Three Dimensional Geometry

The required line passes through the origin. Therefore, its position vector is given by,

$\overrightarrow{a}=0.........(1)$

The direction ratios of the line through origin and $\left(5,-2,3\right)$ are

$\left(5-0\right)=5,\left(-2-0\right)=-2,\left(3-0\right)=3$

The line is parallel to the vector given by the equation,  $\overrightarrow{b}=5\widehat{i}-2\widehat{j}+3\widehat{k}$

The equation of the line in vector form through a point with position vector  $\overrightarrow{a}$ and parallel to  $\overrightarrow{b}$ is,  $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b},\lambda \in R$

$\begin{array}{l}\Rightarrow \overrightarrow{r}=\overrightarrow{0}+\lambda \left(5\widehat{i}-2\widehat{j}+3\widehat{k}\right)\\ \Rightarrow \overrightarrow{r}=\lambda \left(5\widehat{i}-2\widehat{j}+3\widehat{k}\right)\end{array}$

The equation of the line through the point $(x_1, y_1, z_1)$ and direction ratios  $a, b, c$ is given by, 

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

Therefore, the equation of the required line in the Cartesian form is

$\begin{array}{l}\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{3}\\ \Rightarrow \frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\end{array}$

Given,

Cartesian equation,

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

The given line passes through the point $\left(5,-4,6\right)$

i.e. position vector of $\overrightarrow{a}=5\widehat{i}-4\widehat{j}+6\widehat{k}$

Direction ratio are 3, 7 and 2.

Thus, the required line passes through the point $\left(5,-4,6\right)$ and is parallel to the vector $3\widehat{i}+7\widehat{j}+2\widehat{k}$ .

Let $\overrightarrow{r}$ be the position vector of any point on the line, then the vector equation of the line is given by,

$\overrightarrow{r}=\left(5\widehat{i}-4\widehat{j}+6\widehat{k}\right)+\lambda \left(3\widehat{i}+7\widehat{j}+2\widehat{k}\right)$

Given,

The point $\left(-2,4,-5\right)$ .

The Cartesian equation of a line through a point $\left(x_1,y_1,z_1\right)$ and having direction ratios a, b, c is

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

Now, given that

$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ is parallel

to point $\left(-2,4,-5\right)$

Here, the point $\left(x_1,y_1,z_1\right)$ is $\left(-2,4,-5\right)$ and the direction ratio is given by $a=3,b=5,c=6$

$\therefore$ The required Cartesian equation is

$\begin{array}{l}\frac{x-\left(-2\right)}{3}=\frac{y-4}{5}=\frac{z-\left(-5\right)}{6}\\ \Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\end{array}$

The line passes through the point with position vector, $\overrightarrow{a}=2\widehat{i}-\widehat{j}+4\widehat{k}-----(1)$

The given vector: $\overrightarrow{b}=\widehat{i}+2\widehat{j}-\widehat{k}-----(2)$

The line which passes through a point with position vector $\overrightarrow{a}$ and parallel to $\overrightarrow{b}$ is given by,

$\begin{array}{l}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\\ \overrightarrow{r}=2\widehat{i}-\widehat{j}+4\widehat{k}+\lambda \left(\widehat{i}+2\widehat{j}-\widehat{k}\right)\end{array}$

$\therefore$ This is required equation of the line in vector form.

Now,

Let $\begin{array}{l}\overrightarrow{r}=x\widehat{i}-y\widehat{j}+z\widehat{k}\\ \Rightarrow x\widehat{i}-y\widehat{j}+z\widehat{k}=\left(\lambda +2\right)\widehat{i}+\left(2\lambda -1\right)\widehat{j}+\left(-\lambda +4\right)\widehat{k}\end{array}$

Comparing the coefficient to eliminate $\lambda$ ,

$\begin{array}{l}x=\lambda +2,x_1=2,a=1\\ y=2\lambda -1,y_1=-1,b=2\\ z=-\lambda +4,z_1=4,c=-1\end{array}$

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

Given,

The line passes through the point $A\left(1,2,3\right)$ .

Position vector of A,

$\overrightarrow{a}=\widehat{i}+2\widehat{j}+3\widehat{k}$

Let $\overrightarrow{b}=3\widehat{i}+2\widehat{j}-2\widehat{k}$

The line which passes through point $\overrightarrow{a}$ and parallel to $\overrightarrow{b}$ is given by,

$\begin{array}{l}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\\ =\widehat{i}+2\widehat{j}+3\widehat{k}+\lambda \left(3\widehat{i}+2\widehat{j}-2\widehat{k}\right)\end{array}$ , where $\lambda$ is constant

Let AB be the line through the point $\left(4,7,8\right)$ and $\left(2,3,4\right)$ and CD be line through the point $\left(-1,-2,1\right)$ and $\left(1,2,5\right)$

Direction cosine, $a_1,b_1,c_1$ of AB are

$\begin{array}{l}=\left(2-4\right),\left(3-7\right),\left(4-8\right)\\ =\left(-2,-4,-4\right)\end{array}$

Direction cosine, $a_2,b_2,c_2$ of CD are

$\begin{array}{l}=\left(1-\left(-1\right)\right),\left(2-\left(-2\right)\right),\left(5-1\right)\\ =\left(2,4,4\right)\end{array}$

AB will be parallel to CD only

If

$\begin{array}{l}\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\ \Rightarrow \frac{-2}{2}=\frac{-4}{4}=\frac{-4}{4}\\ \Rightarrow -1=-1=-1\end{array}$

Here, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Therefore, AB is parallel to CD.

Let AB be the line joining the points $\left(1,-1,2\right)$ and $\left(3,4,-2\right)$ and CD be the line joining the point $\left(0,3,2\right)$ and $\left(3,5,6\right)$ .

The direction ratios, a, b, c of AB are $\begin{array}{l}\left(3-1\right),\left(4-\left(-1\right)\right),\left(-2-2\right)\\ =\left(2,5,-4\right)\end{array}$

The direction ratios $a_2,b_2,c_2$ of CD are $\begin{array}{l}\left(3-0\right),\left(5-3\right),\left(6-2\right)\\ =\left(3,2,4\right)\end{array}$

AB and CD will be perpendicular to each other, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$\begin{array}{l}=2*3+5*2+\left(-4\right)*4\\ =6+10-16\\ =0\end{array}$

$\therefore$ Therefore, AB and CD are perpendicular to each other.

Two lines with direction cosines l, m, n and l₂, m₂, n₂ are perpendicular to each other, if $l_1{l}_2+m_1{m}_2+n_1{n}_2=0$ .

Now, for the 3 lines with direction cosine,

$\begin{array}{l}\frac{12}{13},\frac{-3}{13},\frac{-4}{13}and\frac{4}{13},\frac{12}{13},\frac{3}{13}\\ l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{12}{13}*\frac{4}{13}+\frac{\left(-3\right)}{13}*\frac{12}{13}+\frac{\left(-4\right)}{13}*\frac{3}{13}\\ =\frac{48}{169}-\frac{36}{169}-\frac{12}{169}\\ =0\end{array}$

Hence, the lines are perpendicular.

For lines with direction cosines,

$\begin{array}{l}\frac{4}{13},\frac{12}{13},\frac{3}{13}and\frac{3}{13},\frac{-4}{13},\frac{12}{13}\\ l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{4}{13}*\frac{3}{13}+\frac{12}{13}*\frac{\left(-4\right)}{13}+\frac{3}{13}*\frac{12}{13}\\ =\frac{12}{169}-\frac{48}{169}+\frac{36}{169}\\ =0\end{array}$

Hence, these lines are perpendicular.

For the lines with direction cosines,

$\begin{array}{l}\frac{3}{13},\frac{-4}{13},\frac{12}{13}and\frac{12}{13},\frac{-3}{13},\frac{-4}{13}\\ l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{3}{13}*\frac{12}{13}+\frac{\left(-4\right)}{13}*\frac{\left(-3\right)}{13}+\frac{12}{13}*\frac{\left(-4\right)}{13}\\ =\frac{36+12-48}{169}\\ =0\end{array}$

Hence, these lines are perpendicular.

Therefore, all the lines are perpendicular.

The vertices of ABC are A (3,5, -4), B (-1,1,2) and C (-5, -5, -2)

Direction ratio of side AB $\begin{array}{l}=\left(-1-3\right)\left(1-5\right)\left(2-(-4)\right)\\ =\left(-4,-4,6\right)\end{array}$

 

Direction cosine of AB,

Direction ratios of BC= $\begin{array}{l}\left(-5-\left(-1\right)\right),\left(-5-1\right),\left(-2-2\right)\\ =\left(-4,-6,-4\right)\end{array}$

Direction cosine of BC =

 

 

 

Direction of CA= $\begin{array}{l}\left(-5-3\right)\left(-5-5\right)\left(-2-\left(-4\right)\right)\\ =\left(-8,-10,2\right)\end{array}$

Direction cosine of CA =

Given,

A (2,3,4), B (-1, -2,1), C (5,8,7)

Direction ratio of AB= $\begin{array}{l}\left(-1-2\right),\left(-2-3\right),\left(1-4\right)\\ =\left(3,-5,-3\right)\end{array}$

Where, a₁=3, b₁=-5, c₁=-3

Direction ratio of BC= $\begin{array}{l}\left(5-\left(-1\right)\right),\left(8-\left(-2\right)\right),\left(7-1\right)\\ =\left(6,10,6\right)\end{array}$

Where, a₂=6, b₂=10, c₂=6

Now,

$\begin{array}{l}\frac{a_2}{a_1}=\frac{6}{-3}=-2\\ \frac{b_2}{b_1}=\frac{10}{-5}=-2\\ \frac{c_2}{c_1}=\frac{6}{-3}=-2\end{array}$

Here, direction ratio of two-line segments are proportional.

So, A, B, C are collinear.