Question

The vector equation of the line whose Cartesian equations are $y = 2$ and $4x - 3z + 5 = 0$ is

• A. $\vec{r} = (2\hat{j} + 5\hat{k}) + \lambda(4\hat{i} - 3\hat{k})$
• B. $\vec{r} = \left(2\hat{j} - \frac{5}{3}\hat{k}\right) + \lambda(3\hat{i} + 4\hat{k})$
• C. $\vec{r} = \left(2\hat{j} - \frac{5}{3}\hat{k}\right) + \lambda(3\hat{i} - 4\hat{k})$
• D. $\vec{r} = \left(2\hat{j} + \frac{5}{3}\hat{k}\right) + \lambda(3\hat{i} + 4\hat{k})$

Hint:

Since $y=2$ is constant throughout, the line's direction vector cannot have any $\hat{j}$ component ($m=0$). This immediately eliminates option (A). Next, look at the equation $4x - 3z + 5 = 0$: if you plug in $x = 0$, you get $z = \frac{5}{3}$. This means the position vector must have a positive coefficient for $\hat{k}$, which uniquely isolates option (D) without any extra algebra!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the Cartesian intersections representing a 3D line: $y = 2$ and $4x - 3z + 5 = 0$. We need to transform these coordinate restrictions into a standard vector equation format, $\vec{r} = \vec{a} + \lambda\vec{b}$.

Step 2: Key Formula or Approach:
To convert a system to the standard symmetric Cartesian form: $$\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$$ We rearrange the terms to isolate the variables with a leading coefficient of 1. Here, since $y = 2$ is a constant plane, its direction component is 0 ($m = 0$). We can rewrite the second equation to link the $x$ and $z$ variables.

Step 3: Detailed Explanation:
Given the equation: $$4x - 3z + 5 = 0$$ Isolate the $x$ term: $$4x = 3z - 5$$ Factor out the coefficient 3 from the right side to ensure $z$ has a leading coefficient of 1: $$4x = 3\left(z - \frac{5}{3}\right)$$ To find a common proportional denominator base, divide both sides of the equation by 12 (the least common multiple of 4 and 3): $$\frac{4x}{12} = \frac{3\left(z - \frac{5}{3}\right)}{12}$$ $$\frac{x}{3} = \frac{z - \frac{5}{3}}{4}$$ We can combine this with the constant condition $\frac{y - 2}{0}$ to write the full symmetric line form: $$\frac{x - 0}{3} = \frac{y - 2}{0} = \frac{z - \frac{5}{3}}{4}$$ From this symmetric form, we can extract: $$\text{Position vector of passing point } \vec{a} = 0\hat{i} + 2\hat{j} + \frac{5}{3}\hat{k} = 2\hat{j} + \frac{5}{3}\hat{k}$$ $$\text{Direction ratios vector } \vec{b} = 3\hat{i} + 0\hat{j} + 4\hat{k} = 3\hat{i} + 4\hat{k}$$ Substituting these vectors into the standard form $\vec{r} = \vec{a} + \lambda\vec{b}$ gives: $$\vec{r} = \left(2\hat{j} + \frac{5}{3}\hat{k}\right) + \lambda(3\hat{i} + 4\hat{k})$$ This matches option (D).

Step 4: Final Answer:
The vector equation of the line is $\vec{r} = \left(2\hat{j} + \frac{5}{3}\hat{k}\right) + \lambda(3\hat{i} + 4\hat{k})$, which corresponds to option (D).