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)

Concept:
The Cartesian form of a line is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. By setting this entire expression equal to a scalar parameter $\mu$, we can write parametric equations for $x, y,$ and $z$. These coordinates can then be combined into a single position vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ to form the vector equation.
Step 1: Set the Cartesian equations equal to a parameter.
$$\frac{x-2}{1} = \frac{y-1}{-1} = \frac{z-1}{-2} = \mu$$
Step 2: Solve for x, y, and z individually.
For $x$: $$\frac{x-2}{1} = \mu \implies x - 2 = \mu \implies x = 2 + \mu$$ For $y$: $$\frac{y-1}{-1} = \mu \implies y - 1 = -\mu \implies y = 1 - \mu$$ For $z$: $$\frac{z-1}{-2} = \mu \implies z - 1 = -2\mu \implies z = 1 - 2\mu$$
Step 3: Construct the final vector equation.
Substitute the parametric expressions into the general position vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$: $$\vec{r} = (2 + \mu)\hat{i} + (1 - \mu)\hat{j} + (1 - 2\mu)\hat{k}$$