 + 144λ = 0 ∴ 212λ – 198 = 0

$\therefore \lambda=\frac{198}{212}=\frac{99}{106}$

Putting $\lambda=\frac{99}{106}$ in (1), the coordinates of $\mathrm{D}$ are

$\left(\frac{198}{106},-11+\frac{792}{106}, 13-\frac{1188}{106}\right)$

i.e. $\left(\frac{198}{106}, \frac{-374}{106}, \frac{190}{106}\right)$, i.e. $\left(\frac{99}{53}, \frac{-187}{53}, \frac{95}{53}\right)$.

$\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}=\lambda_{\ldots}($ Say $)$

The coordinates of any point on the line are given by x = -1 + 2λ, y = 3 + 2λ, z = 8 – λ Let the coordinates of M be (-1 + 2λ, 3 + 3λ, -1 – λ) …..(1) The direction ratios of PM are -1 + 2λ – 2, 3 + 3λ + 3, -1 – λ – 1 i.e. 2λ – 3, 3λ = 6, -λ – 2 The direction ratios of the given line are 2, 3, 8. Since PM is perpendicular to the given line, we get 2(2λ – 3) + 3(3λ + 6) – 1(-λ – 2) = O ∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0 ∴ 14λ + 14 = 0 ∴ λ = -1 Put λ in (1), the coordinates of M are (-1 – 2, 3 – 3, -1 + 1) i.e. (-3, 0, 0). ∴ length of perpendicular from P to the given line = PM

$\begin{aligned} & =\sqrt{(-3-2)^2+(0+3)^2+(0-1)^2} \\ & =\sqrt{(25+9+1)} \\ & =\sqrt{35} \text { units. }\end{aligned}$

Alternative Method : We know that the perpendicular distance from the point

$\sqrt{|\bar{\infty}-\bar{a}|^2-\left[\frac{(\overline{o o}-\bar{a}) \cdot \bar{b}}{|\bar{b}|}\right]^2}$

Here, $\bar{\infty}=2 \hat{i}-3 \hat{j}+\hat{k}, \bar{a}=-\hat{i}+3 \hat{j}-\hat{k}, \bar{b}=2 \hat{i}+3 \hat{j}-\hat{k}$

$\therefore \bar{\infty}-\overline{\mathrm{a}}=(2 \hat{i}-3 \hat{j}+\hat{k})-(-\hat{\mathrm{i}}+3 \hat{j}-\hat{k})$

$=3 \hat{i}-6 \hat{j}+2 \hat{k}$

$\therefore|\bar{\infty}-\bar{a}|^2=3^2+(-6) 2+2^2=9+36+4=49$

Also, $(\bar{\infty}-\overline{\mathrm{a}}) \cdot \overline{\mathrm{b}}=(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})$

Let PM be the perpendicular drawn from the point (1, 0, 0) to the line

$\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}=\lambda_{\ldots}($ Say $)$

The coordinates of any point on the line are given by x = -1 + 2λ, y = 3 + 2λ, z = 8 – λ Let the coordinates of M be (-1 + 2λ, 3 + 3λ, -1 – λ) …..(1) The direction ratios of PM are -1 + 2λ – 2, 3 + 3λ + 3, -1 – λ – 1 i.e. 2λ – 3, 3λ = 6, -λ – 2 The direction ratios of the given line are 2, 3, 8. Since PM is perpendicular to the given line, we get 2(2λ – 3) + 3(3λ + 6) – 1(-λ – 2) = O ∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0 ∴ 14λ + 14 = 0 ∴ λ = -1 Put λ in (1), the coordinates of M are (-1 – 2, 3 – 3, -1 + 1) i.e. (-3, 0, 0). ∴ length of perpendicular from P to the given line = PM

$\begin{aligned} & =\sqrt{(-3-2)^2+(0+3)^2+(0-1)^2} \\ & =\sqrt{(25+9+1)} \\ & =\sqrt{35} \text { units. }\end{aligned}$

Alternative Method : We know that the perpendicular distance from the point

$\sqrt{|\bar{\infty}-\bar{a}|^2-\left[\frac{(\overline{o o}-\bar{a}) \cdot \bar{b}}{|\bar{b}|}\right]^2}$

Here, $\bar{\infty}=2 \hat{i}-3 \hat{j}+\hat{k}, \bar{a}=-\hat{i}+3 \hat{j}-\hat{k}, \bar{b}=2 \hat{i}+3 \hat{j}-\hat{k}$

$\therefore \bar{\infty}-\overline{\mathrm{a}}=(2 \hat{i}-3 \hat{j}+\hat{k})-(-\hat{\mathrm{i}}+3 \hat{j}-\hat{k})$

$=3 \hat{i}-6 \hat{j}+2 \hat{k}$

$\therefore|\bar{\infty}-\bar{a}|^2=3^2+(-6) 2+2^2=9+36+4=49$

Also, $(\bar{\infty}-\overline{\mathrm{a}}) \cdot \overline{\mathrm{b}}=(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})$

$\begin{aligned} & =(3)(2)+(-6)(3)+(2)(-1) \\ & =6-18-2 \\ & =-14\end{aligned}$

$|\overline{\mathrm{b}}|=\sqrt{2^2+3^2+(-1)^2}=\sqrt{14}$

Substitutng tese values in (1), w get length of perpendicular from P to given line = PM

$\begin{aligned} & =\sqrt{49-\left(-\frac{14}{\sqrt{14}}\right)^2} \\ & =\sqrt{49-14} \\ & =\sqrt{35} \text { units }\end{aligned}$

$2 \sqrt{6}$ units, $(3,-4,2)$.

$\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ and $\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$ is

given by

$d=\frac{\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{array}\right|}{\sqrt{\left(m_1 n_2-m_2 n_1\right)^2+\left(l_2 n_1-l_1 n_2\right)^2+\left(l_1 m_2-l_2 m_1\right)^2}}$

The equations of the given lines are

$\begin{aligned} & \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} \text { and } \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1} \\ & \therefore x_1=-1, y_1=-1, z_1=-1, x_2=3, y_2=5, z_2=7, \\ & l_1=7, m_1=-6, n_1=1, l_2=1, m_2=-2, n_2=1 \\ & \left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{array}\right|=\left|\begin{array}{ccc}4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1\end{array}\right|\end{aligned}$

= 4(-6 + 2) – 6(7 – 1) + 8(-14 + 6) = -16 – 36 – 64 = -116

and $\left(m_1 n_2-m_2 n_1\right)^2+\left(l_2 n_1-l_1 n_2\right)^2+\left(l_1 m_2-l_2 m_1\right)^2$

$\begin{aligned} & =(-6+2)^2+(1-7)^2+(-14+6)^2 \\ & =16+36+64=116\end{aligned}$

Hence, the required shortest distance between the given lines $=\left|\frac{-116}{\sqrt{16}}\right|=\sqrt{116}=2 \sqrt{29}$

units

$\bar{r}=\overline{a_2}+\mu \overline{b_2}$ is given by

$d=\left|\frac{\left(\overline{a_2}-\overline{a_1}\right) \cdot\left(\overline{b_1} \times \overline{b_2}\right)}{\left|\overline{b_1} \times \overline{b_2}\right|}\right|$.

Here, $\overline{a_1}=4 \hat{i}-\hat{j}, \overline{a_2}=\hat{i}-\hat{j}+2 \hat{k}$

$\overline{b_1}=\hat{i}+2 \hat{j}-3 \hat{k}, \quad \overline{b_2}=\hat{i}+4 \hat{j}-5 \hat{k}$

$\therefore \overline{b_1} \times \overline{b_2}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5\end{array}\right|$

$\begin{aligned} & =(-10+12) \hat{i}-(-5+3) \hat{j}+(4-2) \hat{k} \\ & =2 \hat{i}+2 \hat{j}+2 \hat{k}\end{aligned}$

and $\begin{aligned} \overline{a_2}-\overline{a_1} & =(\hat{i}-\hat{j}+2 \hat{k})-(4 \hat{i}-\hat{j}) \\ & =-3 \hat{i}+2 \hat{k}\end{aligned}$

$\therefore\left(\overline{a_2}-\overline{a_1}\right) \cdot\left(\overline{b_1} \times \overline{b_2}\right)=(-3 \hat{i}+2 \hat{k}) \cdot(2 \hat{i}+2 \hat{j}+2 \hat{k})$

$\begin{aligned} & =-3(2)+0(2)+2(2) \\ & =-6+0+4=-2\end{aligned}$

and $\begin{aligned}\left|\overline{b_1} \times \overline{b_2}\right| & =\sqrt{2^2+2^2+2^2} \\ & =\sqrt{4+4+4}=2 \sqrt{3}\end{aligned}$

$\therefore$ required shortest distance between the given lines

$=\left|\frac{-2}{2 \sqrt{3}}\right|=\frac{1}{\sqrt{3}}$ units.

$\bar{r}=(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})$.

Let the position vector of the point M be

$\begin{aligned} &(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k}) \\ &=(11+10 \hat{i}) \hat{i}+(-2-4 \hat{k}) \hat{j}+(-8-11 \lambda) \hat{k}\end{aligned}$

Then $\overline{\mathrm{PM}}=$ Position vector of $\mathrm{M}$ - Position vector of $\mathrm{P}$

$\begin{aligned} & =[(11+10 \lambda) \hat{i}+(-2-4 \lambda) \hat{j}+-8-11 \lambda) \hat{k}]-(2 \hat{i}-\hat{j}+5 \hat{k}) \\ & =(9+10 \lambda) \hat{i}+(-1-4 \lambda) \hat{j}+(-13-11 \lambda) \hat{k}\end{aligned}$

Since PM is perpendicular to the given line which is parallel to $\bar{b}=10 \hat{i}-4 \hat{j}-11 \hat{k}$,

$\overline{\mathrm{PM}} \perp^r \bar{b} \therefore \overline{\mathrm{PM}} \cdot \bar{b}=0$

$\begin{aligned} & \therefore[(9+10 \lambda) \hat{i}+(-1-4 \lambda) \hat{j}+(-13-11 \lambda) \hat{k}]-(10 \hat{i}-4 \hat{j}-11 \hat{k})=0 \\ & \therefore 10(9+10 \lambda)-4(-1-4 \lambda)-11(-13-11 \lambda)=0 \\ & \therefore 90+100 \lambda+4+16 \lambda+143+121 \lambda=0 \\ & \therefore 237 \lambda+237=0 \\ & \therefore \lambda=-1\end{aligned}$

Putting this value of $\lambda$, we get the position vector of $M$ as $\hat{i}+2 \hat{j}+3 \hat{k}$.

∴ coordinates of the foot of perpendicular M are (1, 2, 3).

Now, $\overline{\mathrm{PM}}=(\hat{i}+2 \hat{j}+3 \hat{k})-(2 \hat{i}-\hat{j}+5 \hat{k})$

$=-\vec{i}+3 \vec{j}-2 \vec{k}$

$\begin{aligned} \therefore|\overline{\mathrm{PM}}| & =\sqrt{(-1)^2+(3)^2+(-2)^2} \\ & =\sqrt{1+9+4}=\sqrt{14}\end{aligned}$

Hence, the coordinates of the foot of perpendicular are (1,2, 3) and length of

perpendicular $=\sqrt{14}$ units.

$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+1}{-1}=\lambda \ldots($ Say $)$

The coordinates of any point on the line are given by x = -1 + 2λ, y = 3 + 3λ, z = -1 – λ Let the coordinates of M be (-1 + 2λ, 3 + 3λ, -1 – λ) … (1) The direction ratios of PM are -1 + 2λ – 2, 3 + 3λ + 3, -1 – λ – 1 i.e. 2λ – 3, 3λ + 6, -λ – 2 The direction ratios of the given line are 2, 3, -1. Since PM is perpendicular to the given line, we get 2(2λ – 3) + 3(3λ + 6) – 1(-λ – 2) = 0 ∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0 ∴ 14λ + 14 = 0 ∴ λ = -1. Put λ = -1 in (1), the coordinats of M are (-1 – 2, 3 – 3, -1 + 1) i.e. (-3, 0,0). ∴ length of perpendicular from P to the given line

$=\mathrm{PM}=\sqrt{(-3-2)^2+(0+3)^2+(0-1)^2}$

$\begin{aligned} & =\sqrt{25+9+1} \\ & =\sqrt{35} \text { units. }\end{aligned}$

Alternative Method:

We know that the perpendicular distance from the point $\mathrm{P}|\bar{\alpha}|$ to the line $\bar{r}=\bar{a}+\lambda \vec{b}$ is

given by

$\sqrt{|\bar{\alpha}-\bar{a}|^2-\left[\frac{(\bar{\alpha}-\bar{a})-\bar{b}}{|\bar{b}|}\right]^2}$

$\ldots$ (1)

Here, $\bar{\alpha}=2 \hat{i}-3 \hat{j}+\hat{k}, \bar{a}=-\hat{i}+3 \hat{j}-\hat{k}, \bar{b}=2 \hat{i}+3 \hat{j}-\hat{k}$

$\begin{aligned} & \therefore \bar{\alpha}-\bar{a}=(2 \hat{i}-3 \vec{j}+\hat{k})-(-\hat{i}+3 \hat{j}-\hat{k}) \\ & =3 \hat{i}-6 \hat{j}+2 \hat{k} \\ & \therefore|\bar{\alpha}-\bar{a}|^2=3^2+(-6)^2+2^2=9+36+4=49 \\ & \text { Also, }(\bar{\alpha}-\bar{a}) \cdot \bar{b}=(3 \hat{i}-6 \hat{j}+2 \hat{k}) \cdot(2 \hat{i}+3 \hat{j}-\hat{k}) \\ & =(3)(2)+(-6)(3)+(2)(-1) \\ & =6-18-2=-14 \\ & \end{aligned}$

$|\bar{b}|=\sqrt{2^2+3^2+(-1)^2}=\sqrt{14}$

Substituting these values in (1), we get length of perpendicular from P to given line

$\begin{aligned} & =P M=\sqrt{49-\left(\frac{-14}{\sqrt{14}}\right)^2} \\ & =\sqrt{49-14}=\sqrt{35} \text { units. }\end{aligned}$

parallel to the vector $\bar{b}=2 \hat{i}-2 \hat{j}+\hat{k}$ and the line

$\bar{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\mu(\hat{i}-2 \hat{j}+2 \hat{k})$ is parallel to the vector

$\bar{c}=\hat{i}-2 \hat{j}+2 \hat{k}$.

The vector perpendicular to the vectors $\bar{b}$ and $\bar{c}$ is given by

$\vec{b} \times \bar{c}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -2 & 2\end{array}\right|$

$=\hat{i}(-4+2)-\hat{j}(4-1)+\hat{k}(-4+2)$

$=-2 \hat{i}-3 \hat{j}-2 \hat{k}$

Since the required line is perpendicular to the given lines, it is perpendicular to both $\bar{b}$ and

$\bar{C}$

$\therefore$ it is parallel to $\bar{b} \times \bar{c}$

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

$\bar{r}=\bar{a}+\lambda(\bar{b} \times \bar{c})$, where $\lambda$ is a scalar.

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

$\therefore$ the equation of the required line is

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

$\bar{r}=(3 \hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+2 \hat{k})$, where $\mu=-\lambda$

$\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}=\lambda$

$\ldots$ (Say) ... (1)

and $\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}=\mu$

$\ldots$ (Say) ... (2)

From (1), x = -1 -10λ, y = -3 – 2, z = 4 + λ ∴ the coordinates of any point on the line (1) are (-1 – 10λ, – 3 – λ, 4 + λ) From (2), x = -10 – u, y = -1 – 3u, z = 1 + 4u ∴ the coordinates of any point on the line (2) are (-10 – u, -1 – 3u, 1 + 4u) Lines (1) and (2) intersect, if (- 1 – 10λ, – 3 – λ, 4 + 2) = (- 10 – u, -1 – 3u, 1 + 4u) ∴ the equations -1 – 10λ = -10 – u, -3 – 2= – 1 – 3u and 4 + λ = 1 + 4u are simultaneously true. Solving the first two equations, we get, λ = 1 and u = 1. These values of λ and u satisfy the third equation also. ∴ the lines intersect. Putting λ = 1 in (-1 – 10λ, -3 – 2, 4 + 2) or u = 1 in (-10 – u, -1 – 3u, 1 + 4u), we get the point of intersection (-11, -4, 5).

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Here, $\left(x_1, y_1, z_1\right)=(-2,3,4)$ and $\left(x_2, y_2, z_2\right)=(4,-1,0)$

∴ the required cartesian equations of the line AB are

$\frac{x-(-2)}{4-(-2)}=\frac{y-3}{-1-3}=\frac{z-4}{0-4}$

$\therefore \frac{x+2}{6}=\frac{y-3}{-4}=\frac{z-4}{-4}$

$\therefore \frac{x+2}{3}=\frac{y-3}{-2}=\frac{z-4}{-2}$

$\mathrm{C}=(4,-1,0)$

For $x=4, \frac{x+2}{3}=\frac{4+2}{3}=2$

For $y=-1, \frac{y-3}{-2}=\frac{-1-3}{-2}=2$

For $z=0, \frac{z-4}{-2}=\frac{0-4}{-2}=2$

∴ coordinates of C satisfy the equations of the line AB. ∴ C lies on the line passing through A and B. Hence, A, B, C are collinear.