Question:medium

The line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z}{1}$ intersects the XY and YZ planes at A and B. The line through A and B is}

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Points where a line meets coordinate planes are called the traces of the line.
Updated On: Jun 19, 2026
  • $[\vec{r}-(\hat{i}-2\hat{j}+0\hat{k})]\times(-\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{2}\hat{k})=\vec{0}$
  • $[\vec{r}+(\hat{i}-2\hat{j}+0\hat{k})]\times(-\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k})=\vec{0}$
  • $\vec{r}=(-\hat{i}-2\hat{j}+0\hat{k})+\lambda(-\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{2}\hat{k})$
  • $\vec{r}=(\hat{i}+2\hat{j})+\lambda(-\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{2}\hat{k})$
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Question:
Find the intersection points of a line with coordinate planes and then find the equation of the line connecting those two points.

Step 2: Key Formula or Approach:

1. XY plane is $z=0$. YZ plane is $x=0$.
2. Vector equation $(\vec{r} - \vec{a}) \times \vec{b} = \vec{0}$ represents a line passing through $\vec{a}$ in the direction of $\vec{b}$.

Step 3: Detailed Explanation:

General point on the line is $(2\lambda + 1, -\lambda - 2, \lambda)$.
For point A (intersects XY plane $\implies z = 0$):
$\lambda = 0 \implies A = (2(0)+1, -0-2, 0) = (1, -2, 0)$.
For point B (intersects YZ plane $\implies x = 0$):
$2\lambda + 1 = 0 \implies \lambda = -1/2$.
$B = (0, -(-1/2)-2, -1/2) = (0, -3/2, -1/2)$.
Direction vector of line AB:
$\vec{AB} = \vec{b} - \vec{a} = (0-1, -3/2 - (-2), -1/2 - 0) = (-1, 1/2, -1/2)$.
The line passes through $A(1, -2, 0)$, so $\vec{a} = \hat{i} - 2\hat{j} + 0\hat{k}$.
Using the form $(\vec{r} - \vec{a}) \times \vec{d} = \vec{0}$:
$[\vec{r} - (\hat{i} - 2\hat{j} + 0\hat{k})] \times (-\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k}) = \vec{0}$.

Step 4: Final Answer:

The equation is option (A).
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