Question

The Cartesian equation of a line is $3x + 1 = 6y - 2 = 1 - z$, then its vector equation is

• A. $\bar{r} = \left(-\frac{1}{3}\hat{i} + \frac{1}{3}\hat{j} + \hat{k}\right) + \lambda(2\hat{i} - \hat{j} - 6\hat{k})$
• B. $\bar{r} = (-\hat{i} + 2\hat{j} - \hat{k}) + \lambda(3\hat{i} + 6\hat{j} - \hat{k})$
• C. $\bar{r} = \left(-\frac{1}{3}\hat{i} + \frac{1}{3}\hat{j} + \hat{k}\right) + \lambda(2\hat{i} - \hat{j} + 6\hat{k})$
• D. $\bar{r} = \left(-\frac{1}{3}\hat{i} + \frac{1}{3}\hat{j} + \hat{k}\right) + \lambda(2\hat{i} + \hat{j} - 6\hat{k})$

Hint:

You can instantly eliminate options by checking the sign of the components. Notice that the $z$-term in the original equation is $1 - z = -(z - 1)$, meaning the $z$-direction component must carry a negative sign when standardizing. This leaves only options (A) and (D). Checking the $y$-coefficient isolation then gives the final result instantly.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the Cartesian equation of a line in three dimensions. We need to convert this equation into its standard symmetric form to extract a point on the line and its direction ratios, which allows us to formulate the corresponding vector equation.

Step 2: Key Formula or Approach:
The standard symmetric Cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is: $$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$$ Once written in this form, the vector equation is given by $\bar{r} = \bar{a} + \lambda\bar{b}$, where $\bar{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ is the position vector of the point, and $\bar{b} = a\hat{i} + b\hat{j} + c\hat{k}$ is the direction vector.

Step 3: Detailed Explanation:
Given Cartesian equation: $$3x + 1 = 6y - 2 = 1 - z$$ To express this in standard symmetric form, make the coefficients of $x, y,$ and $z$ equal to $1$ by factoring out the constants from each component: $$3\left(x + \frac{1}{3}\right) = 6\left(y - \frac{1}{3}\right) = -1(z - 1)$$ Now, divide all parts of the expression by the least common multiple (LCM) of the coefficients ($3, 6,$ and $-1$), which is $6$, or rewrite them directly as denominators: $$\frac{x + \frac{1}{3}}{\frac{1}{3}} = \frac{y - \frac{1}{3}}{\frac{1}{6}} = \frac{z - 1}{-1}$$ To get integer direction ratios, multiply each denominator by $6$: $$\frac{x + \frac{1}{3}}{2} = \frac{y - \frac{1}{3}}{1} = \frac{z - 1}{-6}$$ From this standard symmetric form, we identify: A passing point coordinates: $(x_1, y_1, z_1) = \left(-\frac{1}{3}, \frac{1}{3}, 1\right)$, so the position vector $\bar{a} = -\frac{1}{3}\hat{i} + \frac{1}{3}\hat{j} + \hat{k}$. Direction ratios: $(a, b, c) = (2, 1, -6)$, so the direction vector $\bar{b} = 2\hat{i} + \hat{j} - 6\hat{k}$. Substituting these vectors into the general vector equation form $\bar{r} = \bar{a} + \lambda\bar{b}$: $$\bar{r} = \left(-\frac{1}{3}\hat{i} + \frac{1}{3}\hat{j} + \hat{k}\right) + \lambda(2\hat{i} + \hat{j} - 6\hat{k})$$

Step 4: Final Answer:
The resulting vector equation matches option (D).