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: Parametrise line 1:
$\dfrac{x-3}{3}=\dfrac{y-3}{-1}=t$ with $z=1$: $(3+3t,\ 3-t,\ 1)$.

Step 2: Impose line 2:
Line 2 requires $y=2$, so $3-t=2$ and $t=1$. The point is $(6,2,1)$.

Step 3: Verify line 2:
$\dfrac{x-6}{2}=0$ and $\dfrac{z-1}{3}=0$, which agree. Option D.

Final Answer:
Using z = 1 and y = 2 fixes x = 6, giving (6, 2, 1). \[ \boxed{\text{(D) }(6,2,1)} \]