Question

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

Answer 1

Let a and b be the position vectors of points (3, -2, - 5) and (3, -2, 6) respectively.
$\therefore a=3\hat{i}-2\hat{j}-5\hat{k}andb=3\hat{i}-2\hat{j}+6\hat{k}$
We know that the vector equation of a line passing through the points having position vectors a and b is $r=a+\lambda (b-a)$
$\therefore r=3\hat{i}-2\hat{j}-5\hat{k}+\lambda \left[\right(3\hat{i}-2\hat{j}+6\hat{k})-(3\hat{i}-2\hat{j}-5\hat{k}\left)\right]\Rightarrow r=3\hat{i}-2\hat{j}-5\hat{k}+\lambda [3\hat{i}-2\hat{j}+6\hat{k}-3\hat{i}+2\hat{j}+5\hat{k}]$
$\Rightarrow r=3\hat{i}-2\hat{j}-5\hat{k}+\lambda \left(11\hat{k}\right)$ (which is vector equation) ...(i)
Putting $r=x\hat{i}+y\hat{j}+z\hat{k}$ in Eq. (i), we have
$x\hat{i}+y\hat{j}+z\hat{k}=3\hat{i}-2\hat{j}+(11\lambda -5)\hat{k}$
Comparing coefficients of $\hat{i},\hat{j}and\hat{k}$ on both sides, we have
x=3, y=-2 and z=11 $\lambda$ -5
$\therefore \frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}[Here,\frac{x-3}{0}\ne \infty ,\frac{y+2}{0}\ne \infty ]$
which is cartesian form of required line.