Question

The vector equation of the line $\frac{x-2}{2}=\frac{2y-5}{-3},z=-1$ is $\overrightarrow{r}=(2\hat{i}+\frac{5}{2}\hat{j}-\hat{k})+\lambda (2\hat{i}-\frac{3}{2}\hat{j}+x\hat{k})$, where $x$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$0$

The given line is $\frac{x-2}{2}=\frac{2y-5}{-3},z=-1$ $\Rightarrow \frac{x-2}{2}=\frac{2y-5}{-3}=\frac{z+1}{0}$

This shows that the given line passes through point $(2,\frac{5}{2},-1)$ and has direction ratios $(2,\frac{-3}{2},0)$

So the direction cosines are

$\frac{2}{\sqrt{2^2+{(-\frac{3}{2})}^2}},\frac{-\frac{3}{2}}{\sqrt{2^2+{(-\frac{3}{2})}^2}},\frac{0}{\sqrt{2^2+{(-\frac{3}{2})}^2}}$ or $\left(\frac{4}{5},-\frac{3}{5},0\right)$

Thus the given line passes through the point having position vector $\overrightarrow{a}=2\hat{i}+\frac{5}{2}\hat{j}-\hat{k}$ and is parallel to the vector $\overrightarrow{b}=2\hat{i}-\frac{3}{2}\hat{j}+0\hat{k}$

So, its vector equations is $\overrightarrow{r}=2\hat{i}+\frac{5}{2}\hat{j}-\hat{k}+\lambda (2\hat{i}-\frac{3}{2}\hat{j}+0\hat{k})$

Hence, $x=0$.