Consider the three planes
P₁ : 3x + 15y + 21z = 9,
P₂ : x - 3y - z = 5, and
P₃ : 2x + 10y + 14z = 5
Then, which one of the following is true?

Solution

Maths sir · 4 steps

Planes are parallel when their normals point the same way, so pull out the common factor from each normal.

1. Read the normals. For a plane ax + by + cz = d, the normal vector is (a, b, c). Two planes are parallel when their normals are scalar multiples of each other.

   Plane	Normal	After removing the common factor
   P₁	(3, 15, 21)	3(1, 5, 7)
   P₂	(1, -3, -1)	(1, -3, -1)
   P₃	(2, 10, 14)	2(1, 5, 7)

2. Compare directions. P₁ and P₃ share the direction (1, 5, 7). The direction of P₂, (1, -3, -1), is not a multiple of (1, 5, 7), because the signs of the y and z parts do not match.
3. Not the same plane. Dividing P₁ by 3 gives x + 5y + 7z = 3, and dividing P₃ by 2 gives x + 5y + 7z = 2.5. The constants differ, so these are two separate parallel planes, not one plane written twice.
4. So. Only P₁ and P₃ are parallel, which is option (c).

The idea to keep. To test planes for parallelism, compare only the coefficients of x, y, z. Equal ratios mean parallel; then check the constants to see whether they are the same plane.

Answer from the NTA final answer key. Solution by PrepWiser.