Question

Find the equation of the bisector which bisects the acute angle of planes 2x - y + 2z + 3 = 0 and 3x - 2y + 6z + 8 = 0.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The given two planes are -
2x - y + 2z + 3 = 0
3x - 2y + 6z + 8 = 0
Where $d_1,d_2$ > $0$
And $a_1{a}_2+b_1{b}_2+c_1{c}_2=\left(2\right).\left(3\right)+(-1)(-2)+\left(2\right).\left(6\right)$
Or 6 + 2 + 12 which is > 0.
So, acute angle bisector will be -
$\frac{a_{1}x+b_{1}y+c_{1}z+d_1}{\sqrt{(a_1^2+b_1^2+c_1^2)}}=\frac{(a_{2}x+b_{2}y+c_{2}z+d_2)}{\sqrt{(a_2^2+b_2^2+c_2^2)}}$
$\frac{2x-y+2z+3}{\sqrt{(4+1+4)}}=\frac{3x-2y+6z+8}{\sqrt{(9+4+36)}}$
$14x-7y+14z+21=9x-6y+18z+24$
Or 5x -y -4z - 3 = 0