Question

Vector equation of line $\frac{3-x}{3}=\frac{2y-3}{5}=\frac{z}{2}$ is __________ $k\in R$.

• A. $\overline{r}=(3,5,2)+k(3,3,0)$
• B. $\overline{r}=(3,\frac{3}{2},0)+k(-6,5,4)$
• C. $\overline{r}=(3,3,0)+k(3,5,2)$
• D. $\overline{r}=(-6,5,4)+$ $k(3,\frac{3}{2},0)$

Answer 1

The correct option is B $\overline{r}=(3,\frac{3}{2},0)+k(-6,5,4)$

$\frac{3-x}{3}=\frac{2y-3}{5}=\frac{z}{2}$
$\therefore \frac{x-3}{-3}=\frac{y-\frac{3}{2}}{\frac{5}{2}}=\frac{z-0}{2}$
Here $A\left(\overline{a}\right)=(3,\frac{3}{2},0)$ and $\left(\overline{l}\right)=(-3,\frac{5}{2},2)$
Vector equation of line : $\overline{r}=\overline{a}+k\left(\overline{l}\right)$
$\therefore r=(3,\frac{3}{2},0)+k^{\prime }(-3,\frac{5}{2},2)$
$=(3,\frac{3}{2},0)+\frac{k^{\prime }}{2}(-6,5,4)$
$=(3,\frac{3}{2},0)+k(-6,5,4)$
$\therefore k=\frac{k^{\prime }}{2}\in R$.