Question

The straight line $\vec{r} = (\hat{i} + \hat{j} + \hat{k}) + \alpha(2\hat{i} - \hat{j} + 4\hat{k})$ meets the $xy$-plane at the point:

• A. $(2, -1, 0)$
• B. $(3, 4, 0)$
• C. $(\frac{1}{2}, \frac{3}{4}, 0)$
• D. $(\frac{1}{2}, \frac{7}{4}, 0)$
• E. $(\frac{1}{2}, \frac{5}{4}, 0)$

Hint:

To find intersection with coordinate planes:
• \( xy \)-plane: set \( z = 0 \)
• \( yz \)-plane: set \( x = 0 \)
• \( zx \)-plane: set \( y = 0 \)

Answer 1

The Correct Option is D

Concept: The equation of the \( xy \)-plane is \( z = 0 \). To find the intersection point, we set the \( z \)-coordinate of the line equal to zero.

Step 1: Write parametric coordinates of the line.
Given: \[ \vec{r} = (1 + 2\alpha)\hat{i} + (1 - \alpha)\hat{j} + (1 + 4\alpha)\hat{k} \] So, \[ x = 1 + 2\alpha, \quad y = 1 - \alpha, \quad z = 1 + 4\alpha \]

Step 2: Apply condition for \( xy \)-plane.
\[ z = 0 \Rightarrow 1 + 4\alpha = 0 \] \[ \Rightarrow \alpha = -\frac{1}{4} \]

Step 3: Find coordinates of intersection point.
\[ x = 1 + 2\left(-\frac{1}{4}\right) = 1 - \frac{1}{2} = \frac{1}{2} \] \[ y = 1 - \left(-\frac{1}{4}\right) = 1 + \frac{1}{4} = \frac{5}{4} \]

Step 4: Final answer.
\[ \boxed{\left(\frac{1}{2}, \frac{5}{4}, 0\right)} \]