Question

If vector equation of the line$\frac{x-2}{2}=\frac{2y-5}{-3}=z+1,$ is
$\vec{r}=\left(2\hat{i}+\frac{5}{2}\hat{j}-\hat{k}\right)+\lambda \left(2\hat{i}-\frac{3}{2}\hat{j}+p\hat{k}\right)$ then $p$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

The given line is $\frac{x-2}{2}=\frac{2y-5}{-3}=z+1,$
$\Rightarrow \frac{x-2}{2}=\frac{y-\frac{5}{2}}{-\frac{3}{2}}=\frac{z+1}{0}$
This shows that the given line passes through the point $\left(2,\frac{5}{2},-1\right)$ and has direction ratios $\left(2,-\frac{3}{2},0\right).$ Thus, given line passes through the point having position vector $\vec{a}=2\hat{i}+\frac{5}{2}\hat{j}-\hat{k}$ and is parallel to the vector
$\vec{b}=\left(2\hat{i}+\frac{3}{2}\hat{j}-0\hat{k}\right)$. So, its vector equation is
$\vec{r}=\left(2\hat{i}+\frac{5}{2}\hat{j}-\hat{k}\right)+\lambda \left(2\hat{i}-\frac{3}{2}\hat{j}-0\hat{k}\right).$
Hence, $p=0$