Question

A line passing through the points \((1,-1,2)\) and \((2,0,1)\) meets the XY plane and the YZ plane at points A and B respectively. The distance AB is equal to...

• A. \(2\sqrt{2}\)
• B. \(2\sqrt{3}\)
• C. \(3\sqrt{2}\)
• D. \(3\sqrt{3}\)

Hint:

Write the line in parametric form and find the parameter values at which x = 0 and z = 0.

Answer 1

The Correct Option is D

Step 1: Use the parameter difference
Along the line each unit of $t$ moves a distance equal to the length of the direction vector, $\sqrt{1 + 1 + 1} = \sqrt{3}$.

Step 2: Parameters
The XY plane is reached at $t = 2$ and the YZ plane at $t = -1$, so the gap in parameter is 3.

Step 3: Distance
$AB = 3\sqrt{3}$.

Step 4: Check
Direct computation of $|A - B|$ from the coordinates also gives $\sqrt{27}$.

Final Answer:
The distance AB is 3 sqrt 3. This is option (D). \[ \boxed{\text{(D) }3\sqrt{3}} \]