Question

Find the vector and the Cartesian equations of the line that passes through the points (3,-2,-5) and (3,-2,6).

Answer 1

Hint: Start by considering the points as some variable and find out the direction ratios of the line joining these two points. Convert this direction ratios into either vector or cartesian format and apply relevant formula for line passing through the point in direction of AB , Convert one form to the other by taking the correct direction ratios and coordinates.

Complete step-by-step answer:
Given,
(3,-2,-5)= A(say)
(3,-2,6)= B(say)
Let the line passing through the points A and B be AB.
Now , we will find out the direction ratios of AB which can be found out by taking the difference of one coordinate to the other.
So, The direction ratios of AB will be
X coordinates=$l$ = $3-3=0$
Y coordinates =$m$ =$-2-(-2)=-2+2=0$
Z coordinates = $n$ =$6-(-5)=6+5=11$
Now , that we know direction ratios of AB ,let us write it in vector form
$\overrightarrow{c}=0\hat{i}+0\hat{j}+11\hat{k}$
Since, AB passes through A(3,-2,-5), the position vector of A will be written as
$\overrightarrow{a}=3\hat{i}-2\hat{j}-5\hat{k}$
The equation of AB in vector form is given by the relation
$\overrightarrow{r}$ = $\overrightarrow{a}$ + $\lambda \overrightarrow{c}$
Substituting the values of $\overrightarrow{a}$and $\overrightarrow{c}$, we get
$\Rightarrow \overrightarrow{r}=(3\hat{i}-2\hat{j}-5\hat{k})+\lambda 11\hat{k}$
The equation of AB in Cartesian form is given by the relation
${\displaystyle \frac{x-x_1}{l}}={\displaystyle \frac{y-y_1}{m}}={\displaystyle \frac{z-z_1}{n}}$ where $x_1,y_1,z_1$are the coordinates of point passing through and $l,m,n$are the direction ratios of the line.
Substituting the values of coordinates of A and direction ratio of AB , we get
$\Rightarrow {\displaystyle \frac{x-3}{0}}={\displaystyle \frac{y-\left(-2\right)}{0}}={\displaystyle \frac{z-\left(-5\right)}{11}}$
${\displaystyle \frac{x-3}{0}}={\displaystyle \frac{y+2}{0}}={\displaystyle \frac{z+5}{11}}$
So, this is the required Cartesian equation.
Therefore , the equation of line AB in vector form is $\overrightarrow{AB}=(3\hat{i}-2\hat{j}-5\hat{k})+\lambda 11\hat{k}$ and in the cartesian form is ${\displaystyle \frac{x-3}{0}}={\displaystyle \frac{y+2}{0}}={\displaystyle \frac{z+5}{11}}$.

Note: Such similar questions can be solved using the above procedure. If the equation is found in one form whether vector or cartesian it can easily be converted into the other easily , by taking correct values. Attention must be given while substituting the values as it may lead to wrong answers.