Solution

The angle $Q$ between the lines with direction cosines, $a, b, c$ and $b-c, c-a$, $a-b$, is given by,

$\cos Q=|\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^{2}+b^{2}+c^{2}}+\sqrt{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}}|$

$\Rightarrow \cos Q=0$

$\Rightarrow Q=\cos ^{-1} 0$

$\Rightarrow Q=90^{\circ}$

Thus, the angle between the lines is $90^{\circ}$.

Solution

The line parallel to $x$-axis and passing through the origin is $x$-axis itself.

Let $A$ be a point on $x$-axis. Therefore, the coordinates of $A$ are given by $(a, 0,0)$, where

$a \in R$.

Direction ratios of $OA$ are $(a, 0,0)$

The equation of $O A$ is given by,

$\frac{x-0}{a}=\frac{y-0}{0}=\frac{z-0}{0}$

$\Rightarrow \frac{x}{1}=\frac{y}{0}=\frac{z}{0}=a$

Thus, the equation of line parallel to $x$-axis and passing through origin is $\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$

Solution

The direction of ratios of the lines, $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}$, are -3 , $2 k, 2$ and $3 k, 1,-5$ respectively.

It is known that two lines with direction ratios, $a_1, b_1, c_1$ and $a_2, b_2, c_2$, are perpendicular, if $a_1 a_2+b_1 b_2+c_1 c_2=0$

$\therefore-3(3 k)+2 k \times 1+2(-5)=0$

$\Rightarrow-9 k+2 k-10=0$

$\Rightarrow 7 k=-10$

$\Rightarrow k=\frac{-10}{7}$

Therefore, for $k=-\frac{10}{7}$, the given lines are perpendicular to each other.

Solution

The given lines are

$\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$

$\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})$

It is known that the shortest distance between two lines, $\vec{r}= \vec{a} _1+\lambda \vec{b} _1$ and $\vec{r}= \vec{a} _2+\lambda \vec{b} _2$, is given by

$d=|\frac{( \vec{b} _1 \times \vec{b} _2) \cdot( \vec{a} _2- \vec{a} _1)}{| \vec{b} _1 \times \vec{b} _2|}|$

Comparing $\vec{r}= \vec{a} _1+\lambda \vec{b} _1$ and $\vec{r}= \vec{a} _2+\lambda \vec{b} _2$ to equations (1) and (2), we obtain

$ \vec{a} _1=6 \hat{i}+2 \hat{j}+2 \hat{k}$

$ \vec{b} _1=\hat{i}-2 \hat{j}+2 \hat{k}$

$ \vec{a} _2=-4 \hat{i}-\hat{k}$

$ \vec{b} _2=3 \hat{i}-2 \hat{j}-2 \hat{k}$

$\Rightarrow \vec{a} _2- \vec{a} _1=(-4 \hat{i}-\hat{k})-(6 \hat{i}+2 \hat{j}+2 \hat{k})=-10 \hat{i}-2 \hat{j}-3 \hat{k}$

$ \begin{aligned} & \Rightarrow \vec{b} _1 \times \vec{b} _2= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} =(4+4) \hat{i}-(-2-6) \hat{j}+(-2+6) \hat{k}=8 \hat{i}+8 \hat{j}+4 \hat{k} \\ & \therefore| \vec{b} _1 \times \vec{b} _2|=\sqrt{(8)^{2}+(8)^{2}+(4)^{2}}=12 \end{aligned} $

$ ( \vec{b} _1 \times \vec{b} _2) \cdot( \vec{a} _2- \vec{a} _1)=(8 \hat{i}+8 \hat{j}+4 \hat{k}) \cdot(-10 \hat{i}-2 \hat{j}-3 \hat{k})=-80-16-12=-108 $

Substituting all the values in equation (1), we obtain

$d=|\frac{-108}{12}|=9$

Therefore, the shortest distance between the two given lines is 9 units.

Solution

Key concept $\rightarrow$ Equation of line passing through $\left(x_1, y_1, z_1\right)$ and having $D R$ ’s $(a, b, c)$ is given by

$ \begin{aligned} & \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \\ \end{aligned} $

$ \begin{aligned} & (1,2,-4) \ \end{aligned} $

$ \begin{aligned} & \frac{x-1}{a}=\frac{y-2}{b}=\frac{z+4}{c} \end{aligned} $

$\begin{aligned} & a_1, b_1, c_1 \text { and } a_2, b_2, c_2 \\ & a_1 a_1+b_1 b_2+c_1 c_2=0 \\ & 3 \hat{i}-16 \hat{j}+7 \hat{k} \text { and } 3 \hat{i}+8 \hat{j}-5 \hat{k} \\ & (3 \hat{i}-16 \hat{j}+7 \hat{k}) \times(3 \hat{\imath}+8 \hat{j}-5 \hat{k})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5\end{array}\right| \\ & =\hat{i}(80-56)-\hat{j}(-15-21)+\hat{k}(24+48) \\ & -24 \hat{i}+36 \hat{j}+72 \hat{k} \rightarrow(24,36,72) \\ & \frac{x-1}{24}=\frac{y-2}{36}=\frac{z+4}{72} \Rightarrow \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6} \\ & \end{aligned}$

$\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$