Question

A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0, passes through the point :0 

• (1) (1, –4, 1) 
• (2) (1, 4, –1) 
• (3) (2, 4, 1) 
• (4) (2, –4, 1) 

Answer 1

Correct option is (4) (2, –4, 1)

Explaination::

From question, the equations of planes given are:

2x – y + 2z – 4 = 0

X + 2y + 2z – 2 = 0

The equation of the planes bisecting the angles between two given planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given as:

$\frac{a_{1}x+b_{1}y+c_{1}z+d_1}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)}}=\pm \frac{a_{2}x+b_{2}y+c_{2}z+d_2}{\sqrt{\left(a_2^2+b_2^2+c_2^2\right)}}$

Now, for this problem, the equation of the planes bisecting the angles is:

$\Rightarrow \frac{2x-y+2z-4}{\sqrt{2^2+{(-1)}^2+2^2}}=\pm \left(\frac{x+2y+2z-2}{\sqrt{1^2+2^2+2^2}}\right)$

$\Rightarrow \frac{2x-y+2z-4}{\sqrt{9}}=\pm \left(\frac{x+2y+2z-2}{\sqrt{9}}\right)$

$\Rightarrow \frac{2x-y+2z-4}{3}=\pm \left(\frac{x+2y+2z-2}{3}\right)$

⇒ 2x – y + 2z – 4 = ±(x + 2y + 2z – 2)

Case I: take positive sign

⇒ 2x – y + 2z – 4 = x + 2y + 2z - 2

⇒ x – 3y – 2 = 0 ----(1)

Case II: take negative sign

⇒ 2x – y + 2z – 4 = -(x + 2y + 2z - 2)

⇒ 2x – y + 2z – 4 = -x – 2y – 2z + 2

⇒ 3x + y + 4z – 6 = 0 ----(2)

Now, we need to find which point in the given options satisfies the obtained equation.

The point (2, -4, 1) satisfies equation (2)