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: Find the line
The points are \((1, -1, 2)\) and \((2, 0, 1)\), so the direction is \((1, 1, -1)\). A point is \((1 + t, -1 + t, 2 - t)\).

Step 2: Meet the XY plane
On the XY plane \(z = 0\), so \(2 - t = 0\) and \(t = 2\). This gives \(A = (3, 1, 0)\).

Step 3: Meet the YZ plane
On the YZ plane \(x = 0\), so \(1 + t = 0\) and \(t = -1\). This gives \(B = (0, -2, 3)\).

Step 4: Distance
\[ AB = \sqrt{(3 - 0)^2 + (1 + 2)^2 + (0 - 3)^2} = \sqrt{27} = 3\sqrt{3} \]
Option (D).

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