Question

The vector form of the straight line $\frac{x-2}{1}=\frac{y-1}{-1}=\frac{z-1}{-2}$ is

• A. $\vec{r}=(2+\mu)\hat{i}+(1-\mu)\hat{j}+(1+2\mu)\hat{k}$
• B. $\vec{r}=(2+\mu)\hat{i}+(1+\mu)\hat{j}+(1-2\mu)\hat{k}$
• C. $\vec{r}=(2+\mu)\hat{i}+(1-2\mu)\hat{j}+(1-2\mu)\hat{k}$
• D. $\vec{r}=(2+3\mu)\hat{i}+(1-\mu)\hat{j}+(1-2\mu)\hat{k}$
• E. $\vec{r}=(2+\mu)\hat{i}+(1-\mu)\hat{j}+(1-2\mu)\hat{k}$

Hint:

Logic Tip: You can bypass the parametric steps entirely. Read the point $(x_1, y_1, z_1)$ from the numerators $\langle 2, 1, 1 \rangle$ and the direction vector $\langle a, b, c \rangle$ from the denominators $\langle 1, -1, -2 \rangle$. The equation is instantly $\vec{r} = (2\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j} - 2\hat{k})$, which groups into Option E.

Answer 1

Answer: (E)