Question

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

Answer 1

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