Question

The point of intersection of the two lines \(\frac{x-3}{3} = \frac{y-3}{-1},z-1 = 0\) and \(\frac{x-6}{2} = \frac{z-1}{3},y-2 = 0\) is....

• A. \((0,0,0)\)
• B. \((1,2,6)\)
• C. \((3,-1,0)\)
• D. \((6,2,1)\)

Hint:

Use the fixed plane coordinates z = 1 and y = 2 to fix the point.

Answer 1

The Correct Option is D

Step 1: Use the given plane conditions:
Line 1 has \(z-1=0\), so every point on it has \(z=1\). Line 2 has \(y-2=0\), so every point on it has \(y=2\).

Step 2: Find x:
For line 2, \(\dfrac{x-6}{2}=\dfrac{z-1}{3}\). With \(z=1\) this gives \(\dfrac{x-6}{2}=0\), so \(x=6\).

Step 3: Check the point (6, 2, 1) on line 1:
\(\dfrac{x-3}{3}=\dfrac{6-3}{3}=1\) and \(\dfrac{y-3}{-1}=\dfrac{2-3}{-1}=1\). The two ratios agree, and \(z=1\). So \((6,2,1)\) is on both lines.

Step 4: Why the other options are wrong.
\((0,0,0)\), \((1,2,6)\) and \((3,-1,0)\) do not have \(z=1\) together with \(y=2\), so they cannot lie on both lines.

Final Answer:
The lines meet at (6, 2, 1). \[ \boxed{\text{(D) }(6,2,1)} \]