Question

The equation of the bisector of the obtuse angle between the planes 3x + 4y – 5z + 1 = 0, 5x + 12y – 13z = 0 is :

• A. 11x + 4y – 3z = 0
• B. 14x – 8y + 13 = 0
• C. x + y + z = 9
• D. 13x – 7z + 18 = 0

Answer 1

The correct option is B 14x – 8y + 13 = 0

Bisectors of given planes are
$\frac{3x+4y-5z+1}{\sqrt{9+16+25}}=\pm \frac{5x+12y-13z}{\sqrt{25+144+169}}\Rightarrow \frac{3x+4y-5z+1}{5\sqrt{2}}=\pm \frac{5x+12y-13z}{13\sqrt{2}}$
Taking positive sign, we got one of the bisecting planes is 14x – 8y + 13 = 0
If $\theta$ is the angle between this plane and the plane 3x + 4y – 5z + 1 = 0, then
$cos\theta =\frac{42-32}{\sqrt{{14}^2+8^2}\sqrt{9+16+25}}=\frac{\sqrt{2}}{\sqrt{260}}<\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\theta }{4}$
Hence, 14x – 8y + 13 = 0 is the bisecting plane of obtuse angle.