Question

Consider the following statements
Statement I The vector equation of a line passing through two points whose position vectors are $a$ and $b$, is $r = a +\lambda( b - a ) \forall \lambda \in R$.
Statement II The cartesian equation of a line passing through two points $\left(x_1, y_1, z_1\right)$ and $\left(x_2, y_2, z_2\right)$ is $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
Choose the correct option.

• A. Statement $I$ is true
• B. Statement II is true
• C. Both statements are true
• D. Both statements are false

Answer 1

Answer: (C)

I. Let $a$ and $b$ be the position vectors of two points $A\left(x_1, y_1, z_1\right)$ and $B\left(x_2, y_2, z_2\right)$, respectively that are lying on a line

Let r be the position vector of an arbitrary point $P(x, y, z)$, then $P$ is a point on the line if and only if $AP = r - a$ and $AB - b - a$ are collinear vectors. Therefore, $P$ is on the line if and only if
$r-a=\lambda(b-a)$
or $ r=a+\lambda(b-a), \lambda \in R$...(i)
This is the vector equation of the line.
II. We have,
$r=x \hat{i}+y \hat{j}+z \hat{k}$
$a=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}$
and $b=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}$
On substituting these values in Eq. (i), we get $x \hat{i}+y \hat{j}+z \hat{k}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}+\lambda\left[\left(x_2-x_1\right) \hat{i}\right.$
$\left.+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}\right]$
Equating the like coefficients of $\hat{i}, \hat{j}$ and $\hat{k}$, we get
$ x=x_1+\lambda\left(x_2-x_1\right)$
$ y=y_1+\lambda\left(y_2-y_1\right) $
$ z=z_1+\lambda\left(z_2-z_1\right)$
On eliminating $\lambda$, we obtain
$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$...(ii)
which is the equation of the line in cartesian form.