Discussion Forum

1)

If vector equation of the line $\frac{x-2}{2}=\frac{2y-5}{-3} = z+1,$ is

$\overrightarrow{r}$= $(2\hat{i}+\frac{5}{2}\hat{j}-\hat{k})+\lambda\left(2\hat{i}-\frac{3}{2}\hat{j}+p\hat{k}\right)$

then p is equal to

Answer:

Option A

Explanation:

The given line is $\frac{x-2}{2}=\frac{2y-5}{-3} = z+1,$

$\frac{x-2}{2} = \frac{y-\frac{5}{2}}{-\frac{3}{2}} = \frac{z+1}{0}$

This shows that the given line passes through the point (2,5/2,-1) and has direction ratios (2,-3/2,0).

Thus. given line passes through the point having Position vector

$\overrightarrow{a}$ = $(2\hat{i}+\frac{5}{2}\hat{j}-\hat{k})$

$\overrightarrow{b}$ = $\left(2\hat{i}-\frac{3}{2}\hat{j}-0\hat{k}\right)$

$\overrightarrow{r}$= $(2\hat{i}+\frac{5}{2}\hat{j}-\hat{k})+\lambda\left(2\hat{i}-\frac{3}{2}\hat{j}-0\hat{k}\right)$

Hence, p = 0