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 need to convert the Cartesian line equations y = 2 and 4x – 3z + 5 = 0 into standard vector form r = a + λb.

Step 2: Key Formula or Approach:
Express the line in symmetric Cartesian form (x–x₁)/l = (y–y₁)/m = (z–z₁)/n, then extract the position vector a and direction vector b.

Step 3: Detailed Explanation:
From 4x – 3z + 5 = 0, rearranging gives 4x = 3(z – 5/3) → x/3 = (z–5/3)/4. With y = 2 constant (m=0), symmetric form is x/3 = (y–2)/0 = (z–5/3)/4, yielding a = 2j + (5/3)k and b = 3i + 4k.

Step 4: Final Answer:
The vector equation is r = (2j + 5/3 k) + λ(3i + 4k), matching option (D).