Let the acute angle bisector of the two planes x-2 y-2 z+1=0 and 2 x-3 y-6 z+1=0 be the plane P. Then which of the following points lies on P ?

Question

Let the acute angle bisector of the two planes x-2 y-2 z+1=0 and 2 x-3 y-6 z+1=0 be the plane P. Then which of the following points lies on P ? (A) (3,1,-(1/2)) (B) (-2,0,-(1/2)) (C) (0,2,-4) (D) (4,0,-2)

Tardigrade Admin · Accepted Answer

P 1: x -2 y -2 z +1=0 P 2: 2 x -3 y -6 z +1=0 |(x-2 y-2 z+1/√1+4+4)|=|(2 x-3 y-6 z+1/√22+32+62)| (x-2 y-2 z+1/3)=± (2 x-3 y-6 z+1/7) Since a1 a2+b1 b2+c1 c2=20>0 ∴ Negative sign will give acute bisector 7 x-14 y-14 z+7 =-[6 x-9 y-18 z+3] ⇒ 13 x-23 y-32 z+10=0 (-2,0,-(1/2)) satisfy it

Q. Let the acute angle bisector of the two planes $x - 2 y - 2 z + 1 = 0$ and $2 x - 3 y - 6 z + 1 = 0$ be the plane P. Then which of the following points lies on $P$ ?

$(3, 1, - \frac{1}{2})$

$(- 2, 0, - \frac{1}{2})$

$(0, 2, - 4)$

$(4, 0, - 2)$

Q. 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 ?

(3,1,−21​)

(−2,0,−21​)

(0,2,−4)

(4,0,−2)

Q. Let the acute angle bisector of the two planes $x - 2 y - 2 z + 1 = 0$ and $2 x - 3 y - 6 z + 1 = 0$ be the plane P. Then which of the following points lies on $P$ ?

$(3, 1, - \frac{1}{2})$

$(- 2, 0, - \frac{1}{2})$

$(0, 2, - 4)$

$(4, 0, - 2)$