Question

The vector equation of line passing through the point $(-1,-1,2)$ and parallel to the line $2x-2=3y+1=6z-2$

• A. $(-\hat{i}-\hat{j}+2\hat{k})+\lambda (3\hat{i}+2\hat{j}+\hat{k})$
• B. $(-\hat{i}-\hat{j}+2\hat{k})+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$
• C. $(-\hat{i}-\hat{j}+2\hat{k})+\lambda (\hat{i}+2\hat{j}+3\hat{k})$
• D. $(-\hat{i}-\hat{j}+2\hat{k})+\lambda (2\hat{i}+3\hat{j}+\hat{k})$

Answer 1

The correct option is A $(-\hat{i}-\hat{j}+2\hat{k})+\lambda (3\hat{i}+2\hat{j}+\hat{k})$

Given

$L_1:2x-2=3y+1=6z-2$

$\Rightarrow 2(x-1)=3(y+\frac{1}{3})=6(z-\frac{1}{3})$

$\Rightarrow \frac{x-1}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{1}{3}}{1}$

Equation of line passing through $(-1,-1,2)$ parallel to $L_1$

$\frac{x+1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$

vector equation is

$(-\hat{i}-\hat{j}+2\hat{k})+\lambda (3\hat{i}+2\hat{j}+\hat{k})$