Question

The plane passing through the point \((4, -1, 2)\) and parallel to the lines
\(\frac{x+2}{3} = \frac{y-2}{-1} = \frac{z+1}{2}\) and \(\frac{x-2}{1} = \frac{y-3}{2} = \frac{z-4}{3}\)
also passes through the point -

• A. \((1, 1, -1)\)
• B. \((1, 1, 1)\)
• C. \((-1, -1, -1)\)
• D. \((-1, -1, 1)\)

Hint:

The plane's normal is the cross product of the two direction vectors; form the plane and test the options.

Answer 1

The Correct Option is B

Direction vectors: \(\vec d_1=(3,-1,2)\) and \(\vec d_2=(1,2,3)\).
Normal \(\vec n = \vec d_1\times\vec d_2 = (-1\cdot3-2\cdot2,\ -(3\cdot3-2\cdot1),\ 3\cdot2-(-1)\cdot1) = (-7,-7,7)\), proportional to \((1,1,-1)\).
Plane through \((4,-1,2)\): \((x-4)+(y+1)-(z-2)=0\), i.e. \(x+y-z-1=0\).
Test the options: \((1,1,-1)\): \(1+1+1-1=2\ne0\). \((1,1,1)\): \(1+1-1-1=0\), yes. \((-1,-1,-1)\): \(-1-1+1-1=-2\). \((-1,-1,1)\): \(-1-1-1-1=-4\).
Only \((1,1,1)\) lies on the plane.
Answer: (2) \((1, 1, 1)\)