Question

The vector equation of the straight line
$\frac{x-2}{1}=\frac{y}{-3}=\frac{1-z}{2}$

• A. $r=2\hat{i}+\hat{k}+t(\hat{i}+3\hat{j}+2\hat{k})$
• B. $r=2\hat{i}-\hat{k}+t(\hat{i}-3\hat{j}+2\hat{k})$
• C. $r=2\hat{i}+\hat{k}+t(\hat{i}-3\hat{j}+2\hat{k})$
• D. $r=2\hat{i}+\hat{k}+t(\hat{i}-3\hat{j}-2\hat{k})$

Answer 1

The correct option is D $r=2\hat{i}+\hat{k}+t(\hat{i}-3\hat{j}-2\hat{k})$

Given equation of straight line is
$\frac{x-2}{1}=\frac{y}{-3}=\frac{1-z}{2}$
$\Rightarrow \frac{x-2}{1}=\frac{y}{-3}=\frac{z-1}{-2}$
$\therefore$ vector equation of line'
$r=(2\hat{i}+0\hat{j}+\hat{k})+t(\hat{i}-3\hat{j}-2\hat{k})$
$r=(2\hat{i}+\hat{k})+t(\hat{i}-3\hat{j}-2\hat{k})$