Question

The vector equation of the straight line $\frac{x-2}{1} = \frac{y}{-3} = \frac{1-z}{2}$ is:

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

Hint:

Always look at the $z$ term carefully. If it is $k-z$, the direction ratio $n$ in the denominator must be multiplied by $-1$ to get the correct direction vector.

Answer 1

Answer: (E)

Concept: The vector equation of a line is $\vec{r} = \vec{a} + t\vec{b}$, where $\vec{a}$ is the position vector of a point $(x_1, y_1, z_1)$ on the line and $\vec{b}$ is the direction vector $(l, m, n)$. These correspond to the Cartesian form $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$.

Step 1: Standardize the Cartesian equation.
The given equation is $\frac{x-2}{1} = \frac{y}{-3} = \frac{1-z}{2}$. Rewrite the $z$ term to ensure the coefficient of $z$ is $+1$: $\frac{x-2}{1} = \frac{y-0}{-3} = \frac{-(z-1)}{2} \Rightarrow \frac{x-2}{1} = \frac{y-0}{-3} = \frac{z-1}{-2}$.

Step 2: Identify the point and direction vector.

• Point $(x_1, y_1, z_1) = (2, 0, 1)$. Thus, $\vec{a} = 2\hat{i} + 0\hat{j} + \hat{k} = 2\hat{i} + \hat{k}$.
• Direction ratios $(l, m, n) = (1, -3, -2)$. Thus, $\vec{b} = \hat{i} - 3\hat{j} - 2\hat{k}$.

Step 3: Formulate the vector equation.
$\vec{r} = (2\hat{i} + \hat{k}) + t(\hat{i} - 3\hat{j} - 2\hat{k})$.