Question

What are the direction ratios of normal to the plane 4x + 2y - 4z + 1 = 0?
1. 〈4, 2, 4〉
2. \(\left\langle 2, \frac{1}{2}, -2 \right\rangle\)
3. 〈 1, -2, 1 〉 
4. \(\rm \left \langle 2 , 1 , -2 \right \rangle\)

Answer 1

Correct Answer - Option 4 : \(\rm \left \langle 2 , 1 , -2 \right \rangle\)

Concept:

General Form of equation of plane: ax + by + cz + d = 0

Calculation:

Given equation of a plane is 4x + 2y - 4z + 1 = 0

 a = 4, b = 2, c = - 4

Direction ratios, \(\rm \left \langle 4, 2, -4 \right \rangle\) i.e., \(\rm \left \langle 2 , 1 , -2 \right \rangle\)

 

Equation of Plane in Different Forms

• General equation of a plane is ax + by + cz + d = 0

• The equation of the plane in Normal form is lx + my + nz = p where p is the length of the normal from the origin to the plane and (l, m, n) be the direction cosines of the normal.

• The equation to the plane passing through P(x₁, y₁, z₁) and having direction ratios (a, b, c) for its normal is a(x – x₁) + b(y – y₁) + c (z – z₁) = 0

• The equation of the plane passing through three non-collinear points (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃ , z₃) is \(\rm \begin{vmatrix} (x - x_{1}) & (y - y_{1}) & (z - z_{1} ) \\ (x_{2} - x_{1}) & (y_{2} - y_{1}) & (z_{2} - z_{1}) \\ (x_{3} - x_{1}) & (y_{3} - y_{1}) & (z_{3} - z_{1}) \end{vmatrix} \) = 0

• The equation of the plane whose intercepts are a, b, c on the x, y, z axes respectively is x/a + y/b + z/c = 1 (a b c ≠ 0)

• The equation of the YZ plane is x = 0, equation of the plane parallel to the YZ plane is x = d.

• Equation of the ZX plane is y = 0, equation of the plane parallel to the ZX plane is y = d.

• Equation of XY plane is z = 0, equation of plane parallel to XY plane is z = d.

• Four points namely A (x₁, y₁, z₁), B (x₂, y₂, z₂), C (x₃, y₃, z₃) and D (x₄, y₄, z₄) will be coplanar if one point lies on the plane passing through other three points.