Question:
Let the acute angle bisector of the two planes \(x-2y-2z+1=0\) and \(2x-3y-6z+1=0\) be the plane \(P\). Then which of the following points lies on \(P\)?
Let the acute angle bisector of the two planes \(x-2y-2z+1=0\) and \(2x-3y-6z+1=0\) be the plane \(P\). Then which of the following points lies on \(P\)?
Show Hint
Bisectors are \(\frac{x-2y-2z+1}{3}=\pm\frac{2x-3y-6z+1}{7}\); since \(\vec n_1\cdot\vec n_2\gt0\) with both constants positive, the acute-angle bisector is the one with the plus sign.
Updated On: Nov 4, 2026
- \(\left(3,1,-\frac12\right)\)
- \(\left(-2,0,-\frac12\right)\)
- \((0,2,-4)\)
- \((4,0,-2)\)
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The Correct Option is B
Solution and Explanation
\(|\vec n_1|=3\), \(|\vec n_2|=7\), \(\vec n_1\cdot\vec n_2=2+6+12=20\gt0\) and both constant terms are positive (origin side), so the origin lies in the obtuse angle; the bisector \(\frac{E_1}{3}=\frac{E_2}{7}\) bisects the obtuse angle, and the acute angle bisector is \(\frac{E_1}{3}=-\frac{E_2}{7}\).
\(7(x-2y-2z+1)+3(2x-3y-6z+1)=0\Rightarrow13x-23y-32z+10=0\).
Test the options: (1) \(39-23+16+10=42\ne0\); (2) \(-26-0+16+10=0\); (3) \(0-46+128+10\ne0\); (4) \(52-0+64+10\ne0\).
Only \(\left(-2,0,-\frac12\right)\) lies on \(P\).
Answer: (2) \(\left(-2,0,-\frac12\right)\)
\(7(x-2y-2z+1)+3(2x-3y-6z+1)=0\Rightarrow13x-23y-32z+10=0\).
Test the options: (1) \(39-23+16+10=42\ne0\); (2) \(-26-0+16+10=0\); (3) \(0-46+128+10\ne0\); (4) \(52-0+64+10\ne0\).
Only \(\left(-2,0,-\frac12\right)\) lies on \(P\).
Answer: (2) \(\left(-2,0,-\frac12\right)\)
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